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viva [34]
3 years ago
10

I need help with the question 8B

Physics
1 answer:
ratelena [41]3 years ago
3 0
The answer of this is C
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An 80kg astronaut traveled to the moon, where gravity is one-sixth (116) as
PIT_PIT [208]

Answer:

Wmoon = 131 [N]

Explanation:

We know that the weight of a body is equal to the product of mass by gravitational acceleration.

Since we are told that the gravitational acceleration of the moon is equal to one-sixth of the acceleration of Earth's gravitation. Then we must multiply the value of Earth's gravitation by one-sixth.

w_{moon}=\frac{1}{6} *m*g\\w_{moon}=\frac{1}{6} *80*9.81\\w_{moon}=130.8 [N] = 131 [N]

7 0
2 years ago
What is the unit for molar mass?
dmitriy555 [2]

Answer:

kilogram per mole

Explanation:

ffdddddfsds

8 0
2 years ago
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How many million kilometers is one astronomical unit.
Andrews [41]

Answer:

150 million kilometres

Explanation:

The astronomical unit (symbol: au, or AU or AU) is a unit of length, roughly the distance from Earth to the Sun and equal to 150 million kilometres (93 million miles) or 8.3 light minutes.

6 0
2 years ago
Two particles in a high-energy accelerator experiment are approaching each other head-on, each with a speed of 0.9520c as measur
Over [174]

Answer:

the magnitude of the velocity of one particle relative to the other is 0.9988c

Explanation:

Given the data in the question;

Velocities of the two particles = 0.9520c

Using Lorentz transformation

Let relative velocity be W, so

v_r = ( u + v ) / ( 1 + ( uv / c²) )

since each particle travels with the same speed,

u = v

so

v_r = ( u + u ) / ( 1 + ( u×u / c²) )  

v_r = 2(0.9520c) / ( 1 + ( 0.9520c )² / c²) )  

we substitute

v_r = 1.904c / ( 1 + ( (0.906304 × c² ) / c²) )  

v_r = 1.904c / ( 1 + 0.906304 )

v_r = 1.904c / 1.906304

v_r = 0.9988c

Therefore, the magnitude of the velocity of one particle relative to the other is 0.9988c

5 0
3 years ago
Your answer should be precise to 0.1 m/s. Use a gravitational acceleration of 10 m/s/s. At it lowest point, a pendulum is moving
saw5 [17]

Explanation:

It is given that,

Speed, v₁ = 7.7 m/s

We need to find the velocity after it has risen 1 meter above the lowest point. Let it is given by v₂. Using the conservation of energy as :

\dfrac{1}{2}mv_1^2=\dfrac{1}{2}mv_2^2+mgh

v_2^2=v_1^2-2gh

v_2^2=(7.7)^2-2\times 10\times 1

v_2=6.26\ m/s

So, the velocity after it has risen 1 meter above the lowest point is 6.26 m/s. Hence, this is the required solution.

4 0
3 years ago
Read 2 more answers
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