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vfiekz [6]
3 years ago
14

a ball at rest starts rolling down a hill with a constant acceleration of 3.2 meters/second2. what is the final velocity of the

ball after 6.0 seconds ​
Physics
1 answer:
Y_Kistochka [10]3 years ago
6 0

Answer:

3.2(6.0) = 19.2 m/s

Explanation:

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Of all the radio waves, these have the shortest wavelengths and the the highest frequency?
Julli [10]
If by chance you mean any wave, the wave with the shortest wavelength/highest frequency is gamma rays.
4 0
3 years ago
Why do nations should establish a set of rules and principles for responsible lunar/moon explortions? ASAP pleaseeee :(
Marianna [84]

So that we do not contaminate it with microorganisms or garbage or other human stuff.

5 0
3 years ago
When might it be harder to stop a vehicle moving at 30 km/h than one moving at 60 km/h?
rodikova [14]

Answer:

when the momentum of the vehicle moving at 30 km/h is higher than the one from the vehicle moving at 60 km/h

Explanation:

It's much harder to stop a freight truck moving at 30 km/h than a hot wheels car moving at 60 km/h.

4 0
2 years ago
A 4.79 g bullet moving at 642.3 m/s penetrates a tree trunk to a depth of 4.35 cm. Use work and energy considerations to find th
vladimir1956 [14]

Answer:

Force, F=2.27\times 10^4\ N

Explanation:

Given that,

Mass of the bullet, m = 4.79 g = 0.00479 kg

Initial speed of the bullet, u = 642.3 m/s

Distance, d = 4.35 cm = 0.0435 m

To find,

The magnitude of force required to stop the bullet.

Solution,

The work energy theorem states that the work done is equal to the change in its kinetic energy. Its expression is given by :

F.d=\dfrac{1}{2}m(v^2-u^2)

Finally, it stops, v = 0

F.d=-\dfrac{1}{2}m(u^2)

F=\dfrac{-mu^2}{2d}

F=\dfrac{-0.00479\times (642.3)^2}{2\times 0.0435}

F = -22713.92 N

F=2.27\times 10^4\ N

So, the magnitude of the force that stops the bullet is 2.27\times 10^4\ N

7 0
3 years ago
As a rocket rises, its kinetic energy changes
Bond [772]
<span>Transformed into potential energy</span>
3 0
4 years ago
Read 2 more answers
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