If the speed of an object in uniform circular motion is constant and the radial distance is doubled, by what factor does the magnitude of the radial acceleration decrease?

Answer:

5,000

Explanation:

Vf = Vi + a * t

Answer:

The resultant velocity has a magnitude of 38.95 m/s

Explanation:

Vector Addition

Given two vectors defined as:

[tex]\vec v_1=(x_1,y_1)[/tex]

[tex]\vec v_2=(x_2,y_2)[/tex]

The sum of the vectors is:

[tex]\vec v=(x_1+x_2,y_1+y_2)[/tex]

The magnitude of a vector can be calculated by

[tex]d=\sqrt{x^2+y^2}[/tex]

Where x and y are the rectangular components of the vector.

We have a plane flying due west at 34 m/s. Its velocity vector is:

[tex]\vec v_1=(-34,0)[/tex]

The wind blows at 19 m/s south, thus:

[tex]\vec v_2=(0,-19)[/tex]

The sum of both velocities gives the resultant velocity:

[tex]\vec v =(-34,-19)[/tex]

The magnitude of this velocity is:

[tex]d=\sqrt{(-34)^2+(-19)^2}[/tex]

[tex]d=\sqrt{1156+361}=\sqrt{1517}[/tex]

d = 38.95 m/s

The resultant velocity has a magnitude of 38.95 m/s

Answer:

b. matter

Explanation:

Waves disturb matter but do not transmit it.

Waves are disturbances that transmit energy from one point to another. Although they cause disturbances, they do not transfer the matters in the medium.

Answer:

599.245km/hr

Explanation:

A plane is heading due west and climbing at the rate of 80 km/hr. If its airspeed is 540 km/hr and there is a wind blowing 80 km/hr to the northwest, what is the ground speed of the plane?

We solve the above question using vectors

In vector form Air speed is -540i + 0j Wind speed is (-80/√2)i + (80/√2)j

Vector notation wind speed is given as: -56.5685 i + 56.5685j

The vector for the ground speed of the plane =

-540i + 0j -56.5685i + 56.5685j

= -596.56854249i + 56.5685j

The the ground speed of the plane √[(596.56854249)² + (56.5685)²]

= √359094.021081 km/hr

= 599.24454197 km/hr

Approximately

= 599.245km/hr

When calcium reacts with chlorine. Electrons move from the calcium atoms to the chlorine atoms. Option D is correct.

When one or more chemicals are transformed into one or more other compounds, a chemical reaction occurs.

In an ionic compound, calcium chloride is present. To build a complete outer shell of electrons, the calcium atom loses two electrons and each chlorine atom obtains one.

When calcium combines with chlorine, it forms calcium chloride. From the calcium atoms to the chlorine atoms, electrons travel.

Hence, option D is correct.

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Answer:

0.585V

Explanation:

Given that:

B = 3.00 × 10-5 T

l = 75.0 m

v = 260 m/s

From Blv = emf between the wing tips

= 3.00 × 10-5 T × 75×260

= 117/200

= 0.585V

Hence, the emf between the wing tips is 0.585V

Answer:

V = 6.65 [volt]

Explanation:

We must first find the power generated, power is defined as the amount of energy consumed or generated in a given time.

[tex]P=\frac{E}{t}[/tex]

where:

P = power [w]

E = energy = 40 [J]

t = time = 30 [s]

[tex]P =40/30\\P = 1.33[w][/tex]

Now we can calculate the voltage or potential drop by means of the power, the power is calculated by means of the product of the voltage by the current.

[tex]P =V*I[/tex]

where:

V = voltage [volts]

I = current = 200mA = 0.2 [A]

[tex]V = P/I\\V = 1.33/0.2\\V = 6.65 [Volt][/tex]

Answer:

= 720000 [k]

Explanation:

The cost is equal to 5 [$/kW-h], kilowatt per hour, this value should be multiplied by the power, and then by the time.

[tex]5[\frac{k}{kw*h}]*200[w]*30[day]*24[\frac{h}{day} ][/tex]

= 720000 [k]

Answer:

Explanation:

F=kx

x=F/k

F=2000 kg

x=100 cm=9*10^-3

effective spring constant=k=F/x

k=2000/9*10^-3=2.2*10^-5

now frequency

f=1/2π√k/m

f=1/2*3.14√2.2*10^-5/310

f=1/6.28√7.097*10^-8

f=1/6.28*2.7*10^-4

f=0.16*2.7*10^-4

f=4.32*10^-5

The effective spring constant of the springs is 33755.55 N/m.

The frequency of the car's vibration is 2.07 Hz.

The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.

A spring balance can be used to calculate the Force. The Newton is the SI unit of force.

Weight of the four people: F = 310 × 9.80 N = 3038 Newton.

The additional compression of the spring: x = 0.90 cm = 0.90 × 10⁻² m.

Hence, the effective spring constant of the springs: k= force/compression

= 3038 N/0.90 × 10⁻² m

= 33755.55 N/m.

The frequency of the car's vibration is: f = 1/2π√(k/m)

=1/2π√(33755.55/2000)

= 2.07 Hz.

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Explanation:

Using the formula;

2x = vt

x is the distance up from the ocean floor the submarine is

v is the speed of sound in water

t is the time

Given

t = 2.3s

v = 1490m/s

Required

how high (approx) up from the ocean floor is the submarine x

From the formula;

x = vt/2

x = 1490(2.3)/2

x = 745(2.3)

x = 1,713.5m

Hence the submarine is 1713.5m high up from the ocean floor

Answer:

Weight = normal force = 24.5 N

Explanation:

Given that,

Mass of a dog, m = 2.5 kg

We need to find its weight and the normal force that it feels.

The weight of an object is given by :

W = mg

Where g is the acceleration due to gravity

[tex]W=2.5\times 9.8\\\\=24.5\ N[/tex]

The normal force is balanced by the weight of an object. So,

Weight = normal force = 24.5 N

Complete Question

A rolling ball moves from [tex]x_1 = 8.0 \ cm[/tex] to [tex]x_2 = - 4.1 \ cm[/tex] during the time from [tex]t_1 = 2.9 s[/tex]  to  [tex]t_2 = 6.0s[/tex]

What is its average velocity over this time interval?

Answer:

The velocity is  [tex]v = 3.903 \ m/s[/tex]

Explanation:

From the question we are told that

    The first position of the ball is  [tex]x_1 = 8.0 \ cm[/tex]

    The second position of the ball is  [tex]x_2 = - 4.1 \ cm[/tex]

Generally the average velocity is mathematically represented as

       [tex]v = \frac{ x_1 - x_2}{t_2 - t_1}[/tex]

=>    [tex]v = \frac{ 8 - -4.1 }{ 6 - 2.9 }[/tex]

=>    [tex]v = 3.903 \ m/s[/tex]

Complete Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer:

The position of the object at t = 10s is  [tex]X = 38.3 \ m[/tex]

Explanation:

From the question we are told that

The acceleration along the x axis is  [tex]a_{x}t = -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)[/tex]

  The position of the object at t = 0 is  x = -14.0 m

  The velocity at t = 0 s is  [tex]v_{0}x = 7.10 m/s[/tex]

Generally from the equation for acceleration along x axis we have that

     [tex]a_x = \frac{dV_{x}}{dt} = -0.032 (15- t)[/tex]

=>   [tex]\int\limits {dV_{x}} \, = \int\limits {-0.032(15- t)} \, dt[/tex]

=>   [tex]V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]

At  t =0  s   and  [tex]v_{0}x = 7.10 m/s[/tex]

=>   [tex]7.10 = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1[/tex]

=>   [tex]K_1 = 7.10[/tex]      

So  

      [tex]\frac{dX}{dt} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]

=>  [tex]\int\limits dX = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}[/tex]

=>  [tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ K_1t +K_2[/tex]

At  t =0  s   and   x = -14.0 m

  [tex]-14 = -0.032 [ 15\frac{0^2}{2} - \frac{0^3 }{6} ]+ K_1(0) +K_2[/tex]

=>   [tex]K_2 = -14[/tex]

So

     [tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ 7.10 t -14[/tex]

At  t = 10.0 s

      [tex]X = -0.032 [ 15\frac{10^2}{2} - \frac{10^3 }{6} ]+ 7.10 (10) -14[/tex]

=>   [tex]X = 38.3 \ m[/tex]

Answer:

authority, accuracy, objectivity, currency, coverage, and appearance.

Explanation:

There are six (6) criteria that should be applied when evaluating any Web site. or each criterion, there are several questions to be asked. The more questions you can answer "yes", the more likely the Web site is one of quality.

As the pushing force x increases, it would be opposed by the static frictional force. As x passes a certain threshold and overcomes the maximum static friction, the block will start moving and will require a smaller magnitude x to maintain opposition to the kinetic friction and keep the block moving at a constant speed. If x stays at the magnitude required to overcome static friction, the net force applied to the block will cause it to accelerate in the same direction.

Let w denote the weight of the block, n the magnitude of the normal force, x the magnitude of the pushing force, and f the magnitude of the frictional force.

The block is initially at rest, so the net force on the box in the horizontal and vertical directions is 0:

n + (-w) = 0

n = w = m g = (5.3 kg) (9.80 m/s²) = 51.94 N

The frictional force is proportional to the normal force, so that f = µ n where µ is the coefficient of static or kinetic friction. Before the block starts moving, the maximum static frictional force will be

f = 0.67 (51.94 N) ≈ 35 N

so for 0 < x < 35 N, the block remains at rest and 0 < f < 35 N as well.

The block starts moving as soon as x = 35 N, at which point f = 35 N.

At any point after the block starts moving, we have

f = 0.48 (51.94 N) ≈ 25 N

so that x = 25 N is the required force to keep the block moving at a constant speed.

As x  is increasing it will be opposed by a static frictional force and for the object to start moving and maintain its acceleration, the magnitude of x must exceed the magnitude of the static frictional force and kinetic frictional force

Given data :

mass of block at rest ( m ) = 5.3 kg

Coefficient of static friction ( μ_s ) =0.67

Coefficient of kinetic friction is ( μ_k ) = 0.48

Horizontal force applied to block = x  

First step : magnitude of normal force ( n ) when object is at rest

n = w            where w = m*g

n - w = 0

n - ( 5.3 * 9.81 ) = 0     ∴  n = 51.94 N

Second step : Required magnitude of x before the movement of object

F =  μ_s * n

F = 0.67 * 51.94  = 34.79 N  ≈ 35 N

∴ The object will start moving once F and x = 35 N

Final step : Magnitude of x  after object start moving

F = μ_k  * n

  = 0.48 * 51.94 = 24.93 N  ≈ 25 N

∴ object will continue to accelerate at a constant speed once F and x = 25N

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Answer:

metals like iron or nickel

Explanation:

[tex] \LARGE{ \underline{ \tt{Required \: answer:}}}[/tex]

We have:

We need to find:

Solution:

According to Newton's 2nd law of motion, or quantitative measure of Force:

Using this,

➝ F = ma

➝ 9N = 3 kg × a

➝ a = 9/3 m/s²

➝ a = 3 m/s²

Hence,

⛱️ [tex] \large{ \blue{ \bf{FadedElla}}}[/tex]

Answer:

I found "As nouns the difference between resultant and equilibrium is that resultant is anything that results from something else; an outcome while equilibrium is the condition of a system in which competing influences are balanced, resulting in no net change. is following as a result or consequence of something."

Explanation:

Explanation:

Given

Velocity v = 23.0m/s

Distance S = 3.45m

Required

Time it will take the skier to reach the ground;

Using the equation of motion;

S = ut + 1/2gt²

3.45 = 23t + 1/2(9.8)t²

3.45 = 23t + 4.9t²

4.9t²+23t-3.45 = 0

Factorize;

t = -23 ±√23²-4(4.9)(-3.45)/2(4.9)

t = -23 ±√529+67.62/9.8

t = -23±√596.62/9.8

t = -23±24.43/9.8

t = 1.43/9.8

t = 0.146 secs

Hence take the skier 0.146 secs to reach the ground.

b) Horizontal distance covered is the range;

Range = U√2H/g

Range = 23√2(3.45)/9.8

Range = 23√6.9/9.8

Range = 23√0.7041

Range = 23(0.8391)

Range = 19.29m

Hence the horizontal distance travelled in air is 19.29m

(a).The time taken by the skier to reach the ground is 0.145 second.

(b).The skier travel in the air before landing is 19.29 meter.

a. Given that A skier leaves the end of a horizontal ski jump at 23.0 m/s and falls through a vertical distance of 3.45 m before landing.

Using equation of motion.

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

Where S is vertical distance , u is initial velocity and g is gravitational acceleration.

Substitute S = 3.45 m, u = 23m/s  and g = 9.8 in above equation.

        [tex]3.45=23t+\frac{1}{2} (9.8)t^{2}\\\\4.9x^{2} +23t-3.45=0\\\\t=0.145,-4.83[/tex]

Since, time can not be negative.

So that,  [tex]t=0.145s[/tex]

b. The horizontal distance travel before landing is known as Range.

           Horizontal distance  ,

                                             [tex]=v*\sqrt{\frac{2S}{g} }[/tex]

Substitute v = 23m/s , S = 3.45 and g = 9.8 meter per second square.

                         [tex]Distance=23*\sqrt{\frac{2*3.45}{9.8} } \\\\Distance=23*\sqrt{0.7041} \\\\Distance=23*0.8391=19.29m[/tex]

Thus, The skier travel in the air before landing is 19.29 meter.

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Answer:

1000x Newton

Explanation:

Step one

given data

acceleration= 1000 m/s²

The question did not specify the mass of the mass of the space ship.

So, let's assume the mass is x kg

Step two:

Required is the force F in Newton

From Newtons first law, it states that a body will continue to be at rest or uniform motion unless acted upon by a force.

F=mass x Acceleration

F=ma

Substituting our given data we have

F=1000x Newton

Answer: 1.14 kg*m/s

Explanation:

The first person explained everything right, they just forgot to convert the angular acceleration to rads/sec^2 from revs/sec^2. Once that is converted, your answer should come out right.

Another small thing, the answer there has an extra unnecessary step. It tells you to find the square root of w^2 to find w but that is unnecessary since the final equation asked for w^2. Hope this helps! :)

The moment of inertia I of the flywheel about its rotation axis is

Given

Angular displacement,

[tex]\theta = 30rev \\\\\theta = (30) * 2\pi rad \\\\\theta = 188.495rad[/tex]

Therefore, Final angular velocity (w) will be:

[tex]w^2 = 2\alpha\theta\\\\w^2 = 2 * (0.200 * 2\pi) * (188.49)\\\\w^2 = 473.73\\\\w = 21.76 rad/s[/tex]

Therefore,

moment of inertia

[tex]I = 2 * K / w^2[/tex]

[tex]I = 2 * 330 / 473.73[/tex]

[tex]I = 1.39kgm^2[/tex]

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Answer:

-7/19

Explanation:

Answer:

[tex]0.00005202\ \text{rad}=0.003^{\circ}[/tex]

Explanation:

d = Diameter of typical neutron star = 20 km = 20000 m

D = Distance between Earth and Moon = [tex]384.4\times 10^6\ \text{m}[/tex]

Here, [tex]D>>d[/tex] so we use small angle approximation

[tex]\delta=\dfrac{d}{D}\\\Rightarrow \delta=\dfrac{20000}{384.4\times 10^6}\\\Rightarrow \delta=0.00005202\ \text{rad}=\dfrac{0.00005202\times 180}{\pi}=0.003^{\circ}[/tex]

The angular diameter of the neutron star would be [tex]0.00005202\ \text{rad}=0.003^{\circ}[/tex] from Earth.

Answer:

5 million miles

Explanation:

Answer:

A radar gun is a device for measuring the speed of moving objects. ... The radar gun is a Doppler radar unit that can be static, vehicle-mounted or hand-held. It measures the

Explanation:

Answer:

The displacement of the car after 6s is 43.2 m

Explanation:

Given;

velocity of the car, v = 12 m/s

acceleration of the car, a = -1.6 m/s² (backward acceleration)

time of motion, t = 6 s

The displacement of the car after 6s is given by the following kinematic equation;

d = ut + ¹/₂at²

d = (12 x 6) + ¹/₂(-1.6)(6)²

d = 72 - 28.8

d = 43.2 m

Therefore, the displacement of the car after 6s is 43.2 m

Answer:

The maximum speed of the car is 35 m/s

The total distance traveled by the car is 658.33 m

Explanation:

Given;

initial velocity of the car, u = 15 m/s

acceleration of the car, a = 2 m/s²

time of car motion, t = 10 s

(i)

Initial distance traveled by the car is given by;

d₁ = ut + ¹/₂at²

d₁ = (15 x 10) + ¹/₂(2)(10)²

d₁ = 150 + 100

d₁ = 250 m

The maximum speed of the car during this is given by;

v² = u² + 2ad₁

v² = (15)² + (2 x 2 x 250)

v² = 1225

v = √1225

v = 35 m/s

(ii)

The final distance cover by the car during the deceleration of 1.5 m/s².

Note: the final or maximum speed of the car becomes the initial velocity during deceleration.

v² = u² + 2ad₂

where;

v is the final speed of the car when it stops = 0

0 = u² + 2ad₂

0 = (35²) + (2 x - 1.5 x d₂)

0 = 1225 - 3d₂

3d₂ = 1225

d₂ = 1225 / 3

d₂ = 408.33 m

The total distance traveled by the car is given by;

d = d₁ + d₂

d = 250 m + 408.33 m

d = 658.33 m

Answer:

50,000N

Explanation:

According to Newton's second law of motion;

Net Force = Mass * acceleration

Given

Mass = 5000kg

Let the acceleration = 10m/s²

Net force = 5000 * 10

Net force = 50,000N

Hence the net force acting on the bus is 50000N

Answer:

Yeah it's ok I think. Also, I can't see the answer you gave so maybe updating the question would be nice.