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3 years ago

14

A 60-kg skier starts at the top of a 10-meter high slope. At the bottom, she is traveling 10 m/s. How much energy does she lose

to friction?

1 answer:

Karo-lina-s [1.5K]3 years ago

4 0

The potential energy of the skier at the top is
60 kg (9.81 m/s2) (10 m) = 5886 J

Her kinetic energy at the bottom is
(1/2) (60 kg) (10 m/s)^2 = 3000 J

The energy lost to friction is
5886 J - 3000 J = 2886 J

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shutvik [7]

<h2>Answer:2.65 seconds</h2>

Explanation:

Let $a$ be the acceleration.

Let $u$ be the initial velocity.

Let $v$ be the final velocity.

Let $t$ be the time taken.

As we know from the equations of motion,

$v=u+at$

Given, $v=0\\u=8.12ms^{-1}\\a=-3.06ms^{-2}$

$0=8.12-3.06t$

$t=2.65sec$

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3 years ago

What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.

valkas [14]

<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (1)

Where;

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$M=6.39(10)^{23}kg$ is the mass of Mars

$a=3.4(10)^{6}m$ is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}$ (2)

$T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}$ (3)

$T=\sqrt{11581157.44 s^{2}}$ (4)

Finally:

$T=3403.1099s=56.718min$ This is the orbital period of a spacecraft in a low orbit near the surface of mars

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3 years ago

Radar is used to determine distances to various objects by measuring the round-trip time for an echo from the object. (a) How fa

zloy xaker [14]

Answer:

150000000000 m

0.0000005 seconds

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Explanation:

Speed of electromagnetic waves through vacuum = $3\times 10^8\ m/s$

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The accuracy by which I will be able to measure the echo time is 33.33 ns

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3 years ago

Which is a complete sentence?

8_murik_8 [283]

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3 years ago

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Vlada [557]

Answer: The work done in J is 324

Explanation:

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P = pressure = 732 torr = 0.96 atm (760torr =1atm)

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Putting values in above equation, we get:

$W=-0.96atm\times (2.35-5.68)L=3.20L.atm$

To convert this into joules, we use the conversion factor:

$1L.atm=101.33J$

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The positive sign indicates the work is done on the system

Hence, the work done for the given process is 324 J

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3 years ago