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2 years ago

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If a weightlifter used 1,200 newtons of force to lift a barbell to a height of 2.5 meters in 3 seconds, how much work did he do?

A. 500 J B. 1,000 J C. 3,000 J D. 3,600 J

1 answer:

Blizzard [7]2 years ago

3 0

C.) Work done = Force x displacement = 1200 x 2.5 = 3000 J. Hope this helps!

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During a total lunar eclipse, the moon

Anit [1.1K]

A, C, and D all happen at different stages
of a total lunar eclipse.

I'll describe the stages of the eclipse, but before I do, I just need
to clarify: The Earth doesn't have an umbra or a penumbra, but
its shadow does.

-- the eclipse begins when the first edge of the moon
   moves into the penumbra of Earth's shadow; ( C )
   this part of the moon grows steadily.

-- After a while, the first edge of the moon begins to move
   into the umbra of Earth's shadow ( A ), and gets very dark.

-- The total phase of the eclipse begins when the ENTIRE
    moon is in the umbra of Earth's shadow.

Then everything happens in reverse.

-- Eventually, the leading edge of the moon moves out
   of the shadow's umbra, into the penumbra. This part
   steadily grows.

-- After a while, none of the moon is in the umbra, and
   the whole thing is in the penumbra. The moon is
   fully illuminated, but not quite as bright as it should be.

-- Soon, the leading edge of the moon leaves the penumbra
of Earth's shadow, and gets brighter. This portion of the moon
    steadily grows, until ...

-- the moon completely leaves the penumbra, all of it is as bright
as it's supposed to be. The eclipse is completely over. ( B )

==> The whole process lasts several hours.

==> Everybody on the night side of the Earth sees the same thing
   at the same time. It doesn't matter WHERE you are on the night
   side ... if you can see the moon in the sky, you see the present
   phase of the eclipse.

==> The lunar eclipse can only happen at the Full Moon. In fact, the
   mid-point of the total phase is the exact moment of Full Moon.

8 0

3 years ago

Read 2 more answers

A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mas

konstantin123 [22]

Answer:

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

Explanation:

The expression for conservation of the angular momentum (L) is

$L_{i} = L_{f} I_{i}\times\omega_{i} = I_{f}\times\omega_{f}$

Where

$I_{i}\ and \ \omega_{i}$ initial moment of inertia and angular velocity

$I_{f}\ and \ \omega_{f}$ is the final moment of inertia and angular velocity

The expression of moment of inertia of the satellite (a solid sphere) is

$I_{i} = \frac{2}{5}m_{s}r^{2}$

Where $m_{s}$ is the satellite mass

r is the radus of the sphere

Substititute 1900kg for m and 4.6m for r

$I_{i} = \frac{2}{5}m_{s}r^{2}\\\\ = \frac{2}{5}\times1900 kg\times (4.6 m)^{2} \\\\= 1.61 \cdot 10^{4} kgm^{2}$

The final moment of inertia of the satellite about the centre of mass

$I_{f} = I_{i} + 2\timesI_{x} \\\\= 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}m_{x}l^{2}$

Where $m_{x}$ is the antenna's mass and

I is the length of the antenna

$I_{f} = 1.61 \cdot 10^{4} kgm^{2} + 2\times\frac{1}{3}150.0 kg\times(6.6 m)^{2} \\\\= 2.05 \cdot 10^{4} kgm^{2}$

So, the Final rotation rate of the satellite is:

$I_{i}\times\omega_{i} = I_{f}\times\omega_{f} \\\\\omega_{f} = \frac{I_{i}\times\omega_{i}}{I_{f}} \\\\= \frac{1.61 \cdot 10^{4} kgm^{2}\times8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kgm^{2}} \\\\= 6.3 rev/s$

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

5 0

3 years ago

Find the velocity in m/s of a swimmer who swims 110m toward the shore in 72 s

Sladkaya [172]

1.5 m/s toward shore

8 0

2 years ago

A 23 g bullet traveling at 230 m/s penetrates a 2.0 kg block of wood and emerges cleanly at 170 m/s. If the block is stationary

Ann [662]

The distance traveled by the wood after the bullet emerges is 0.16 m.

The given parameters;

<em>mass of the bullet, m = 23 g = 0.023 g</em>
<em>speed of the bullet, u = 230 m/s</em>
<em>mass of the wood, m = 2 kg</em>
<em>final speed of the bullet, v = 170 m/s</em>
<em>coefficient of friction, μ = 0.15</em>

The final velocity of the wood after the bullet hits is calculated as follows;

$m_1u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\0.023(230) + 2(0) = 0.023(170) + 2v_2\\\\5.29 = 3.91 + 2v_2\\\\2v_2 = 1.38\\\\v_2 = \frac{1.38}{2} = 0.69 \ m/s$

The acceleration of the wood is calculated as follows;

$\mu = \frac{a}{g} \\\\a = \mu g\\\\a = 0.15 \times 9.8\\\\a = 1.47 \ m/s^2$

The distance traveled by the wood after the bullet emerges is calculated as follows;

$v^2 = v_0^2 + 2as\\\\v^2 = 0 + 2as\\\\v^2 = 2as\\\\s = \frac{v^2}{2a} \\\\s = \frac{(0.69)^2}{2(1.47)} \\\\s = 0.16 \ m$

Thus, the distance traveled by the wood after the bullet emerges is 0.16 m.

Learn more here:brainly.com/question/15244782

7 0

3 years ago

A student read his right ear against his desk while Teacher talks loudly on the desk and can hear the tapping sound only through

sladkih [1.3K]

Answer:

The student hears the wave that is transmitted by the desk

Explanation:

Mechanical waves need a material medium to be able to be transmitted, in the case of sound waves, one of the most common media is air, but it is also transmitted in other media in this case, stationery is transmitted.

The student hears the wave that is transmitted by the desk

The speed of the wave is proportional to the density of the material, so the wave that the student hears arrives much faster through the desk than through the air

6 0

2 years ago