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Monica [59]
3 years ago
14

A rock is dropped from rest. How fast is it going after it has been falling for 9.2 s?

Physics
1 answer:
natka813 [3]3 years ago
6 0

90.2 m/s. A rock dropped from rest after it has falling for 9.2 seconds will have a velocity of 90.2m/s.

The bodies left in free fall increase their speed (downwards) by 9.8 m/s² every second. The acceleration of gravity is the same for all objects and is independent of the masses of these. In the free fall the air resistance is not taken into account.

v = g*t

v = (9.8m/s²)(9.2s) = 90.2 m/s

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HELP PLEASE !!
Umnica [9.8K]
If a car crashes into another car like this, the wreck should go nowhere. Besides this being an unrealistic question, the physics of it would look like this:

Momentum before and after the collision is conserved.

Momentum before the collision:
p = m * v = 50000kg * 24m/s + 55000kg * 0m/s = 50000kg * 24m/s
Momentum after the collision:
p = m * v = (50000kg + 55000kg) * v

Setting both momenta equal:
50000kg * 24m/s = (50000kg + 55000kg) * v

Solving for the velocity v:
v = 50000kg * 24m/s/(50000kg + 55000kg) = 11,43m/s

3 0
3 years ago
Which of the following is the best definition of a closed system?
lisov135 [29]

The answer is either

b A system in which Newton's Laws are valid

or

c A system in which there are no external forces.

Explanation:

not a, and not d

There are energy changes in a closed system.

A closed system obeys the conservation laws in its physical description.

4 0
3 years ago
The battery of a flashlight develops 3 V, and the current through the bulb is 200 mA.What power is absorbed by the bulb?
Verizon [17]

Answer : The power absorbed by the bulb is, 0.600 W

Explanation :

As we know that,

Power = Voltage × Current

Given:

Voltage = 3 V

Current = 200 mA = 0.200 A

Conversion used : (1 mA = 0.001 A)

Now put all the given values in the above formula, we get:

Power = Voltage × Current

Power = 3V × 0.200 A

Power = 0.600 W

Thus, the power absorbed by the bulb is, 0.600 W

3 0
3 years ago
An object has a height of 0.066 m and is held 0.210 m in front of a converging lens with a focal length of 0.140 m. (Include the
Anvisha [2.4K]

Explanation:

Given that,

Size of object, h = 0.066 m

Object distance from the lens, u = 0.210 m (negative)

Focal length of the converging lens, f = 0.14 m

If v is the image distance from the lens, we can find it using lens formula as follows :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{0.14 }+\dfrac{1}{(-0.21)}\\\\v=0.42\ m

(a) Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{0.42}{(-0.21)}\\\\m=-2

(b) Magnification, m=\dfrac{h'}{h}

h' is image height

-2=\dfrac{h'}{0.066}\\\\h'=-2\times 0.066\\\\h'=-0.132\ m

Hence, this is the required solution.

4 0
3 years ago
What is the density to the object g/cm3
alexdok [17]
Should be 1.4, I hope this helps you out
6 0
2 years ago
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