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seraphim [82]
2 years ago
13

Consider to cars traveling directly towards each other. Car A has twice the mass of Car B. In order for the cars to completely s

top when they collide, which of the following is true? (Remember, Momentum = mass * velocity)
Car A should have twice the velocity of Car B


Car B should have twice the velocity of Car A


Car A should have 4x the velocity of Car B


Car B should have 4x the velocity of Car A
Physics
1 answer:
Greeley [361]2 years ago
7 0

Please mark brainliest you answer is B).

Have a good night!

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A bicycle rider pushes a 13kg bicycle up a steep hill. the incline is 24 degree and the road is 275m long. the rider pushes the
Digiron [165]

Answer:

A. W = 6875.0 J.

B. W = -14264.6 J.

Explanation:

A. The work done by the rider can be calculated by using the following equation:

W_{r} = |F_{r}|*|d|*cos(\theta_{1})

Where:                

F_{r}: is the force done by the rider = 25 N

d: is the distance = 275 m

θ: is the angle between the applied force and the distance

Since the applied force is in the same direction of the motion, the angle is zero.

W_{r} = |F_{r}|*|d|*cos(0) = 25 N*275 m = 6875.0 J

Hence, the rider does a work of 6875.0 J on the bike.

B. The work done by the force of gravity on the bike is the following:

W_{g} = |F_{g}|*|d|*cos(\theta_{2})  

The force of gravity is given by the weight of the bike.

F_{g} = -mgsin(24)     

And the angle between the force of gravity and the direction of motion is 180°.

W_{g} = |mgsin(24)|*|d|*cos(\theta_{2})  

W_{g} = 13 kg*9.81 m/s^{2}*sin(24)*275 m*cos(180) = -14264.6 J  

The minus sign is because the force of gravity is in the opposite direction to the motion direction.

Therefore, the magnitude of the work done by the force of gravity on the bike is 14264.6 J.  

I hope it helps you!                                                                                          

3 0
2 years ago
Consider a particle on which several forces act, one of which is known to be constant in time: Fi = 3.00 i +4.00 ) N. As a resul
Leviafan [203]

Given that.

F=3•i+4•j

And it from point (0,0)m to (5,6)m

dx=final position - initial position

dx=(5,6)-(0,0)

dx=(5,6)m

dx=5•i +6•j

Work done by the force is give by

W = F•dx

W=F•dx

Note that i•i=j•j=1 and i•j=j•i=0

Then,

W=(3i+4j)•(5i+6j)

Therefore,

W=3i•(5i+6j)+4j•(5i+6j)

W=15i•i+18i•j+20j•i+24j•j

W=15+0+0+24

W=39J

Then the work done by the force is 39 Joules

4 0
3 years ago
How does the work required to accelerate a particle from 10 m/s to 20 m/s compare to that required to accelerate it from 20 m/s
poizon [28]

To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore

W = \Delta KE

\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2

Here,

m = mass

v_{f,i} = Velocity (Final and initial)

First case) When the particle goes from 10m/s to 20m/s

\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2

\Delta W = \frac{1}{2} (m)(20)^2 -\frac{1}{2} (m)(10)^2

W_1 = 150(m) J

Second case) When the particle goes from 20m/s to 30m/s

\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2

\Delta W = \frac{1}{2} (m)(30)^2 -\frac{1}{2} (m)(20)^2

W_1 = 250(m) J

As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.

5 0
3 years ago
What minimum number of cycles are necessary for the engine to lift a rock of mass 570 kg through a height of 150 m?
gogolik [260]
5 is the minimum number of cycles necessary for the engine to lift a rock of mass 570 kg
6 0
3 years ago
The three major scales used to measure earthquakes are Mercalli Scale, Richter Scale and Magnitude Scale.
tester [92]

The statement ‘The three major scales used to measure earthquakes are Mercalli Scale, Richter Scale and Magnitude Scale’ is true. These three scales used to measure the seismic waves released by the earthquake.

7 0
3 years ago
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