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Strike441 [17]
3 years ago
5

A 55 kg skier is at the top of a slope. The vertical distance between start and finish is 10m. (a) What is her Potential Energy?

(b) How fast is she moving at the bottom of the hill?
Physics
1 answer:
Blababa [14]3 years ago
6 0

Explanation:

Given that,

Mass of a skier, m = 55 kg

The vertical distance between start and finish is 10 m.

(a) The potential energy of a skier is given by :

E=mgh\\\\E=55\times 9.8\times 10\\\\=5390\ J

(b) When she reaches the bottom of the hill, the potential energy is converted to kinetic energy. So,

5390=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2\times 5390}{55}} \\\\v=14\ m/s

Hence, her potential energy is 5390 J and at the bottom of the hill she is moving with a speed of 14 m/s.

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What does the rotational speed of the ring module have to be so that an astronaut standing on the outer rim of the ring module f
Usimov [2.4K]

Complete question:

In the movie The Martian, astronauts travel to Mars in a spaceship called Hermes. This ship has a ring module that rotates around the ship to create “artificial gravity” within the module. Astronauts standing inside the ring module on the outer rim feel like they are standing on the surface of the Earth. (The trailer for this movie shows Hermes at t=2:19 and demonstrates the “artificial gravity” concept between t= 2:19 and t=2:24.)

Analyzing a still frame from the trailer and using the height of the actress to set the scale, you determine that the distance from the center of the ship to the outer rim of the ring module is 11.60 m

What does the rotational speed of the ring module have to be so that an astronaut standing on the outer rim of the ring module feels like they are standing on the surface of the Earth?

Answer:

The rotational speed of the ring module have to be 0.92 rad/s

Explanation:

Given;

the distance from the center of the ship to the outer rim of the ring module r, = 11.60 m

When the astronaut standing on the outer rim of the ring module feels like they are standing on the surface of the Earth, then their centripetal acceleration will be equal to acceleration due to gravity of Earth.

Centripetal acceleration, a = g = 9.8 m/s²

Centripetal acceleration, a = v²/r

But v = ωr

a = g = ω²r

\omega = \sqrt{\frac{g}{r}} = \sqrt{\frac{9.8}{11.6} } = 0.92 \ rad/s

Therefore, the rotational speed of the ring module have to be 0.92 rad/s

3 0
4 years ago
Which diagram represents an open circuit?
Kruka [31]
Answer : D) Circuit A

This circuit is the only circuit where it is not complete, having and open spot towards the bottom of it, making it and open circuit.
6 0
3 years ago
Read 2 more answers
A sample of ore containing radioactive strontium 38Sr90 has an activity of 8.2 × 105 Bq. The atomic mass of strontium is 89.908
ahrayia [7]

From the activity values and the decay constant, the mass of of Strontium in the sample is:

1.62 × 10^{-7}g

<h3>What is the decay constant of an element?</h3>

The decay constant of an element is the probability of decay of a nucleus per unit time.

{λ = ln 2 / t1/2 

where;

t1/2 is the half-life of the isotope.

The half-life is converted to seconds since the decay constant is asked in per seconds.

28.8 years = 28.8 × 3.156 × 10^{7} = 908928000 s \\

Hence;

λ = \frac{ln2}{90892800s} = 7.626 s^{-1}

                                       

The activity of the element, A, the decay constant, λ and the number of nuclei, N are related as follows:

  • A = (–) λN

N = \frac{8.25 ×10^{5}}{7.626×10^{-10}} = 1.082 × 10^{15}              

Molar mass of Strontium-90 is 90 g.

1 mole of Strontium-90 contains 6.02×10^23 nuclei.

The mass, m of Strontium in the sample is calculated:

m = 1.082 × 10^{15} × \frac{90 g}{6.02 × 10^{23}} = 1.62 × 10^{-7}g \\

Therefore, the mass of of Strontium in the sample is:

1.62 × 10^{-7} \: g

Learn more about decay constant at: brainly.com/question/17159453

5 0
3 years ago
A horizontal force of 14.0N is applied to a box of m=32.5kg with Vo=0. Ignoring friction, how far does the crate travel in 10.0s
Alex Ar [27]
I’m going to assume initial velocity is 0.

Use Newton’s second law:

F = m•a

F/m = a

14.0/32.5kg= 28/65 m/s^2

Use constant SUVAT acceleration formulae:

S- displacement - what we need to find out

U - initial velocity - 0

V

A - 28/65 m/s^2

T - 10 seconds

S = ut + 1/2at^2

Since u = 0

S = 1/2at^2

1/2• 28/65 • 10^2 = 21.5metres~

Answer is 21.5 metres

~Hoodini, here to help.
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In which approximate direction is the warm front in the upper left of the weather map moving?
Pie

Answer:

East

Explanation:

I did the test

8 0
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