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Blizzard [7]
3 years ago
11

Wo people sit on a balanced seesaw. when one person leans toward the center of the seesaw, that person's end of the seesaw will

_____.
a. fall
Physics
1 answer:
hram777 [196]3 years ago
4 0
Go upwards(i.e Rise)
This will be due to the Effects known as 'Moments'
The person who is moving towards the centre will distribute its weight towards the centre causing the end of the see saw to be lighter.

HOPE IT HELPS
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A planet has a circular orbit around a star. It is a distance of 53,000,000 km from the centre of the star.
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it takes 365 days to revolve around the star(sun)

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What will a spring scale read for the weight of a 57.0-kg woman in an elevator that moves upward with constant speed of 5.0 m/s
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4 0
3 years ago
A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule
ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

p_i = 0

The final total momentum is instead:

p_f = m_a v_a + m_c v_c

where

m_a = 125 kg is the mass of the astronaut

v_a = 2.50 m/s is the velocity of the astronaut

m_c = 1900 kg is the mass of the capsule

v_c is the velocity of the capsule

Since the total momentum must be conserved, we have

p_i = p_f = 0

so

m_a v_a + m_c v_c=0

Solving the equation for v_c, we find

v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

F \Delta t = \Delta p

The change in momentum of the astronaut is

\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s

And the duration of the push is

\Delta t = 0.600 s

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J

3 0
3 years ago
Just the answer (PLSS)
Salsk061 [2.6K]

The miracle year for Albert Einstein  was the year 1905 within which he published so many renowned papers.

<h3>When was Einstein miracle year?</h3>

The miracle year for Albert Einstein  was the year 1905 within which he published so many renowned papers in a short time and became very popular.

His mindset in that year was one that challenged the orthodox explanations and sought to think outside the box.

Learn more about Albert Einstein:brainly.com/question/2964376

#SPJ1

8 0
2 years ago
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