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larisa [96]
2 years ago
8

Please !!! I really need help !!! How do I understand these ?!!!!

Physics
2 answers:
trasher [3.6K]2 years ago
8 0

Answer

The answer for the first one I think is false.

The second one would be true i think. I hope i got it right and have a wonderful day

Ymorist [56]2 years ago
5 0

Answer:

True

False

Explanation:

From 0 to E, the train moves a distance of 55 m.

From F to J, the train moves a distance of 59 m.

The total distance is 55 + 59 = 114 m.

The displacement is the difference between the final position and initial position.  Here, the distance between J and 0 is -4 m.

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Please help with this and explain it,if you can.
vladimir2022 [97]

Answer:

displacement at 45 s  =  30

                           65 s  = 50

So the average speed over the interval from 45 s to 65 s is

(50 - 30) cm / 20 s = 1 cm / sec

As a check an average speed of 1 cm / sec for 20 sec will produce a

displacement of 1 cm / sec * 20 sec = 20 cm  or from 30 to 50 cm

4 0
3 years ago
Acceleration is equal to the initial velocity minus the final velocity, then divided by time.true or false
tatiyna

Almost true but not quite.

That would give you the negative of the actual acceleration.

It should be the other way around:

(final v) minus (initial v), then divide by time.

5 0
2 years ago
Consider a 100 g object dropped from a height of 1 m. Assuming no air friction (drag), when will the object hit the ground and a
Katyanochek1 [597]

Answer:

speed and time are Vf = 4.43 m/s and  t = 0.45 s

Explanation:

This is a problem of free fall, we have the equations of kinematics

      Vf² = Vo² + 2g x

As the object is released the initial velocity is zero, let's look at the final velocity with the equation

      Vf = √( 2 g X)

      Vf = √(2 9.8  1)

      Vf = 4.43 m/s

This is the speed with which it reaches the ground

 

Having the final speed we can find the time

      Vf = Vo + g t

       t = Vf / g

       t = 4.43 / 9.8

       t = 0.45 s

This is the time of fall of the body to touch the ground

3 0
3 years ago
Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at
Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

v = \dfrac{2\pi R}{T}

v = \dfrac{2\pi\times 8}{4.30}

  v = 11.69 m/s

now, Force does the ring push on her at the top

- N - m g = \dfrac{-mv^2}{R}

N + m g = \dfrac{mv^2}{R}

N = \dfrac{mv^2}{R}- m g

N = m(\dfrac{v^2}{R}- g)

N = 58\times (\dfrac{11.69^2}{8}- 9.8)

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

6 0
2 years ago
Why are there only two elements in the first period of the periodic table?(1 point)
tester [92]

Answer:

because each row increases in atomic mass by a specific number, so anything over five is in the second row.

8 0
2 years ago
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