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jekas [21]
3 years ago
5

Which of the following is an example of a force inertia push pull

Physics
1 answer:
Whitepunk [10]3 years ago
8 0
A force can be considered a push or pull

hope this helps :)


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Using the right amount of significant figures, calculate the answer to the following problem, 215.5+101.02555
cestrela7 [59]

Answer:

This is how I figured it out:

  1. 215.5 rounded to one significant figure is 200
  2. 101.02555 rounded to one significant figure is 100.
  3. 200 + 100 = 300.

Hope this helps!

Explanation:

7 0
3 years ago
True or false. when objects collide , some momentum is lost
ycow [4]

Answer:

It is neither false nor true. When they collide some of one of the objects goes to the other object.

Explanation:

5 0
2 years ago
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A man of 60kg moves in a lift of constant velocity 5m/s .What is the reactive force acting on the man by the elevator?
Flauer [41]

Answer:

588 N

Explanation:

Since the 60 kg is moving at a constant velocity there is no acceleration. In order for the system to be balanced, both the normal force and the force of gravity must be equal. In this case the man has a mass of 60 kg. So to find the force you multiply mass by gravitys constant (9.81). And you end up with an answer of 588.6 but I rounded to 588.

4 0
3 years ago
Read 2 more answers
When non-metric units were used in the United Kingdom, a unit of mass called the pound-mass (lbm) was employed, where 1lbm=0.453
Drupady [299]

Answer:

a) 0.022%

b) 10014.32 lb

Explanation:

a) Percentage uncertainty would be

0.0001\times \frac{100}{0.4539}=0.022%

Percent uncertainty is 0.022%

b) For 1 kg uncertainty mass in kg would be

\frac{1}{0.022}\times {100}=4545.5\ kg

Mass in pounds would be

\frac{4545.5}{0.4539}=10014.32\ lb

Mass in pound-mass is 10014.32 lb

8 0
3 years ago
A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed on 10 m/s. If air
fiasKO [112]

The distance of the rock from the base of the cliff is C) 20 m

Explanation:

The motion of the rock in this problem is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion to find the time of flight of the rock (the time it takes to reach the ground). We can do it by using the suvat equation:

s=u_y t+\frac{1}{2}at^2

where, taking downward as positive direction,

s = 20 m is the vertical displacement of the rock

u_y=0 is the initial vertical velocity

t is the time of flight

a=g=9.8 m/s^2 is the acceleration of gravity

Solving for t,

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s

Now we can analzye the horizontal motion: the rock moves horizontally with a constant velocity of

v_x = 10 m/s

Therefore, the horizontal distance covered after a time t is

d=v_x t

and substituting t = 2.02 s, we find the final distance of the rock from the base of the cliff:

d=(10)(2.02)=20 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0
3 years ago
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