a box of mass 50.0 kg fell from a shelf to the floor 2.00 m below. when the box hit the floor, one corner was crushed while the box came to rest in 0.05 s. what was the average net force applied to the box while it came to rest?

Answers

Answer 1

The net force applied by to the box is 6264.2 N, if the mass of the box is 50.0 kg and it comes to rest in 0.05 second.

Free fall height of the box, h = 2.0 m

Free fall velocity of the box just before touching the ground, v = √(2gh)

= √(2×9.81×2) = 6.264 m/s

Final velocity or the box after 0.05 sec, = 0

Time to stop, t = 0.05 sec

We know that the applied force is equal to the rate of change of momentum.

F = m(v-u)/t

F = 50 × (6.264-0)/0.05 = 6264.2

So, F = 6264.2 N

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Related Questions

a 3500 pf air-gap capacitor is connected to a 22-v battery. if a piece of mica is placed between the plates, how much charge will flow from the battery?

Answers

The amount of charge that will flow from the battery is 77000pC

A capacitor is a device that stores electrical charge. The amount of charge stored in a capacitor is proportional to the voltage applied across it, according to the equation Q = C*V.

where Q is the charge in coulombs, C is the capacitance in farads, and V is the voltage in volts.

In this case, the capacitance of the air-gap capacitor is 3500pF (pico-farads) and the voltage is 22V, so the charge stored in the capacitor is:

Q = 3500pF * 22V = 77000pC (pico-coulombs)

When a piece of mica is placed between the plates, it reduces the space between the plates, which will increase the capacitance of the capacitor. The increase in capacitance will result in an increase in the amount of charge stored in the capacitor.

It's worth noting that the charge that flows from the battery is equal to the initial charge stored in the capacitor. Since the initial charge stored in the capacitor is 77000pC, this is the amount of charge that will flow from the battery.

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Mice can hear sounds with frequencies between 1 kHz and 70 kHz but are there frequencies which mice can hear but humans cannot?

Answers

In contrast to the human hearing range of 20 Hz to 20 kHz, mouse hearing extends into the ultrasonic region and ranges from 1 to approximately 100 kHz7,9.

What frequency is too high for human hearing?Our hearing is most sensitive in the frequency range between 2000 and 5000 Hz, however the range of absolute human hearing is 20 to 20,000 Hz. Humans normally have a hearing range of 0 dB and above in terms of loudness.In contrast to the human hearing range of 20 Hz to 20 kHz, mouse hearing extends into the ultrasonic region and ranges from 1 to approximately 100 kHz7,9.The bulk, elasticity, muscles, ligaments, bones, fluid, and inner ear structures all play a role in determining how the ear reacts to sounds. Low frequency noises are resisted by elastic forces, whereas high frequency sounds are resisted by greater masses of force. In this way, frequency range is constrained by mass and elasticity.          

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a saturn v rocket is launched straight up with a constant acceleration of 218 m/s . after 150 s, how fast is the rocket moving and how far has it traveled?

Answers

After 150 seconds, the rocket is moving at a velocity of 32700 m/s and has traveled 1,623,500 meters.

A constant acceleration is a change in velocity that does not change over time. If a car accelerates by 20 mph one minute and then another 20 mph the next, its average acceleration remains constant at 20 mph per minute.

We can use the equations of motion to calculate the speed and distance traveled of the rocket. The equation for velocity as a function of time is:

v = at

where v is the velocity, a is the acceleration, and t is the time. Substituting in the given values, we get:

v = 218 m/s * 150 s = 32700 m/s

To find the distance traveled, we can use the equation:

d = 1/2 * at^2

Substituting in the given values, we get:

d = 1/2 * 218 m/s^2 * 150 s^2 = 1,623,500 m

So after 150 seconds, the rocket is moving at a velocity of 32700 m/s and has traveled 1,623,500 meters.

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Find the tension in the two ropes that are holding the 4. 2 kg object in place.


Rope 1 : θ = 57. 6° with respect to the vertical. Rope 2: θ = 33. 8° with respect to the vertical

Answers

The tension in the rope at the 57. 6° angle is 48.58 N, and the tension in the rope at the 33. 8° angle is 36.28 N.

What is meant by angle?

An angle is a figure in plane geometry that is created by two rays or lines that have a shared endpoint. The Latin word "angulus," which meaning "corner," is the source of the English term "angle." The shared terminus of two rays is known as the vertex, and the two rays are referred to as sides of an angle.A pair of rays (half-lines) with a shared terminal are combined to form an angle. The latter is referred to as the angle's vertex, and the rays are sometimes referred to as the angle's legs and other times as its arms.Angles with a measurement of zero degrees or zero radians are referred to as zero angles. No of the unit of measurement, the zero angle has a measure of.

The tension force exerted by the first rope, T1

The tension force exerted by the second rope, T2

and the force of gravity, mg

T1x = −T1 cos 57.6°

T2x = T2 cos 33.8°

mgx = 0

T1y = T1 sin 57.6°

T2y = T2 sin 33.8°

mgy = −mg

Thus, the x and y components of the resultant force will be:

x:

Rx = T1x + T2x + mgx

Rx = −T1 cos 57.6° + T2 cos 33.8° + 0

Rx = T2 cos 33.8° − T1 cos 57.6° (1)

y:

Ry = T1y + T2y + mgy

Ry = T1 sin57.6°+ T2 sin 33.8° + (−mg)

Ry = T1 sin 57.6° + T2 sin33.8° − mg (2)

The next step is to substitute Rx and Ry in Eq. (1) and Eq. (2) with 0:

Rx = T2 cos 33.8° − T1 cos 57.6°

Ry = T1 sin57.6° + T2 sin 33.8° − mg

0 = T2 cos33.8°− T1 cos 57.6° (3)

0 = T1 sin57.6°+ T2 sin 33.8°− mg (4)

Using these two equations we can easily find T1 and T2.

There are multiple ways in which we can do this. One way would be to first solve Eq. (3) for T2:

0 = T2 cos33.8°− T1 cos 57.6°

T2 cos 33.8° − T1 cos 57.6° = 0

T2 cos 33.8° = T1 cos 57.6°

T2 = T1  cos57.6°/ cos 33.8°

T2 = 0.747T1 (5)

hen, we substitute T2 with 0.735T1 in Eq. (4):

0 = T1 sin 57.6°+ T2 sin 33.8° − mg

0 = T1 sin 57.6° + 0.735T1 sin 33.8°− mg

And we solve it for T1:

0 = T1 (sin 57.6° + 0.747 sin 33.8°) − mg

0 = T1 (0.848) − mg

mg = 0.848T1

0.848T1 = mg

T1 =  mg /0.848

T1 =  (4.2kg) (9.81 N/kg) / 0.848T

T1 = 48.58 N

Finally, we substitute the value of T1 in Eq. (5) to find T2:

T2 = 0.747T1

T2 = (0.747) (48.58 N)

T2 = 36.28 N

Therefore, the tension in the rope at the 57. 6°angle is 48.58 N, and the tension in the rope at the 33. 8°angle is 36.28 N.

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how much natural gas (1019 btu/ ft3) is needed to generate the electricity needed to keep a 100- watt (w) light bulb lit for one (1) year if the overall efficiency of the generating station is 40%?

Answers

Only about 40 percent of the thermal energy in coal is converted to electricity, so the electricity generated per ton of coal is 0.4 x 6,150 kWh or 2,460 kWh/ .

Define  natural gas?

Natural gas is an odorless, gaseous mixture of hydrocarbons—predominantly made up of methane (CH4). It accounts for about 30% of the energy used in the United States.It is, as the name suggests, a naturally occurring flammable gas that can be used as a fuel or source of energy for a wide range of purposes like cooking, heating, transportation, and power generation.It is formed when layers of decomposing plants and animals are subject to intense heat from the Earth and pressure from rocks. All this pressure, heat and millions of years turned the natural material into coal, petroleum and natural gas.

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two point charges placed 5 cm aparnt on the xa axis. at what points along the x axis is the potential zero?

Answers

The point where the potential is zero is located at [tex]x = 5 cm * k * Q1/(Q1 + Q2)[/tex]

The potential at a point in space due to a point charge is given by the equation:

V = k * Q/r

where V is the potential, k is the Coulomb constant  [tex](8.99 x 10^9 N * m^2/C^2)[/tex], Q is the charge, and r is the distance from the point charge to the point where the potential is being calculated.

In this problem, we have two point charges placed 5 cm apart on the x-axis. How about we call the charges Q1 and Q2, and we should assume that Q1 is located at the origin (x = 0) and Q2 is located at x = 5 cm.

The potential at a point along the x-axis is the amount of the potentials due to each of the point charges.

V = k * Q1/r1 + k * Q2/r2

where r1 is the distance from the point to Q1 and r2 is the distance from the point to Q2.

In the event that the potential at a point is zero, the amount of the potential due to each of the point charges is zero.

So, k * Q1/r1 + k * Q2/r2 = 0

To find the points along the x-axis where the potential is zero, we can set the equation above equal to zero and address for x.

r1 = x and r2 = |x - 5 cm|

k * Q1/x + k * Q2/|x - 5 cm| = 0

x = k * Q1/Q2 * |x - 5 cm|

x = k * Q1 * (5 cm - x)/Q2

x = 5 cm * k * Q1/(Q1 + Q2)

In this way, the point where the potential is zero is located at

x = 5 cm * k * Q1/(Q1 + Q2)

This is valid for any point in the x-axis, and these points are called equipotential points.

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if 300 waves pass a point in one minute, the frequency is

Answers

when 300 waves pass a point in one minute, the frequency is 5 Hz.

What is frequency?

Frequency can be defined as, the no.of waves that pass in a unit of time through a fixed point. frequency is measured hertz (Hz)

Define wave and wavelength.

Wave: The propagation of disturbances that have place in a medium  is referred to as a wave.

There are two types of waves.

1.Longitudinal waves

2.Transverse waves

Wavelength(λ): The distance between two peaks or two troughs can be used to define a wavelength. It is measured in meters (m), centimeters (cm) or millimeters (mm).

Wavelength Formula: λ=ν/f,

where ν is velocity, f is frequency .

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A frequency of 5 Hz is achieved when 300 waves pass a place in one minute.

How can one determine a wave's frequency? A frequency of 5 Hz is achieved when 300 waves pass a place in one minute.By counting how many wave crests—or high points—pass a particular place in a second or other time interval, one may determine the frequency of the waves.More specifically, the frequency of the waves increases with increasing quantity.The number of waves that pass by in a second makes up a wave's frequency, which is expressed in Hertz (Hz).A sound wave, for instance, might be 450 Hz in frequency.Frequency is the measure of how many waves traverse a spot in one second.The total number of occurrences at the conclusion of the observation period serves as the frequency.Count the instances of the behavior that took place throughout the observation period.By how long the activity was noticed, divide the count by that amount.

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a car drives straight off a flat cliff that is 57 m high (its initial velocity is purely horizontal). the point of impact is 130 m from the base of the cliff. how fast was the car traveling when it went over the cliff?

Answers

The speed of the car traveling when it went over the cliff is 38.12 m/s.

The equation of motion for calculating the displacement of a moving object is given as

                                           [tex]S = ut + \frac{1}{2} at^{2}[/tex]

u = initial velocity

t = time taken

a = acceleration of the object

S = displacement of the object

The above equation can be used to find out the speed of the car.

In this case, the initial velocity, u, is zero because the car falls under gravity. The acceleration, a, will be considered as acceleration due to gravity (g). The time taken is the 'time of flight' taken as 't' and S is taken as the height (h) of the flat cliff.

        Therefore, the above equation can be written as

                                               [tex]h = \frac{1}{2} gt^{2}\\t = \sqrt[]{\frac{2h}{g} } \\t = \sqrt[]{\frac{2(57)}{9.8} } \\\\t = 3.41 s[/tex]

Therefore, the time of flight = 3.41 s.

If the point of impact from the base of the cliff is 130 m and the time of flight is 3.41 s.

Then, the velocity of the car traveling over the cliff is calculated as

                                            [tex]v = \frac{point of impact}{t} \\\\v = \frac{130}{3.41} \\\\v = 38.12 m/s[/tex]

The velocity of the car traveling when it went over the cliff is 38.12 m/s.

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calculate the wavelength of light that has its third minimum at an angle of 30 degrees when falling on double slits separated by 3

Answers

The wavelength of light that is inclined at a specified angle when falling on double slits is calculated to be 428.57 nm.

Angle = 30°

Order of interference m = 3

Distance between the slits d = 3 μ m = 3 × 10⁻⁶ m

We know the relation between wavelength, order of interference and distance as,

λ = d sinθ/(m+1/2)

By putting in the values, we have,

⇒ (3 × 10⁻⁶× sin30°)/(3+1/2) = 4.2857 × 10⁻⁷ m

By converting it to nm, we have,

⇒ 4.2857 × 10⁻⁷ × 10⁹ = 428.57 nm

Thus, the required wavelength of light is calculated to be 428.57 nm.

The question is incomplete. The complete question is 'Calculate the wavelength of light that has its third minimum at an angle of 30.0∘ when falling on double slits separated by 3.00 μm. Explicitly, show how you follow the steps in Problem-Solving Strategies for Wave Optics.'

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at what angular velocity is the centripetal acceleration 10.5 g if the rider is 12.5 m from the center of rotation? give your answer in rad/s.

Answers

The angular velocity of the rider is 2.9 rad/s. The result is obtained by using the formulas in circular motion.

What is circular motion?

Circular motion is a motion of an object along a circular path. When an object moves with an acceleration, it has a centripetal acceleration calculated by

ac = v²/r

Where

ac = centripetal acceleration (m/s²)v = linear velocity (m/s)r = radius (m)

In circular motion, an object moves with linear and angular velocity. The relationship between both velocities is

v = ωr

Where

ω = angular velocity (rad/s)

A rider moves in a circular motion with

Centripetal acceleration, ac = 10.5gRadius, r = 12.5 m

Find the angular velocity! (ω = ?)

We find the centripetal acceleration in m/s².

ac = 10.5g

ac = 10.5 × 9.8 m/s²

ac = 102.9 m/s²

The linear velocity is

ac = v²/r

102.9 = v²/12.5

v² = 102.9 × 12.5

v² = 1,286.25

v ≈ 35.9 m/s

Use the equation of relationship between linear and angular velocity!

v = ωr

35.9 = ω (12.5)

ω ≈ 2.9 rad/s

Hence, the rider has the angular velocity of 2.9 rad/s.

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Help!!! A heat engine receives 1240 J of heat and exhausts 910 J per cycle of the engine what net work does the engine perform during the cycle?

a) 0 J

b) 330 J

c) 910 J

d) 2150 J

Answers

Net work the engine perform during the cycle is b) 330 J.

The net work done by a heat engine is the difference between the heat absorbed by the engine and the heat exhausted. Therefore, the net work done by the engine during a cycle is

1240 J - 910 J = 330 J.

How does the temperature of the hot reservoir affect the efficiency of a heat engine?

The efficiency of a heat engine is directly related to the temperature of the hot reservoir. The higher the temperature of the hot reservoir, the more thermal energy is available for conversion into work. This means that the efficiency of the heat engine increases as the temperature of the hot reservoir increases. The maximum efficiency of a heat engine is given by the Carnot efficiency, which is only achievable when the temperature of the cold reservoir is 0 K (absolute zero) and the temperature of the hot reservoir is infinite.

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a small source of light at the bottom face of a rectangular glass slab 2.25 cm thick is viewed from above. rays of light totally internally reflected at the top surface outline a circle of 7.60 cm in diameter on the bottom surface. determine the refractive index of the glass.

Answers

The refractive index of the glass is 1.16, if the thickness of the glass slab is 2.25 cm, and the diameter of the circle is 7.60 cm.

incident angle of the light,θ = 90°

Let the angle of internal refraction is θ₁

Let the refractive index of the glass slab, = μ₁

Refractive index of vacuum, μ₂ = 1

By Snell's law, μ₁sinθ₁ = μ₂sin90

Radius of the circle, R = 7.6/2 = 3.8 cm

Thickness of the slab, and radius of the circle, R, makes the right angle triangle with hypotenuse(AC). So applying the pythagorus theorem,

AC² = 2.25² + 3.8²

AC = √(2.25² + 3.8²)

AC = 4.42

So sinθ₁ = 3.8/4.42

sinθ₁ = 0.86

μ₁ × 0.86 = μ₂×sin90

μ₁ × 0.86 = 1 × 1

μ₁ = 1/0.86

μ₁ = 1.16

So the refractive index of the glass slab, μ₁ = 1.16

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two narrow slits are separated by a distance d. their interference pattern is to be observed on a screen a large distance l away. (a) calculate the spacing between successive maxima near the center fringe for light of wavelength 500. nm, when l is 1.00 m and d is 1.00 cm. (b) would you expect to be able to observe the interference of light on the screen for this situation? (c) how close together should the slits be placed for the maxima to be separated by 1.00 mm for this wavelength and screen distance?

Answers

The spacing between successive maxima near the center fringe for light of wavelength 500. nm, when l is 1.00 m and d is 1.00 cm is 510^-7 m, Yes, observe the interference of light, distance between the slits be placed for the maxima to be separated by 1.00 mm is 0.5 mm.

Describe wavelength?

Wavelength is a measure of the distance between consecutive peaks (or troughs) of a wave, such as a sound wave or an electromagnetic wave (like light). It is typically measured in meters (m) or nanometers (nm). The wavelength of a wave determines its color, with shorter wavelengths appearing as blue or violet and longer wavelengths appearing as red. In the case of electromagnetic waves, the wavelength also determines the type of radiation, such as radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays.

(a) The spacing between successive maxima near the center fringe for light of wavelength 500 nm is given by the formula dsin(θ) = (λ)/d.

Plugging in the given values, we get: dsin(θ) = [tex]\frac{50010^{-91} }{0.01} =510^{-7} m[/tex]

(b) Yes, it would be possible to observe the interference pattern on the screen for this situation.

(c) To find the distance between the slits, we use the same formula and solve for d: d = (λ)/(dsin(θ)) = [tex]\frac{50010^{-91} }{1*10^{-3} } =0.5mm[/tex]

The slits should be placed 0.5 mm apart to have the maxima separated by 1.00 mm for this wavelength and screen distance.

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A skier with mass of 62kg is sliding down a snowy slope. Find the coefficient of kinetic friction for the skier if friction is known to be 45. 0​

Answers

A skier with mass of 62 kg is sliding down a snowy slope. The coefficient of kinetic friction is 0.0738 N

Calculating the problem:

The coefficient of kinetic friction can be found using the formula:

Friction = coefficient of kinetic friction × Normal force

We know that the friction force is 45 N and the normal force is equal to the skier's weight, which can be found using the formula

Weight = mass × gravity

Where the mass is 62 kg and the acceleration due to gravity is approximately :9.8 m/s²

So the normal force is: 62 kg × 9.8 = 607.6 N

Therefore, the coefficient of kinetic friction is: Friction / Normal force  or approximately

= [tex]\frac{45}{607.6}[/tex] = 0.0738

How does kinetic friction work?

The force that opposes the motion of two surfaces in contact as they slide against one another is known as kinetic friction, but other names for it include sliding friction and dynamic friction. The properties of the two surfaces in contact, such as the roughness of their surfaces and the materials they are made of, determine its strength. It acts in the opposite direction of the force applied to an object.

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a photographer takes a picture with a f/16 stop at 1/30 of a second (as per indicated by her camera). she notices that the subject is a bit blurred so wants to increase the speed to 1/500 of a second. what should the new stop be to have the same exposure level?

Answers

To maintain an exposure level equivalent to f/16 at 1/30 of a second, the new stop should be f/5.6. Due to their inverse relationship, the aperture and shutter speed must be adjusted to account for changes in either one by photographer.

When the shutter speed is increased from 1/30 of a second to 1/500 of a second, the aperture must likewise be changed in order to keep the same exposure level. The amount of light entering the camera, which is managed by the aperture and shutter speed, determines the exposure level. The "exposure triangle" describes the connection between aperture, shutter speed, and exposure. The "reciprocal rule," which dictates that if the shutter speed is increased by a factor of 16, the aperture should be reduced by a factor of 4, should be followed by the photographer in order to maintain the same exposure level. In this instance, the new stop should be f/4 since the photographer wishes to raise the shutter speed from 1/30 to 1/500 while maintaining the same exposure level.

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The half-life of francium-223 is 20 minutes. If a sample originally contains 10 g of the isotope, how much remains after an hour?

Answers

The amount of the mass of the radioactive isotope remaining after 1 hour is 1.25 g.

What is the mass of the sample remaining after an hour?

The mass of the sample remaining after an hour is calculated from the half life of the isotope as shown below.

The half life of the radioactive isotope = 10 minutes

The amount of the mass of the radioactive isotope remaining after 1 hour is calculated as follows;

1 hour = 60 minutes

0 half life ---------------------------> 10 g

20 minutes ------------------------> 5 g

40 minutes -----------------------> 2.5 g

60 minutes -----------------------> 1.25 g

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Korbel pushes a 15 kg chair for a distance of 30 meters. How much
work did he do?

Answers

The work that has been done by the Korbel is  4.41kJ.

What is the work done?

We have to note that we define work in physics as the product of the force and the distance. In the case of the chair in the question, we have been told that Korbel pushes a 15 kg chair for a distance of 30 meters.

The force in this case would be the weight of the chair that we have and we can write that;

W = mgh

m = mass

g = acceleration due to gravity

h = height

Thus

W = 15 * 9.8 * 30

W = 4.41kJ

Thus a work of about  4.41kJ is done.

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Help please I've been stuck on this for so long.​

Answers

Answer:

Explanation:

I think it is the top one as the formula for density is d = M/V, where d is density, M is mass, and V is volume. Density is commonly expressed in units of grams per cubic centimetre.

But I'm not 100% sure

a 6 kg object starts from rest. it is then pushed by a 10 n force for 8 seconds. how much momentum does it have afterward?

Answers

The momentum of the object afterwards will be 80 Kgms-1.

Given the following data as per the question:

Mass of the object is given as 6 Kg

Force is given as 10 N in the question

Time is given as the following 8 sec

The value of the Momentum is same as impulse.

So, the formula of impulse is given as:

I = Ft

Where

I is the impulse

F is the force

and t is the time

From the given information,

F = 10 N

t = 8 secs

Therefore,

I = 10 × 8

I = 80 Ns.

Momentum is 80 Kgms-1.

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a small metal sphere has a mass of 0.15 g and a charge of - 23.0 nc. it is 10.0 cm directly above an identical sphere that has the same charge. this lower sphere is fixed and cannot move. if the upper sphere is released, it will begin to fall. what is the magnitude of its initial acceleration?

Answers

The magnitude of the acceleration is |8.1 * 10^5| m/s^2 = 8.1 * 10^5 m/s^2

What is meant by acceleration?

In mechanics, rate of change of velocity of an object with respect to the time is called as acceleration.

As it is known, F = k * (q1 * q2)/r^2

where F is force, k is Coulomb's constant (9 x 10^9 N * m^2/C^2), q1 and q2 are charges of the two spheres, and r is distance between the two spheres.

As we know that , F = m * a

a = F/m

a = (k * (q1 * q2))/r^2) / m

a = (9 * 10^9 N * m^2/C^2 * (-23 * 10^-9 C)^2) / (10^-2 m)^2

Acceleration of the upper sphere is a = -8.1 * 10^5 m/s^2. The negative sign indicates that acceleration is in opposite direction to the force, as expected, since the force is repulsive.

The magnitude of the acceleration is |8.1 * 10^5| m/s^2 = 8.1 * 10^5 m/s^2

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35. an object of mass 3.00 kg, moving with an initial velocity of 5.001 m/s, collides with and sticks to an object of mass 2.00 kg with an initial velocity of -3.00j m/s. find the fi- nal velocity of the composite object. brainly

Answers

The final velocity of the composite object after collision is 1.8002 m/s in the direction of the first object.

Initial velocity of the first object, v₁ = 5.001 m/s

Mass of the first object, m₁ = 3.0 kg

Initial velocity of the second object, v₂ = -3.001 m/s (In opposite direction)

Mass of the second object, m₂ = 2.0 kg

Let the final velocity of the composite object, = V

Mass of the composite object, = (3.0 + 2.0) = 5.0 kg

By the law of conservation of momentum,

m₁v₁ + m₂v₂ = (m₁ + m₂) × V

V = (m₁v₁ + m₂v₂)/(m₁ + m₂)

V = (3 × 5.001 + 2 × (-3.001))/(2 + 3)

V = 9.001/5

V = +1.8006 m/s

Positive sign indicates that the direction of the velocity will be the same as that of the first object.

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an object travels in a circular orbit. if the speed of the object is doubled, its centripetal acceleration will be

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If an object is traveling in a circular orbit where its speed is doubled, then its centripetal acceleration will be quadrupled.

Centripetal acceleration is the acceleration of an object traversing a circular path. Because velocity is a vector quantity, i.e., it has both a magnitude ( the speed and a direction), so when the object travels on a circular path, its direction changes constantly and thus as a result of this its velocity changes, producing acceleration,

For instance, when an object in a circular orbit is moved with double doubled, then it has quadrupled centripetal acceleration.

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if a spring is released x meters from equilibrium, at what distance from equilibrium will it have acceleration equal to half its maximum acceleration?

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When the stretch distances are at half of their maximum displacement, the acceleration will be half of its maximum value by Hooke's Law.

F= -kx

In accordance with Hooke's law, a principle of elasticity, for relatively minor deformations of an object, the displacement or size of the deformation is directly proportional to the deforming force or load The elasticity of an object is likewise governed by Hooke's law; for example, a metal spring can only be stretched so far before being broken by too much effort. A mass that we already have is then expanded out by x. Displacement and acceleration are inversely related. When the stretch distances are at half of their maximum displacement, the acceleration will be half of its maximum value.

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if the average speed of a helium atom at a certain temperature is 1100 m/s, what is the speed in miles per hour? (1 mi

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The average speed of a helium atom at a certain temperature is 1100 m/s expressed in miles per hour is: 2460.63 mi/h

What is unit conversion?

It is the transformation of a value expressed in one unit of measurement into an equivalent value expressed in another unit of measurement of the same nature.

To solve this problem the we have to convert the units with the given information.

Information about the problem:

v= 1100 m/sv= mi/h?1 mi= 1609.34 m1 h= 3600 s

By converting the speed units from (m/s) to (mi/h) we have:

v= 1100 m/s * 1 mi/1609.34 m * 3600 s/1 h

v= 2460.63 mi/h

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a ringing alarm clock is put under a glass jar. the air is slowly removed from the space around it. what will happen as the air is removed?

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The level of the sound can be decreased by gradually draining the air from a plastic bottle. The sound about an alarm clock cannot be heard at all since sound cannot travel through a vacuum.

What will occur once the air is drained away?

A ringing alarm clock will gradually grow quieter when air is gently drawn out of the area around it and placed beneath a glass jar. Because sound is indeed a mechanical wave that moves across a medium, this is the case.

We can hear the sound because the air molecules in the jar shake as sound waves pass through them. The amount of air particles available to vibrate and convey sound waves declines when air is eliminated, which lessens the sound's strength.  There will be so little remaining air to vibrate due to the almost full removal of the air that the sound will be almost undetectable.

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A positive charge is brought close to a fixed neutral Which of the figures best represents the charge distribution on the inner and outer walls of the conductor that has a cavity. The cavity is neutral; that is, there is no net charge inside the cavity. Which of the figures best represents the change distribution on the inner and outer walls of the conductor?

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The correct option: (e) There is a negative charge on the surface of the cavity, and a positive charge on the outer surface of the conductor.

State the Gauss law for the conductor?

Any extra charge on the a conductor will be located on its surface, according to Gauss' Law.

According to Gauss' Law, the surface's electric field and internal charge are related. The charge encapsulated by any surface on the inside of a conductor equals zero since there is no electric field there. Any charge here on conductor must therefore be located on its surface. There wouldn't be an electric field inside the conductor's vicinity if a cavity was cut out of it, and there wouldn't be any charge inside the cavity either. Therefore, the hollow wouldn't have an electric field. Because of electric field should be perpendicular to its surface and the conductor's internal electric field is zero, no flux can penetrate the conductor.

Thus, on the cavity's surface there is a negative charge, while the conductor's outer surface has a positive charge.

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The complete question is-

A positive point charge is placed in a cavity inside a neutral conductor without transferring charge to the conductor. What is the sign of the charge on the surface of the cavity and the outer surface of the conductor?

(a) There is no charge on the surface of the cavity or the outer surface of the conductor.

(b) There is a positive charge on the surface of the cavity, but no charge on the outer surface of the conductor.

(c) There is a negative charge on the surface of the cavity, but no charge on the outer surface of the conductor.

(d) There is no charge on the surface of the cavity, but a positive charge on the outer surface of the conductor.

(e) There is a negative charge on the surface of the cavity, and a positive charge on the outer surface of the conductor.

what is the maximum amount of trans fat as a percentage that could be in this cereal?physics uncertainty and error

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The maximum amount of trans fat as a percentage that could be in this cereal should be 0.5%.

This cereal is formulated to be as healthy as possible, which is why we limit the amount of trans fat to 0.5% or less. Trans fat is an unhealthy form of fat that has been linked to several health conditions, so we strive to keep our trans fat content as low as possible. By limiting trans fats to 0.5% or less, we are confident that this cereal is a healthy choice.

This cereal contains no more than 0.5% trans fat, as mandated by the FDA. Trans fat is an unhealthy type of fat that can increase bad cholesterol levels and the risk of heart disease and stroke. To ensure optimal health and safety, this cereal only contains the amount of trans fat that is legally allowed.

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Betty was enjoying the warm sunshine in a park. After a few minutes, it became windy. Betty felt cold and put on her jacket. Which component helps a robot detect such a change in its environment?


A. Sensor

B. Motor

C. Actuator

D. Manipulator

Answers

The component of robot which helps detect the change in environment is a sensor. The correct option is A.

Robotics is the study area devoted to creating robots and automation, while a robot is a programmable machine that can execute a task. Different robots have varying degrees of autonomy. These levels range from totally autonomous bots that operate independently to bots that are controlled by humans and accomplish tasks.

Robots consist of some sort of mechanical construction. The mechanical part of a robot helps it perform tasks in the environment for which it was designed.

Robots have some degree of computer programming. If a robot didn't have a set of instructions telling it what to perform, it would just be another piece of simple hardware. Programming a robot can teach it when and how to perform a task.

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A new planet is discovered orbiting a star with a mass 3. 5 × 10^31 kg at a distance of 1. 2x10^11 m. Assume that the orbit is circular. What is the gravitational acceleration on this new planet?​

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The gravitational acceleration of this new planet is [tex]24.3 * 10^{9} G m/s^{2}[/tex].

Any object located in the field of the planet experiences a gravitational pull. Gravitational acceleration is described as the object receiving an acceleration due to the force of gravity acting on it. It is represented by ‘g’ and its unit is m/s2. Gravitational acceleration is a quantity of vector, that is it has both magnitude and direction.Using the following equation, the gravitational acceleration acting on anybody can be explained. When the object is on or near the surface of the body, the force of gravity acting on the object is almost constant and the following equation can be used:

                                        g = GM/R^2

Here,

G = the universal gravitational constant (G = gravitational constant of planet)

M = Mass of planet = [tex]3.5*10^{31} kg[/tex]

R = Radius = 1.2*10^11 m

Putting these values in above equation we get: g=  [tex]24.3 * 10^{9} G m/s^{2}[/tex].

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a car moves with a total distance of 45.0 meters in 10 seconds while slowing down smoothly with a final speed of 3.00 m/s. a) what is the original speed of the car? (2 points) b) what is the car's acceleration? (2 points)

Answers

Answer : Original Speed is 6m/s while acceleration is 0.3m/s^2.

What is speed?

--In everyday use and in kinematics, the speed of an object is the magnitude of the change of its position over time or the magnitude of the change of its position per unit of time;

-- it is thus a scalar quantity.

Here in this question we have given final speed , we have to find the intial speed as well as the acceleration of the body.

d = V1 + V2 /2 × t

45 = V1 + 3/2 × 10

V1 = 6 m/s.

Hence initial speed of body is 6m/s .

By using first equation of motion we find the acceleration

we get acceleration is 0.3 .

What is velocity?

-- Velocity is the directional speed of an object in motion as an indication of its rate of change in position as observed from a particular frame of reference and as measured by a particular standard of time .

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