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2 years ago

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3. An astronaut on a space-walk swings a 0.5 kg metal tool in a circular path at the end of a cable 3.6 m long. The astronaut th

en reels in the tool to a 1.2 m distance. Show that the speed of the tool will be three times as great at the reduced distance.

1 answer:

Semenov [28]2 years ago

6 0

Answer:

Explanation:

2 km a hour

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On a part time job you are asked to bring a cylindrical iron rod of density 7800 kg/m^3, length 81.2 cm and diameter 2.75 cm fro

Hitman42 [59]

Answer:

3.76188 kg

36.9040428 N

Yes

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

L = Length of rod = 81.2 cm

d = Diameter of rod = 2.75 cm

r = Radius = $\frac{d}{2}=\frac{2.75}{2}=1.375\ cm$

A = Area of the cylinder = $\pi r^2$

v = Volume = $L\times A$

m = Mass

$\rho$ = Density of rod = 7800 kg/m³

Density

$\rho=\dfrac{m}{v}\\\Rightarrow m=\rho\times v\\\Rightarrow m=\rho\times L\times A\\\Rightarrow m=7800\times 0.812\times \pi\times 0.01375^2\\\Rightarrow m=3.76188\ kg$

The mass of the rod is 3.76188 kg.

As the the mass of the rod is 3.76188 kg I will be able to carry the rod.

Weight

$W=mg\\\Rightarrow W=3.76188\times 9.81\\\Rightarrow W=36.9040428\ N$

The weight of the rod is 36.9040428 N

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2 years ago

Hằng số phổ biến chất khí

padilas [110]

Answer:

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Explanation:

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2 years ago

A rod 16.0 cm long is uniformly charged and has a total charge of -25.0 µC. Determine the magnitude and direction of the electri

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Answer:

-1.4x10^6N/C

Explanation:

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3 years ago

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.40 m/s

tekilochka [14]

Answer:

a) 378Ns

b) 477.27N

Explanation:

Impulse is the defined as the product of the applied force and time taken. This is expressed according to the formula

I = Ft = m(v-u)

m is the mass = 70kg

v is the final velocity = 5.4m/s

u is the initial velocity = 0m/s

Get the impulse

I = m(v-u)

I = 70(5.4-0)

I = 70(5.4)

I = 378Ns

b) Average total force is expressed as

F = ma (Newton's second law)

F = m(v-u)/t

F = 378/0.792

F = 477.27N

Hence the average total force experienced by a 70.0-kg passenger in the car during the time the car accelerates is 477.27N

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2 years ago

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3 years ago

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