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Scrat [10]
3 years ago
6

How much gravitational potential energy does a 45.2 kg object have when it is 21.9m above the ground?

Physics
1 answer:
Blizzard [7]3 years ago
6 0

Answer:

Explanation:

The formula for gravitational potential energy is

Ep = m · g · h   Assuming that the acceleration is g = 10m/s²

Ep = 45.4 · 10 · 21.9 = 9,942.6 J

God is with you!!!

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You are trying to overhear a juicy conversation, but from your distance of 20.0 m , it sounds like only an average whisper of 30
12345 [234]

Answer:

r₂ = 0.2 m

Explanation:

given,

distance = 20 m

sound of average whisper = 30 dB

distance moved closer = ?

new frequency = 80 dB

using formula

\beta = 10 log(\dfrac{I_1}{I_0})

   I₀ = 10⁻¹² W/m²

now,

30 = 10 log(\dfrac{I_1}{10^{-12}})

\dfrac{I_1}{10^{-12}}= 10^3

I_1= 10^{-8}\ W/m^2

to hear the whisper sound = 80 dB

80 = 10 log(\dfrac{I_2}{10^{-12}})

\dfrac{I_2}{10^{-12}}= 10^8

I_2= 10^{-4}\ W/m^2

we know intensity of sound is inversely proportional to square of distances

\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}

\dfrac{10^{-8}}{10^{-4}}=\dfrac{r_2^2}{20^2}

10^{-4}=\dfrac{r_2^2}{20^2}

  r₂ = 0.2 m

6 0
3 years ago
Define relative density​
weeeeeb [17]

Hello There!

Relative density is another word for specific gravity. Relative density is comparing the density of one thing to the density of another. Most of the time, you look at the density of water so we look at the density of an object relative to the density of water.

5 0
4 years ago
A large, metallic, spherical shell has no net charge. It is supported on an insulating stand and has a small hole at the top. A
cluponka [151]

Answer:

D) -Q

Explanation:

The charge inserted will induce -Q charge on the inner surface and + Q on the outer surface of the shell . This charge is called bound charge because it remained attached with opposite charge inserted inside.

5 0
4 years ago
An object moving in the xy-plane is subjected to the force f⃗ =(2xyı^ 3yȷ^)n, where x and y are in m
ryzh [129]

The work done by the applied force on the object is (2ab²i + 3b²j) J.

<h3>Magnitude of the force on the object</h3>

The magnitude of the force on the object is calculated as follows;

f = (2xyi + 3yj)

when;

x = a, and y = b

f = (2abi + 3bj)

<h3>Work done by the force</h3>

The work done the applied force is the product of force and displacement of the object.

W = fΔs

where;

  • Δs is displacement of the object

Δx = a - a = 0

Δy = 0 - b = -b

Δs = √(Δx² + Δy²)

Δs = √(-b)²

Δs = b

W = (2abi + 3bj) x b

W = (2ab²i + 3b²j) J

Thus, the work done by the applied force on the object is (2ab²i + 3b²j) J.

The complete question is below;

An object moving in the xy-plane is subjected to the force f = (2xyi + 3yj),  where x and y are in m. The particle moves from the origin to the point with coordinates (a, b) by moving first along the x-axis to (a, 0), then parallel to the y-axis. How much work does the force do?

Learn more about work done here: brainly.com/question/8119756

6 0
2 years ago
Sanjeev noticed his plants were wilting and turned on the sprinklers in his garden. An hour later, he notices that the plant's l
alexira [117]

Answer:

B

Explanation:

3 0
3 years ago
Read 2 more answers
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