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Rainbow [258]
3 years ago
6

Two objects collide and bounce apart. Assuming no outside forces act on the system, which best describes the total momentum afte

r the collision?
It is always greater than it was before the collision.
It is often greater than it was before the collision.
It is always the same as it was before the collision.
It is often the same as it was before the collision.
Physics
2 answers:
Yuliya22 [10]3 years ago
8 0

Answer:

It is always the same as it was before the collision.

wariber [46]3 years ago
8 0

Answer:

The answer is C on Edge

Explanation:

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Delvig [45]
The kinetic energy will increase
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3 years ago
A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.6 multiply 1010 m (inside t
Doss [256]

We will apply the concepts related to energy conservation to develop this problem. In this way we will consider the distances and the given speed to calculate the final speed on the path from the sun. Assuming that the values exposed when saying 'multiply' is scientific notation we have the following,

d_1 = 4.6*10^{10}m

v_i = 9.3*10^4m/s \rightarrow \text{Initial velocity comet}

d_2 = 6*10^{12}m

The difference of the initial and final energy will be equivalent to the work done in the system, therefore

E_f = E_i +W

K_f +U_f = K_i +U_i + 0

\frac{1}{2} mv_f^2+\frac{-GMm}{d_2} = \frac{1}{2} mv_i^2+\frac{-GMm}{d_1}

Here,

m = Mass

v_f = Final velocity

G = Gravitational Universal Constant

M = Mass of the Sun

m = Mass of the comet

v_i = Initial Velocity

Rearranging to find the final velocity,

v_f = \sqrt{v_i^2+2GM(\frac{1}{d_2}-\frac{1}{d_1})}

Replacing with our values we have finally,

v_f = \sqrt{(9.3*10^4)+2(6.7*10^{-11})(1.98*10^{30})(\frac{1}{6*10^{12}}-\frac{1}{4.6*10^{10}})}

v_f = 75653.9m/s

Therefore the speed is 75653m/s

8 0
4 years ago
How large must the coefficient of static friction be between the tires and the road if a car is to round a level curve of radius
Korvikt [17]

Answer:

<h2>The coefficient of static friction will be 0.7</h2>

Explanation:

Given data

the radius of curve= 90m

speed v= 90 km/h to m/s = (90*100)/60*60=  25 m/s

we know that the expression for the centripetal force acting on the car

Fc= \frac{mv^2}{r}-------1

we also know that the expression for the frictional force between road and tire.

Ff= μmg--------2

Equating equation 1 and 2 we have

μmg= mv^2/r

μ= v^2/gr

substituting the values of speed and radius we have (assuming g= 9.81m/s^2)

μ= 25^2/9.81*90

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μ= 0.7

4 0
3 years ago
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Explanation:

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3 years ago
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If a bird collides with the windshield of a fast-moving bus, Which one experiences and impact force with larger magnitude? Which
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