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Marizza181 [45]
4 years ago
14

An orange of mass 3kg falls from a height 50m. When its height is H, its speed is

Physics
1 answer:
MArishka [77]4 years ago
8 0

Answer:

(10 m/s²) (50 m) = (10 m/s²) H + ½ (4 m/s)²

Explanation:

Initially, the orange has only gravitational potential energy.

As the orange falls, it has a combination of potential energy and kinetic energy.

Energy is conserved, so initial energy = final energy:

PE₀ = PE + KE

mgH₀ = mgH + ½ mv²

gH₀ = gH + ½ v²

Plugging in values:

(10 m/s²) (50 m) = (10 m/s²) H + ½ (4 m/s)²

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Pedro picked up and held a cube of ice in his hand and noticed that he no longer felt a difference in temperature after the ice
nataly862011 [7]

Answer:

the answers is 1.

Explanation:

this is because heat always travel's from the warmest object to the coldest. His hands thermal energy is transferred to the ice making the ice round to the same temperature of his hand. This can also be an example of equalibrum.

4 0
3 years ago
A 500kg elevator is raised to a height of 10 m in 10 seconds. Calculate the power of the motor.
Hunter-Best [27]

Answer:

Correct answer: Third statement  P = 4900 W

Explanation:

Given:

m = 500 kg  the mass of the elevator

h = 10 m  reached height after t = 10 seconds

P = ? power of the motor

The formula for the calculating power of the motor is:

P = W / t

since work is a measure of change in this case of potential energy then it is:

W = ΔEp = Ep - 0 = Ep

In this case we must take g = 9.81 m/s²

Ep = m g h = 500 · 9.81 · 10 = 49,050 W ≈ 49,000 W

Ep ≈ 49,000 W

P = Ep / t = 49,000 / 10 = 4,900 W

P =4,900 W

God is with you!!!

3 0
4 years ago
Estimate the acceleration due to gravity at the surface of Europa (one of the moons of Jupiter) given that its mass is 4.9×1022k
Taya2010 [7]

Answer:g=1.97 m/s^2

Explanation:

Given

mass of Jupiter is M=4.9\times 10^{22} kg

Density of Jupiter is same as Earth

density\ of\ Earth=density\ of\ Jupiter=5510 kg/m^3

mass=volume\times density

considering Jupiter to be sphere of radius r

M=\frac{4}{3}\times \pi r^3\times \rho

r^3=\frac{3M}{\rho \times 4\pi}

r^3=\frac{3\times 4.9\times 10^{22}}{5510\times 4\pi}

r=(\frac{3\times 4.9\times 10^{22}}{5510\times 4\pi})^{\frac{1}{3}}

r=1.28\times 10^6 m

acceleration due to gravity is given by

g=\frac{GM}{r^2}

g=\frac{6.67\times 10^{-11}\times 4.9\times 10^{22}}{(1.28\times 10^6)^2}

g=1.97 m/s^2

3 0
4 years ago
Calculate how fast the ball would be moving at the instant it leaves the projectile launcher of the spring is compressed by 3.75
Mnenie [13.5K]

Answer:

V = 8.34m/s

Explanation:

Given that

1/2ke^2 = 1/2mv^2 ......1

Where e = 3.75cm = (3.75/100)m

e = 0.0375m

K = 500 N/m

m = 10g = 10/1000

= 0.01kg

Substitute the values into equation 1

0.5×500×(0.0375)^2 = 0.5×0.01×v^2

250×0.001395 = 0.005v^2

0.348 = 0.005v^2

v^2 = 0.348/0.005

v^2 = 69.6

V = √69.6

V = 8.34m/s

The ball launches at the speed of V = 8.34m/s

6 0
3 years ago
The international space station makes 15.65 revolutions per day in its orbit around the earth. assuming a circular orbit, how hi
sweet-ann [11.9K]
<span>373.2 km The formula for velocity at any point within an orbit is v = sqrt(mu(2/r - 1/a)) where v = velocity mu = standard gravitational parameter (GM) r = radius satellite currently at a = semi-major axis Since the orbit is assumed to be circular, the equation is simplified to v = sqrt(mu/r) The value of mu for earth is 3.986004419 Ă— 10^14 m^3/s^2 Now we need to figure out how many seconds one orbit of the space station takes. So 86400 / 15.65 = 5520.767 seconds And the distance the space station travels is 2 pi r, and since velocity is distance divided by time, we get the following as the station's velocity 2 pi r / 5520.767 Finally, combining all that gets us the following equality v = 2 pi r / 5520.767 v = sqrt(mu/r) mu = 3.986004419 Ă— 10^14 m^3/s^2 2 pi r / 5520.767 s = sqrt(3.986004419 * 10^14 m^3/s^2 / r) Square both sides 1.29527 * 10^-6 r^2 s^2 = 3.986004419 * 10^14 m^3/s^2 / r Multiply both sides by r 1.29527 * 10^-6 r^3 s^2 = 3.986004419 * 10^14 m^3/s^2 Divide both sides by 1.29527 * 10^-6 s^2 r^3 = 3.0773498781296 * 10^20 m^3 Take the cube root of both sides r = 6751375.945 m Since we actually want how far from the surface of the earth the space station is, we now subtract the radius of the earth from the radius of the orbit. For this problem, I'll be using the equatorial radius. So 6751375.945 m - 6378137.0 m = 373238.945 m Converting to kilometers and rounding to 4 significant figures gives 373.2 km</span>
7 0
3 years ago
Read 2 more answers
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