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slega [8]
3 years ago
9

Suppose a skydiver (mass = 80kg) is falling toward the Earth. Calculate the skydiver’s gravitational potential energy at a point

when the skydiver is 100m above the Earth.
Physics
1 answer:
Vlad [161]3 years ago
4 0
Ep is gravitational potential energy
m is mass (kg)
g is gravitational field strength (N/kg)
h is height (m)

Ep= mgh
= 80kg*9.8N/kg*100m
= 78 400 J
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Weather balloons appear only partially inflated when launched into the air. Why would scientists do this?
zvonat [6]

The reason weather balloons are only partially inflated is to allow room for the helium in the balloon to expand.

7 0
3 years ago
How much work does the charge escalator do to move 2.30 μC of charge from the negative terminal to the positive terminal of a 3.
loris [4]

Answer:

6.9\times 10^{-6} J

Explanation:

We are given that

Charge=q=2.3\mu C=2.3\times 10^{-6} C

1\mu C=10^{-6} C

Potential difference=V=3 V

We know that

Work done=V\times q

Using the formula

Work done by charge to move from the negative terminal to the positive terminal of battery=2.3\times 10^{-6}\times 3 J

Work done by charge to move from the negative terminal to the positive terminal of battery=6.9\times 10^{-6} J

6 0
4 years ago
Your electricity bill says 834-kWh of consumption. With this amount of energy, how many 59-W light bulbs can be powered for an a
kvv77 [185]

Answer:

2019 light bulbs

Explanation:

So the electrical energy consumed by each 59-W light bulb within 7 hours is the product of the power and time duration

E = Pt = 59 * 7 = 413 Wh or 0.413 kWh

If each light bulb consumes 0.413 kWh, then the total number of light bulbs needed to consume 834 kWh would be

834 / 0.413 = 2019 light bulbs

5 0
4 years ago
Which of the following is not a region of the brain?
lorasvet [3.4K]
Definitely the last one, brain stem
7 0
3 years ago
A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i
erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

\mu_k = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

or

u₁ = √(2 × 9.8 × 5)

or

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

or

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem,  we have

Work done by the friction force = change in kinetic energy of the mass 2

or

0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)

or

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring =  Kinetic energy of the mass 2

or

\frac{1}{2}kx^2=\frac{1}{2}mv_3^2

on substituting the values, we get

\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2

or

x = 2.86 m

8 0
3 years ago
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