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Anon25 [30]
3 years ago
11

An object is dropped onto the moon (gm = 5 ft/s2). How long does it take to fall from an elevation of 250 ft.?

Physics
2 answers:
anyanavicka [17]3 years ago
8 0

Answer:

10 seconds

Explanation:

x = x₀ + v₀ t + ½ at²

250 = 0 + (0) t + ½ (5) t²

250 = 2.5 t²

t² = 100

t = 10

It takes 10 seconds to land from a height of 250 ft.

Papessa [141]3 years ago
3 0

Answer:

10 seconds

Explanation:

It takes 10 seconds for an object to fall from an elevation of 250 ft.

250 = 0 + (0) t + ½ (5) t²

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8. A car initially has 100 J of total energy. After driving down the road, the car's
sveta [45]

The work done on the car is -20 J.

Work done on the car is negative, meaning that the car actually does work on the external system.

<h3>Energy and law of conservation of energy</h3>
  • Energy is the ability to do work
  • the law of conservation of energy states that the total energy in a system is conserved

From the law of conservation of energy, the initial energy of the car before it moves down the road remains constant or unchanged.

  • Initial energy = 100 J
  • Initial energy = Final energy - work done on car
  • Final Energy = Work done on car + initial energy

80J = Work done on car + 100 J

Work done on car = 80 - 100J

Work done on car = -20 J

Hence, the work done on the car is -20 J

Work done on car is negative.

Since work done on the car is negative, it means that the car actually does work on the external system. Hence, the decrease in the energy of the car.

Learn more about energy and work at: brainly.com/question/13387946

8 0
2 years ago
The Earth revolves around the Sun once a year at an average distance of 1.50×1011m. Find the orbital radius that corresponds to
DedPeter [7]

Answer:

9.4\cdot 10^{10} m

Explanation:

We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write

\frac{r_a^3}{T_a^2}=\frac{r_e^3}{T_e^2}

where

r_o is the distance of the new object from the sun (orbital radius)

T_o=180 d is the orbital period of the object

r_e = 1.50\cdot 10^{11} m is the orbital radius of the Earth

T_e=365 d is the orbital period the Earth

Solving the equation for r_o, we find

r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m

3 0
3 years ago
On Earth, a spring stretches by 5.0 cm when a mass of 3.0 kg is suspended from one end.
Neko [114]

Answer:

Mass = 18.0 kg

Explanation:

From Hooke's law,

F = ke

where: F is the force, k is the spring constant and e is the extension.

But, F = mg

So that,

mg = ke

On the Earth, let the gravitational force be 10 m/s^{2}.

3.0 x 10 = k x 5.0

30 = 5k

⇒ k = \frac{30}{5} ................ 1

On the Moon, the gravitational force is \frac{1}{6} of that on the Earth.

m x \frac{10}{6} = k x 5.0

\frac{10m}{6} = 5k

⇒ k = \frac{10m}{30} ............. 2

Equating 1 and 2, we have;

\frac{30}{5}  = \frac{10m}{30}

m = \frac{900}{50}

    = 18.0

m = 18.0 kg

The mass required to produce the same extension on the Moon is 18 kg.

8 0
2 years ago
Read 2 more answers
Please help, no links please! I dont understand stand this question and im going to cry
Vaselesa [24]

Answer:

I'm pretty sure its helium

3 0
2 years ago
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Determine the kinetic energy of a 1000 kg roller coaster car that is moving with speed of 40.0 m/s​
ololo11 [35]

Answer:

KE=800,000

Explanation:

The formula for kinetic energy is KE=1/2mv^2 or Kinetic Energy= 0.5*mass*velocity^2

so 1000 is the mass and 40 is the velocity

KE=0.5*1000*40^2

KE=0.5*1,000*1,600

KE=800,000 Joules

8 0
2 years ago
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