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BartSMP [9]
3 years ago
9

An astronaut weighing 190 lbs on Earth is on a mission to the Moon and Mars.

Physics
1 answer:
dusya [7]3 years ago
3 0

Answer:

The weight is defined as:

W = m*g

where:

m = mass

g = gravitational acceleration.

We know that in Earth the astronaut weights 190 lb-f (this is force, not mass, the correct unit here is 190 lb*m/s^2)

then:

190 lb*m/s^2 = m*9.8m/s^2

(190 lb*m/s^2)/(9.8m/s^2) = 19.39 lb

Now we know the mass of the astronaut.

a) wieght on the moon in Newtons.

Newtons uses kilograms as the units of mass, then we need to rewrite the mass of the astronaut in kg.

we know that 1lb = 0.454 kg

Then 19.39 lb is equal to: 19.39*0.454 kg = 8.8 kg

We know that the acceleration due to gravity on the Moon is one-sixth that on Earth.

then: g = (9.8m/s^2)/6

And the weight of the astronaut in the moon will be:

W = 8.8 kg*(9.8m/s^2)/6 = 14.37 N

b) The weight on mars, where the acceleration due to gravity is 0.38 times that on Earth, we have:

g = (9.8m/s^2)*0.38

then the weight will be:

W = 8.8kg*(9.8m/s^2)*0.38 = 32.77 N

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Ilia_Sergeevich [38]

The sprinter’s average acceleration is 1.98 m/s²

The given parameters;

  • initial velocity of the sprinter, u = 18 km/h
  • final velocity of the sprinter, v = 27 km/h
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Convert the velocity of the sprinter to m/s;

initial \ velocity, u = 18 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 5 \ m/s\\\\final \ velocity, v =27 \frac{km}{h} \times \frac{1000 \ m}{1 \ km} \times \frac{1 \ h}{3600 \ s} = 7.5 \ m/s\\\\

The time of motion is seconds;

t = 3.5 \times 10^{-4} \ h \times \frac{3600 \ s}{1 \ h} = 1.26 \ s

The sprinter’s average acceleration is calculated as follows;

a = \frac{v- u}{t} \\\\a = \frac{7.5 \ m/s \ - \ 5 \ m/s}{1.26 \ s} \\\\a = 1.98 \ m/s^2

Thus, the sprinter’s average acceleration is 1.98 m/s²

Learn more here:brainly.com/question/17280180

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3 years ago
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Explanation:

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We can say that it moves 8 meters every second or 800 cm every second.

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