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Rashid [163]
2 years ago
13

N

Physics
1 answer:
9966 [12]2 years ago
7 0

Answer: de?

Explanation:

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A penny is dropped from the top of a building that is 300.0 m tall. Calculate the speed of the penny as it hits the ground. (met
Sauron [17]

We have the equation of motion s = ut + \frac{1}{2} at^2, where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.

Here s = 300 m, u = 0 m/s, a = 9.81 m/s^2

Substituting

   300 = 0*t+\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2 = 300\\ \\ t =7.82 seconds

Now we have v = u+at, where v is the final velocity

Here u = 0 m/s, a= 9.81 m/s^2 and t = 7.82 seconds

Substituting

     v = 0+9.8*7.82 = 76.68 m/s

The speed with which the penny strikes the ground = 76.68 m/s.

3 0
3 years ago
A square sheet of rubber has sides that are 20 cm long. What is the area of the square of rubber in cm squared?
Alinara [238K]

Answer:

400cm^2

Explanation:

sides are 20cm long Area for a square is a squared

since all the lides are of equal length you can just choose one side.

20squared is 400

20 x 20 = 400cm squared

Hope this helps :)

6 0
3 years ago
Which of the following describes the charge of an atom before any electrons are transferred?
Nadya [2.5K]
I am pretty sure that<span> the following which describes the charge of an atom before any electrons are transferred is </span>neutral charge. According to the fact that <span> atom is always like that before ionization, this answer is definitely correct. Hope it helps!</span>
7 0
3 years ago
How to tell how much work gravity does on something?
Grace [21]
By dropping a ball and seeing how long it takes to hit the ground or throw a ball up and time it as well
4 0
2 years ago
If a galaxy has an apparent velocity of 2300 km/s, what is its distance if the Hubble constant is assumed to be 70 km/s/Mpc
prohojiy [21]

The distance of the galaxy is 32.86 Mpc.

Using the hubble law, v = H₀D where v = apparent velocity of galaxy = 2300 km/s, H = hubble constant = 70 km/s/Mpc and D = distance of galaxy.

Since we require the distance of the galaxy, we make D subject of the formula in the equation. So, we have

D = v/H₀

Substituting the values of the variables into the equation, we have

D = 2300 km/s ÷ 70 km/s/Mpc

D = 32.86 Mpc

So, the distance of the galaxy is 32.86 Mpc

Learn more about hubble law here:

brainly.com/question/18484687

4 0
2 years ago
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