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Bond [772]
2 years ago
10

Find the weight of an astronaut whose mass is 75 kg on the moon

Physics
1 answer:
gladu [14]2 years ago
8 0
The formula for weight is always weight=mass X gravitational field strength.
We already know the mass is 75kg.
The gravitational field strength on the moon is 1.6N. To find out the weight, we can substitute these values in to the formula.
Weight=75 X 1.6
Weight= 120N
Weight is measured on Neutons as it is a force.
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You throw a glob of putty straight up toward the ceiling, which is 3.50 mm above the point where the putty leaves your hand. The
LuckyWell [14K]

Answer: 4.65\ m/s

Explanation:

Given

Distance putty has to travel is 3.5 m

The initial speed of putty is 9.50 m/s

Using equation of motion to determine the velocity of putty just before it hits ceiling

v^2-u^2=2as

\Rightarrow v^2-(9.5)^2=2(-9.8)(3.5)\\\\\Rightarrow v^2=9.5^2-68.6\\\Rightarrow v=\sqrt{90.25-68.6}\\\Rightarrow v=4.65\ m/s

So, the velocity of putty just before hitting is 4.65\ m/s

5 0
3 years ago
Pre-Test Active
vlada-n [284]

n is the answer I guess

not so sure

7 0
2 years ago
How many neutrons are in an atom of carbon with a mass number of 13
BabaBlast [244]

C is the answer to your question.

4 0
3 years ago
What should you do if you get caught in quicksand
balandron [24]

Answer:

-you have to make yourse;f as light as possible so toss your bag, jacket, and shoes.

-Try to take a few steps backwards.

-Keep your arms up and out of the quicksand.

-Try to reach for a branch or person's hand to pull yourself out.

-Take deep breaths.

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8 0
3 years ago
(1 point) A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 new
Nonamiya [84]

Answer:

x(t)=0.337sin((5.929t)

Explanation:

A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2 m/sec, find the position of the mass after t seconds.

Solution. Let x(t) denote the position of the mass at time t. Then x satisfies the differential equation  

m \frac{d^{2}x}{dt^{2}} +kx=0

Definition of parameters  

m=mass 3kg

k=force constant

e=extension ,m

ω =angular frequency

k=90/1.6=56.25N/m

ω^2=k/m= 56.25/1.6

ω^2=35.15625

ω=5.929

General solution will be

x(t)=c1cos(ωt)+c2Sin(ωt)

x(t)=c1cos(5.929t)+c2Sin(5.929t)

differentiating x(t)

dx(t)=-5.929c1sin(5.929t)+5.929c2cos(5.929t)

when x(0)=0, gives c1=0

dx(t0)=2m/s gives c2=0.337

Therefore, the position of the mass after t seconds is  

x(t)=0.337sin((5.929t)

6 0
3 years ago
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