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2 years ago

Please help me with the following question:

Physics

1 answer:

pishuonlain [190]2 years ago

3 0

Answer: a. 667N

b. 665N

c. 54.5N

Explanation:

a) on the surface of the earth

W = mg

W = 68 × 9.81

= 667N

b) at the top of Everest (8848 m above sea level).

W =mg × R²/(R + H)²

W = 667 × [6378²/(6378 + 8.848)²

W = 665N

c) has 2 1/2 times the radius of the earth

W = mg × R²/(R + H)²

W = 667 × R²/(R + 2.5R)²

W = 54.5N

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A comet is traveling through space with speed 3.01 ✕ 104 m/s when it encounters an asteroid that was at rest. The comet and the

Tcecarenko [31]

Answer: $8.493(10)^{-3} m/s$

Explanation:

According to the conservation of linear momentum principle, the initial momentum $p_{i}$ (before the collision) must be equal to the final momentum $p_{f}$ (after the collision):

$p_{i}=p_{f}$ (1)

In addition, the initial momentum is:

$p_{i}=m_{1}V_{1}+m_{2}V_{2}$ (2)

Where:

$m_{1}=1.71(10)^{14} kg$ is the mass of the comet

$m_{2}=6.06(10)^{20} kg$ is the mass of the asteroid

$V_{1}=3.01(10)^{4} m/s$ is the velocity of the comet, which is positive

$V_{2}=0 m/s$ is the velocity of the asteroid, since it is at rest

And the final momentum is:

$p_{f}=(m_{1}+m_{2})V_{f}$ (3)

Where:

$V_{f}$ is the final velocity

Then :

$m_{1}V_{1}+m_{2}V_{2}=(m_{1}+m_{2})V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1}V_{1}}{m_{1}+m_{2}}$ (5)

$V_{f}=\frac{(1.71(10)^{14} kg)(3.01(10)^{4} m/s)}{1.71(10)^{14} kg+6.06(10)^{20} kg}$

Finally:

$V_{f}=8.493(10)^{-3} m/s$ This is the final velocity, which is also in the positive direction.

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3 years ago

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3 years ago

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3 years ago

An 81.5-kg man stands on a horizontal surface.

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Answer:

a)V= 0.0827 m³

b)P=181.11 x 10² N/m²

Explanation:

Given that

m = 81.5 kg

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Mass = Volume x Density

81.5 = V x 985

V= 0.0827 m³

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3 years ago

Two pulses move in opposite directions on a string and are identical in shape except that one has positive displacements of the

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The correct option is (c)

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1 year ago