The force might make the object move or bring the object to rest or change in direction of motion
A 1000 kg car and a 1500 kg car are driving in opposite directions at 20 m/s and 10 m/s respectively. If the cars collide head-on and stick together, determine their combined speed after the collision.
Hi there!
Recall the conservation of momentum for an inelastic collision:
[tex]\large\boxed{m_1v_1 + m_2v_2 = (m_1 + m_2)v_f}}[/tex]
Remember, velocity is a VECTOR and direction must be accounted for. Let the 1000 kg car have a positive velocity and the 1500 kg car have a negative velocity (opposite direction).
Plug in the given values:
[tex](1000)(20) + (1500)(-10) = (1000 + 1500)v_f}}[/tex]
Solve:
[tex]20000 -15000 = 2500v_f}}\\\\5000 = 2500v_f\\\\v_f =\boxed{ 2 m/s}[/tex]
A massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface. The mass hangs over a
frictionless pulley. When the mass is released, the cartaccelerates to the right
2.45 m/s2
4.90 m/s
9.80 m/s?
19.6 m/s
Both masses will have the same acceleration. The cart accelerates to the right with a magnitude of 4.9 m/[tex]s^{2}[/tex]. The correct answer is 4.90 m/[tex]s^{2}[/tex]
Given that a massless string connects a 1.00 kg mass to a 3.00 kg cart which is resting on a frictionless horizontal surface.
Let M = 1kg and m = 3 kg
Since the horizontal surface is frictionless, the tension in the string will be the same. when the mass is hanged over a frictionless pulley, the tension will also be the same.
When the mass is released, the cart accelerates to the right can be calculated from Newton' second law of motion. That is,
M( g + a) = m(g - a)
1(9.8 + a) = 3( 9.8 - a)
9.8 +a = 29.4 - 3a
collect the like terms
4a = 19.6
a = 19.6/4
a = 4.9 m/[tex]s^{2}[/tex]
Therefore, the cart accelerates to the right with a magnitude of 4.9 m/[tex]s^{2}[/tex]. The correct answer is 4.90 m/[tex]s^{2}[/tex]
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Help me answer this question please!
why is acceleration not constant near the speed of light
Answer:
because when an object approaches the speed of light, it's mass starts to increase exponentially, and would be infinite at the speed of light. It would therefore require MORE than an infinite amount of energy to accelerate even a single electron to the speed of light
A hockey player hits a rubber puck from one side of the rink to the other. It has a mass of .170 kg, and is hit at an initial speed of 6 m/s. If the rink is 61 meters long, how fast is the puck moving when it hits the far wall?
By using third law of equation of motion, the final velocity V of the rubber puck is 8.5 m/s
Given that a hockey player hits a rubber puck from one side of the rink to the other. The parameters given are:
mass m = 0.170 kg
initial speed u = 6 m/s.
Distance covered s = 61 m
To calculate how fast the puck is moving when it hits the far wall means we are to calculate final speed V
To do this, let us first calculate the kinetic energy at which the ball move.
K.E = 1/2m[tex]U^{2}[/tex]
K.E = 1/2 x 0.17 x [tex]6^{2}[/tex]
K.E = 3.06 J
The work done on the ball is equal to the kinetic energy. That is,
W = K.E
But work done = Force x distance
F x S = K.E
F x 61 = 3.06
F = 3.06/61
F = 0.05 N
From here, we can calculate the acceleration of the ball from Newton second law
F = ma
0.05 = 0.17a
a = 0.05/0.17
a = 0.3 m/[tex]s^{2}[/tex]
To calculate the final velocity, let us use third equation of motion.
[tex]V^{2}[/tex] = [tex]U^{2}[/tex] + 2as
[tex]V^{2}[/tex] = [tex]6^{2}[/tex] + 2 x 0.3 x 61
[tex]V^{2}[/tex] = 36 + 36
[tex]V^{2}[/tex] = 72
V = [tex]\sqrt{72}[/tex]
V = 8.485 m/s
Therefore, the puck is moving at the rate of 8.5 m/s (approximately) when it hits the far wall.
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Guys No one's answering my question so sad! Once again I'm asking the same question –Here
Chapter Name :- Vector
Question :- A car is moving at Vc speed. Rain is falling vertically at a speed of 10 m / s. What is the value of Vc in this case,if the front glass of the car will get wet? Answer :- Vc ≥ 10
Once again, I need Explanation! If you don’t know then no need to answer!
For the front glass of the car to get wet, [tex]V_c \geq 10 \ m/s[/tex].
The given parameters:
Speed of the car, = VcSpeed of the rain, = 10 m/sThe relative velocity of the car with respect to the falling rain is calculated as;
[tex]V_{C/R} = V_C- V_R[/tex]
If the speed of the car equals the speed of the rain, the rain will fall behind the car.If the speed of the rain is greater than speed of the car, the rain will fall far in front of the car.If the speed of the car is greater than speed of the rain, the rain will fall on the car.Thus, for the front glass of the car to get wet, [tex]V_c \geq 10 \ m/s[/tex].
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Two carts, A and B, are connected by a spring and sitting at rest on a track. Cart A has a mass of 0.4 kg and Cart B has a mass of 0.8 kg. The spring is released and causes the two carts to push off from each other. Which of the following correctly compares the motion and forces acting on the two carts?
A. Cart A experiences a greater force but has the same speed as Cart
B. Cart A experiences a greater force and has a greater speed after the recoil.
C. Both carts experience the same force but Cart A has a greater speed after the recoil.
D. Both carts experience the same force and have the same speed after the recoil.
Both carts experience the same force but Cart A has a greater speed after the recoil.
The given parameters;
Mass of the cart A = 0.4 kgMass of the cart B = 0.8 kgApply the principle of conservation of linear momentum to determine the velocity of the carts after collision;
[tex]m_Av_0_A\ + m_Bv_0_B = m_Av_f_A \ + m_Bv_f_B\\\\the \ initial \ velocity \ of \ both \ carts = 0\\\\0.4(0) + 0.8(0) = 0.4v_f_A + 0.8v_f_B\\\\0 = 0.4v_f_A + 0.8v_f_B\\\\0.4v_f_A = -0.8v_f_B\\\\v_f_A= \frac{-0.8 v_f_B}{0.4} \\\\v_f_A = - 2 \ v_f_B \ \ m/s[/tex]
According to Newton's third law of motion, action and reaction are equal and opposite. The force exerted on cart A is equal to the force exerted on cart B but in opposite direction.
[tex]F_A = -F_B[/tex]
Thus, the correct statement that compares the motion and forces acting on the two carts is "Both carts experience the same force but Cart A has a greater speed after the recoil."
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the system shown above is released from rest. if friction is negligible, the acceleration of the 4.0 kg block sliding on the table shown above is most nearly
The acceleration of the first block (4 kg) is -9.8 m/s².
The given parameters:
Mass of the first block, m₁ = 4.0 kgMass of the second block, m₂ = 2.0 kgThe net force on the system of the two blocks is calculated as follows;
[tex]m_2 g - T = m_1 a[/tex]
where;
T is the tension in the connecting string due weight of the first block[tex]m_2 g - m_1 g = m_1 a\\\\a = \frac{m_2 g - m_1g}{m_1} \\\\a = \frac{g(m_2 - m_1)}{m_1} \\\\a = \frac{9.8(2-4)}{2} \\\\a = -9.8 \ m/s^2[/tex]
Thus, the acceleration of the first block (4 kg) is -9.8 m/s².
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A passenger of mass 90kg is involved in a minor car crash. The car approaches a solid barrier at 32m/s. It crashes into the barrer and stops in 0.2s. Determine the impulse that must be applied to the car to bring it to rest
Answer:
I hope this helps:)
Explanation:
good luck:)
What is the energy of a 5 kg object that is held at a height of 3 m above the ground?
I really need the Formula, substitute, answer
Answer:
a 5 kilogram mass, at a height of 3 meters, while acted on by Earth's gravity would have 147.15 Joules of potential energy, PE = 3kg * 9.81 m/s2 * 5m = 147.15 J.
Explanation:
1. A ball rolls off a desk at a speed of 3.0 m/s, lands 0.40 seconds later, and lands 1.2 m away from the desk
b) how high is the desk
Answer:
0.784m
Explanation:
This is a projectile motion problem so we analyze the horizontal and vertical motions separately. Since the question is asking for the height of the desk, it shows we need to analyze the vertical motion.
1. A block with a mass of 5.0 kg is pushed on a frictionless surface by applying a horizontal force of 80.0 N. The block starts from rest, and its final velocity is 12.6 m/s.
Answer:
397 j
Explanation:
Because 5.0kg yuh
The kinetic energy of the block with a mass of 5.0 kg is pushed on a frictionless surface by applying a horizontal force of 80.0 N. The block starts from rest, and its final velocity is 12.6 m/s is 396.9 J.
What is kinetic energy?A moving item or particle has a certain kind of power called kinetic energy. When an item exerts a net force to perform work, which involves the transfer of energy, the object accelerates and as a result, gains kinetic energy.
The amount of kinetic energy that a moving object or particle possesses is determined by both its mass and rate of motion. Any combination of vibration, axis rotation, translation (or movement along a path from one location to another), and translation are all possible motion types.
Given:
The mass of the block, m = 5 kg,
The horizontal force, F = 80 N,
The velocity of the block, v = 12.6 m / s,
Calculate the kinetic energy by the formula given below,
[tex]KE = 1/2mv^2[/tex]
KE = 1 / 2 × 5 × 12.6²
KE = 1 / 2 × 5 × 158.6
KE = 396.9 J
Therefore, the kinetic energy of the block with a mass of 5.0 kg is pushed on a frictionless surface by applying a horizontal force of 80.0 N. The block starts from rest, and its final velocity is 12.6 m/s is 396.9 J.
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The complete question is A block with a mass of 5.0 kg is pushed on a frictionless surface by applying a horizontal force of 80.0 N. The block starts from rest, and its final velocity is 12.6 m/s. Find its kinetic energy?
The average distance between the Earth and the Moon is 384,000,000 meters. What is the distance in kilometers?
Answer:
384000
Explanation:
it would be 384000 in km
Before a rifle is fired, the linear momentum of the bullet-rifle system is zero.
After the rifle is fired:
I. kinetic energy of the system is zero
II. linear momentum of the system is zero
Hi there!
II. Linear momentum of the system is zero.
This is an example of a RECOIL collision. With the Law of Conservation of Momentum, momentum remains constant before and after the collision.
Thus, the total momentum would also be equivalent to zero after the collision.
which is the following is a modern method used to pressure food
1. sprey drying
2.salting
3. smoking
4. Honey
Answer:
1. Spray Drying
Explanation:
it is the most modern
you push on a 24.5kg bag of sand and give it an acceleration of 2.30 m/s ^s. what is the magnitude of the force the bag returns …
Answer:
Below in bold.
Explanation:
Newton's Second Law:
Force = mass * acceleration
So
Force = 24.5 * 2.3
= 56.35 Newtons.
Can someone please give me the (Answers) to this? ... please ...
Answer:
1. 60 kg m/s
2. 2.4 kg
3. none they both have same momentum
help please I really need this
Answer:
Explanation:
Height(h) = 4m; depth of the half filled tank 4m/2 =2m
Acceleration due to Gravity (g)=9.8m/s^2
Density (d)=1000kg/m^3
Pressure = hdg
Pressure = 2m × 9.8m/s^2 × 1,000kg/m^3
Pressure = 19,600Pa
what is the cost of monthly (30 days) electric bill of ana if her city's cost of electricity is 0.05$ per kwh and she uses three refrigerators running in 600-watt power rating and open 24 hours
Answer:
1kW = 1000W
600W = 0.6kW
Cost of electric bill = 0.6kWh × 24 × 30 × $0.05
= $21.60
a 25 kg cart has 125 kg*m/s of momentum, how fast is the car going
Answer:
5m/s
Explanation:
equation for momentum is mass times velocity
25 x 5 = 125
Shanika is an engineer at an amusement park who is experimenting with changes to the setup for a magnetic roller coaster ride. In one ride, there are two identical roller coaster cars (orange and green) that start on opposite sides of a large magnet located at the center of a station. Shanika wants to get the largest increase in potential energy she can by moving one car one space to the left or the right.
Shanika can move the orange car to point A or point B, or she can move the green car to point C or point D. Which movement should she make? Why will that movement result in the largest increase in potential energy? Describe the magnetic force that will act on the roller coaster car she moves.
Answer:
I believe it might be point A since the question ask what will result in the ln a largest increase in potential energy
Explanation:
point A since the question ask what will result in the ln a largest increase in potential energy
What are the types of energy ?The energy can be defined as ability to work or produce action and / or movement and manifests itself in many different ways, such as body movement, heat, electricity, etc.
The different types of energy include Kinetic energy which is associated with the movement of bodies. Potential energy which is stored by virtue of a body's position relative to its surface is also called gravitational potential energy.
Thermal Energy or Heat can be defined as the energy associated with the kinetic energy of the molecules that make up an element, it can be manifested if there is a temperature difference between two bodies.
Chemical energy released or formed from chemical reactions, Solar energy from sunlight. This form of energy is used to generate electricity through photovoltaic plates, for example.
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what is the magnitude of the force exerted on the bottom of the bottle?
Under the Big Top elephant. Ella [2500 kg]. is attracted to Phant, the 3,000 kg elephant. They are separated by 8 m. What is the gravitational attraction between them? G=6.67×10^-11 (-11 is an exponent)
Hi there!
We can use the same equation for Gravitational Force:
[tex]\large\boxed{F_g = G\frac{m_1m_2}{r^2}}[/tex]
Fg = force due to gravity (N)
G = gravitational constant
m1,m2 = masses of objects (kg)
r = distance between objects (m)
Plug in the values provided:
[tex]F_g = (6.67*10^{-11})\frac{(2500)(3000)}{8^2} = \large\boxed{7.814 * 10^{-6}N }[/tex]
[tex]\huge\bf\underline{\underline{\pink{A}\orange{N}\blue{S}\green{W}\red{E}\purple{R:-}}}[/tex]
Here we've been given,
Universal gravitational constant (G) = [tex] \sf{6.67 \times {10}^{ - 11} }[/tex]Mass of object 1 (m1) = 2500 kg Mass of object 2 (m2) = 3000 kg Distance between two objects (r) = 8 mWe have to find the gravitational attraction force (Fg) = ?
The standard formula to solve is given by,
[tex]:\implies\tt{F_g = g \frac{m_1m_2}{ {r}^{2} } } [/tex]
[tex]:\implies\tt{F_g = 6.67 \times {10}^{ - 11} \times \frac{(2500 )(3000)}{ {8}^{2} } }[/tex]
[tex]:\implies\tt{F_g = 6.67 \times {10}^{ - 11} \times \frac{7500000}{64} }[/tex]
[tex]:\implies\tt{F_g = 7.814 \times {10}^{ - 6} }[/tex]
Gravitational force of attraction is 7.814 × 10^-6 N.State one use and one disadvantage of the expansion of materials when they are heated.
use
disadvantage
An example of the use of expansion of materials when they are heated is in
Thermometers. Thermometer is an instrument which is used to measure the
temperature of a place.
It contains mercury which expands as temperature increases. This is used
to accurately determine the temperature of an object.
The disadvantage of expansion of materials when they are heated can be
seen in roads as there is expansion of the surface as a result of heat
exposure leads to roads being rough thereby incurring more costs for
repairs.
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What is the radius of a circular space station that spins with a linear speed of 90.0 m/s I’m order for its walls to supply a centripetal acceleration equal to the acceleration provided by gravity on the surface of the earth?
Answer:
a = v^2 / R centripetal acceleration
R = v^2 / a a at the surface of earth is 9.8 m/s^2
R = (90 m/s)^2 / 9.8 m/s^2 = 827 m
What is the net force on a car with a mass of 1000 kg if its
acceleration is 35 m/s^2?
Answer:
3000N
Explanation:
divided to get answer
the force needed to accelerate the 1000kg car by 3m/s2 is 3000N
Which model below shows the positions of the Sun, Moon, and Earth that have the greatest effect on ocean tides?
Answer:
Below!
Explanation:
Referring to the picture, we can conclude that the picture J will have the most effect on ocean tides.
Whichever model is similar to model J in the picture will be your answer.
Hoped this helped.
1. An 80 kg skydiver uses a parachute to produce an applied force of 700 N while falling with an initial
velocity of 40 m/s
Time taken by the skydiver to land on the ground is 4.57 seconds
The applied force, F = 700 N
The mass of the skydiver, m = 80 kg
The velocity, v = 40 m/s
To calculate the time taken by the skydiver to fall to the ground will be calculated using the formula
[tex]F=\frac{mv}{t}[/tex]
Substitute F = 700, m = 80, and v = 40 to solve for t
[tex]700=\frac{80(40)}{t}\\\\700t=3200\\\\t=\frac{3200}{700}\\\\t=4.57 s[/tex]
Time taken by the skydiver to land on the ground is 4.57 seconds
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An EMT raceway contains two 3-phase, 480-volt branch circuits. What minimum size THHN branch circuit conductor is required for one of the circuits when it supplies a noncontinuous load of 27 kW?
Answer:8 AWG
Explanation: 6 current carrying conductors with 2 - 3 phase branch circuits. Divide 27000 by (480v × 1.732 or square root of 3) and you get 32.46 amps. Now using the code book, divide 32.46 by the adjustment factor of 6 CCC's which is .8 and you get 40.61 amps. Now go to Table 310.16 in the 2020 NEC book and you'll see that in the 90° column that 40.61 amps is above #10 AWG so you size up to #8 AWG.
The mass of a sample of sodium bicarbonate is 2. 1 kilograms (kg). There are 1,000 grams (g) in 1 kg, and 1 Times. 109 nanograms (ng) in 1 g. What is the mass of this sample in ng? 2. 1 Times. 103 ng 2. 1 Times. 106 ng 2. 1 Times. 109 ng 2. 1 Times. 1012 ng.
2.1 kg of sodium bicarbonate is equal to the 2.1 x 10¹² ng of sample. Option D is correct.
Mass is the quantity of the substance in the body or object. The SI unit of mass is Kilogram.
There are other units of measure,
Milligram: 1 g is equal to the [tex]\bold {10^3 \ mg}[/tex]Micro-gram: 1 g is equal to [tex]\bold {10^{6} \ \mu g}[/tex]Nano-gram: 1 g is is equal to[tex]\bold {10^{9} \ ng}[/tex]First convert kg to gram,
Since, 1 Kg = 1000 g
2.1 kg = grams of sample
So,
Do the cross multiplication,
[tex]\rm mass\ of\ sample = \dfrac {2.1\ kg \times 1000\ g }{ 1 kg}\\\\\rm mass\ of\ sample =2100 g[/tex]
Now, convert 2100 g to nano-grams
Since, 1 g = 1 x 10⁹ ng
2100 g = ng of sample
So,
Do the cross multiplication,
[tex]\rm mass\ of\ sample = \dfrac {2100 g \times 1 \times 10^9 ng }{1\ g}\\\\\rm mass\ of\ sample = 2.1 \times 10^1^2 ng[/tex]
Therefore, 2.1 kg of sodium bicarbonate is equal to the 2.1 x 10¹² ng of sample.
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