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notsponge [240]
3 years ago
7

A stalled car is being pushed up a hill at constant velocity by three people. the net force on the car is

Physics
1 answer:
frosja888 [35]3 years ago
6 0
I think you need more information like the force of gravity and the force of the three people. Once you combine the two, however, you should get the net force.
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A car (mass = 1090 kg) is traveling at 30.4 m/s when it collides head-on with a sport utility vehicle (mass = 2880 kg) traveling
Thepotemich [5.8K]

Answer:

The sport utility vehicle was traveling at V2= 11.5 m/s.

Explanation:

m1= 1090 kg

V1= 30.4 m/s

m2= 2880 kg

V2= ?

m1*V1 = m2*V2

V2= (m1*V1)/m2

V2= 11.5 m/s

7 0
3 years ago
9. How much work is done when a 15kg box is lifted to a height of 2 meters?
torisob [31]

Answer: W = 294 J

Explanation: Solution:

Work is expressed as the product of force and the distance of the object.

W = Fd        where F = mg

W= Fd

   = mg d

   = 15 kg ( 9.8 m/s²) ( 2m )

   = 294 J

5 0
3 years ago
A small, positively charged ball is moved close to a large, positively charged ball. which describes how the small ball likely r
NNADVOKAT [17]

Answer;

-it will move away from the large ball because like charges repel.

Explanation;

-Electric force is the force that pushes apart two like charges, or that pulls together two unlike charges. The basic law of electrostatics Like charges of electricity repel each other, whereas unlike charges attract each other.

When small, positively charged ball is moved close to a large, positively charged ball it would be pushed away from the large positively charged ball since they are both positively charged. One has to put in energy to try to move the small ball closer to the large ball. The closer one try to move it to the large ball, the more energy one has to put in, so the more electrical potential energy the small ball would have.

6 0
3 years ago
Read 2 more answers
block with of mass m is at rest on horizontal frictionless surface at time t=0. A force given by F=Bt+C is applied horizontally
Alexeev081 [22]

Answer:

v_{2} =\frac{1}{2}

Explanation:

From the second law of Newton movement laws, we have:

F=m*a, and we know that a is the acceleration, which definition is:

a=\frac{dv}{dt}, so:

F=m*\frac{dv}{dt}\\\frac{dv}{dt}=\frac{F}{m}=\frac{\frac{1}{2}(t+1)}{4}=\frac{t+1}{8}

The next step is separate variables and integrate (the limits are at this way because at t=0 the block was at rest (v=0):

dv=\frac{1}{8}(t+1)dt\\\int\limits^{v_{2}}_0 \, dv=\int\limits^{2}_{0} {\frac{1}{8}(t+1)} \, dt

v_{2}=\frac{1}{8}*(\frac{t^{2}}{2}+t) (This is the indefinite integral), the definite one is:

v_{2}=\frac{1}{8}*(2+2)=\frac{1}{2}

3 0
3 years ago
What is the equation for an inelastic collision
abruzzese [7]
M1 v1 = (m1 + m2)v2.

All of the exponents should be lowered to the bottom right of the letters.
7 0
3 years ago
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