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Vesnalui [34]
2 years ago
9

An 80 kg skier stands at the top of a 40-meter slope. She then skis down the slope. What is her approximate speed at the bottom

of the slope if friction is negligible and her kinetic energy is 31360 J?
Physics
1 answer:
irinina [24]2 years ago
5 0

Thank you for that information.  If we know her mass and kinetic energy, then we don't need any of that other business about the height of the slope, her skiing ability, the friction of the snow, what she uses to wax her skis, the color of her parka, or what she had for lunch at the chalet.  The mass and kinetic energy are enough to answer the question.

Kinetic Energy = (1/2) · (mass) · (speed)²

31,360 J = (1/2) · (80 kg) · (speed)²

speed² = (31,360 J) / (40 kg)

speed² = 784 m²/s²

speed = <em>28 m/s</em>

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A charged particle moves at 2.5 × 104 m/s at an angle of 25° to a magnetic field that has a field strength of 8.1 × 10–2 T. If t
nekit [7.7K]

the magnitude of charge=q=8.76 x 10⁻⁵C

Explanation:

the magnetic force Fm is given by

Fm= q V B sinθ

q= charge

v= velocity= 2.5 x 10⁴ m/s

B= magnetic field strength= 8.1 x 10⁻²T

Fm= magnetic force= 7.5 x 10⁻² N

θ=25°

so 7.5 x 10⁻² =q (2.5 x 10⁴ ) (8.1 x 10⁻²) sin25

q=8.76 x 10⁻⁵C

4 0
3 years ago
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What is the orbital velocity of Jupiter around the sun
jekas [21]
13.1 km/s, that is the mean orbital velocity of Jupiter around the sun
3 0
3 years ago
Can someone help me?
Cloud [144]

What do we know that might help here ?

-- Temperature of a gas is actually the average kinetic energy of its molecules.

-- When something moves faster, its kinetic energy increases.

Knowing just these little factoids, we realize that as a gas gets hotter, the average speed of its molecules increases.

That's exactly what Graph #1 shows.

How about the other graphs ?

-- Graph #3 says that as the temperature goes up, the molecules' speed DEcreases.  That can't be right.

-- Graph #4 says that as the temperature goes up, the molecules' speed doesn't change at all.  That can't be right.

-- Graph #2 says that after the gas reaches some temperature and you heat it hotter than that, the speed of the molecules starts going DOWN.  That can't be right.  

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4 0
3 years ago
Rick shoots a basketball at an angle of 35' from the horizontal. It leaves his hands 7 feet from the ground with a velocity of 2
Korvikt [17]

Given:

The angle of projection of the basketball, θ=35°

The height at which the ball leaves the hand, h=7 ft

The initial velocity of the basketball, v=20 ft/s

To find:

The parametric equations describing the shot.

Explanation:

The range, x of the basketball is given by,

x=v\cos\theta t

On substituting the known values,

\begin{gathered} x=20\times\cos35\degree\times t \\ \implies x=16.4t \end{gathered}

The change in the height, y of the basketball is given by,

y=-v\sin\theta t+\frac{1}{2}gt^2

Where g is the acceleration due to gravity.

On substituting the known values,

\begin{gathered} y=-20\times\sin35\degree\times t+\frac{1}{2}\times32\times t^2 \\ \implies y=-11.5t+16t^2 \end{gathered}

Final answer:

The parametric equations describing the shot are

\begin{gathered} \begin{equation*} x=16.4t \end{equation*} \\ \begin{equation*} y=-11.5t+16t^2 \end{equation*} \end{gathered}

8 0
1 year ago
An astronaut lands on a new, recently discovered planet in a different star system. The astronaut measures the acceleration due
Nezavi [6.7K]

Answer:

The radius of the new planet is ~2.04 * 10⁶ m, or 2,041,752 m.

Explanation:

We can use Newton's Law of Universal Gravitation:

  • \displaystyle F_g=G\frac{Mm}{r^2}

Let's look at Newton's 2nd Law:

  • F=ma

We can set these equations equal to each other:

  • \displaystyle G\frac{Mm}{r^2} =ma

The mass of the second mass (astronaut) cancels out. We are left with:

  • \displaystyle G\frac{M}{r^2} =a

We are solving for the radius of the new planet, so we can rearrange the equation:

  • \displaystyle r=\sqrt{\frac{GM}{a} }

Substitute in our known values given in the problem (<u><em>G = 6.67 * 10⁻¹¹ </em></u><em> ; </em><u><em>M = 7.5 * 10²³</em></u><em> ; </em><u><em>a = 12</em></u>).

  • \displaystyle r =\sqrt{\frac{(6.67\times 10^{-11})(7.5 \times 10^{23}}{12} }
  • r=2.04 \times 10^6

The radius of the new planet is ~2.04 * 10⁶ m.

3 0
2 years ago
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