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Sergeu [11.5K]
3 years ago
14

.Why does the tension in each chain of a

Physics
1 answer:
anyanavicka [17]3 years ago
7 0
Because if your putting tension on something tensions obviously going to increase with more pressure and weight on it
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When starting a foot race, a 70.0kg sprinter exerts an average force of 650 N backward on the ground for 0.800 s. How far does h
melomori [17]

Answer:

Answer is option b) 2.97m

Explanation:

With the relationship between the force exerted by the runner and the mass that it has, I can determine the acceleration it will have:

F= m × a ⇒ a= (650 kg ×(m/s^2)) / (70kg)= 9.286 (m/s^2)

With the acceleration that prints the force exerted and the time I can determine the distance traveled in the interval:

Distance= (1/2) × a × t^2 = (1/2) × 9.286 (m/s^2) × ((0.8s)^2)= 2.97m

8 0
3 years ago
Pls help with either of these I will give up brainliest
Contact [7]

Answer:

1. The bird close to the center

2. 4/25 of the original force.

Explanation:

1. Tangential velocity is v=w*d (in m/s), where w is the rotational speed, commonly denoted as the letter omega (in radians per second). d is the distance from the center of the rotating object to the position of where you would like to calculate the velocity (in meters).

As we can note, the furthest from the center we are calculating the velovity the higher it is, because the rotational velocity is not changing but the distance of the object with respect to the center is. If v=w*d, then the lower the d (distance) the lower the tangential velocity.

2. Take a look at the picture:

We have the basic equation for the gravitational force.

We have to forces: Fg1, which is the original force, and Fg2, the force when the mass and the distance changes.

If we consider that mass 2 didn't change (m2'=m2), mass 1 is four times its original (m1'=4*m1) and distance is 5 times the original (r'=5*r), then next step is just plugging it into the equation for Fg2.

Dividing the original force Fg1 by the new force Fg2 (notice you can just as well do the inverse, Fg2 divided by Fg1) gives us the relation between the forces, cancelling all the variables and being left only with a simple fraction!

6 0
3 years ago
A car could move at constant speed on an icy curve which is banked for _______________ (all, one, no) speed(s) of the car.
son4ous [18]
All is the answer I believe.
4 0
3 years ago
Learning Task No. 5 Identify the word or words being described by each statement. Choose y box below. 1. It is the process of ch
Rashid [163]

Identifying the word that is described by each statements in the question is listed below:

  1. Evaporation
  2. Transpiration
  3. Hydrosphere
  4. Water Cycle
  5. Condensation
<h3>Meaning of Water </h3>

Water can be simply defined as any fluid or substance that is odurless, colorless  and tasteless.

Water is a very important factor and substance that sustains life in humans and plants on earth.

Water tends to undergo different changes and processes in plant, animals and even as a substance on its own

In conclusion the above listed processes are all related to water.

Learn more about Water: brainly.com/question/1313076

#SPJ1

4 0
2 years ago
If an astronaut can throw a certain wrench 10.0 m vertically upward on earth, how high could he throw it on our moon if he gives
Jobisdone [24]

I think you forgot to include the acceleration due to gravity of astronauts. I assume that it is = 0.170 g. To get the answer we have to use the formula s = v0t – (1/2) At². Where s is the altitude, A is the acceleration of gravity, t is the time after throwing.

v = v0 –At

v = 0 at max altitude so v0 – At = 0

t = v0/A at max altitude

Using the formula above for the altitude:

s = v0t – (1/2) At²

s = v0(v0/A) – (1/2) A (v0/A)²

s = v0²/A – (1/2) v0²/A

s = (1/2) v0²/A

The earth: E = (1/2) v0²/g

The moon: M = (1/2)v0²(0.17g)

So, take the ratio of M/E = g/0.17g = 1/0.17 = 588

M = 5.88 E

He can throw the wrench 5.88 times higher on the moon

<span>M =5.88 (10 m) = 58.8 meters that the can throw the wrench a little over on the moon.</span>
7 0
4 years ago
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