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True [87]
2 years ago
8

What test are included in the Physical Fitness Test?

Physics
1 answer:
strojnjashka [21]2 years ago
7 0
The first option is right
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There are competitions in which pilots fly small planes low over the ground and drop weights, trying to hit a target. A pilot fl
nikdorinn [45]

Answer:

Also 3s.

Explanation:

Each component is independent in two dimensional motion. This means that <em>how much time does something take to reach the ground when dropped is independent from any horizontal velocity</em>. If at one run a drop lasts 3s, at another run with twice the (horizontal) velocity and same height will also last 3s, no matter what.

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3 years ago
Newton’s second law of motion addresses the relationship between what two variables that influence the force on a body?
Helen [10]

Answer:

A. Mass and acceleration

Explanation:

  • According to Newton's second law of motion, the resultant force is directly proportional to the rate of change in momentum
  • Therefore; F = ma , where F is the resultant force, m is the mass, and a is the acceleration of the body.
  • <u>Resultant force depends on the acceleration and the mass of a body in motion, an increase in acceleration causes a corresponding increase in resultant force.</u> A body with higher mass will have a larger strong force if the acceleration is kept constant.
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2 years ago
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A student heats a piece of aluminum, with specific heat 0.900 J/gºC, in
tankabanditka [31]

Answer:

my bad i need points good luck

Explanation:

adawd

8 0
2 years ago
Brainliest!! Please help with number 4 and please give me the equation too!! Brainliest!!!
Thepotemich [5.8K]

Answer:

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Explanation:

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2 years ago
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One hundred jumping beans are placed along the center line of a gymnasium floor at six-inch intervals. Twelve hours later, the d
stira [4]

Answer:

a) Diffusion  coefficient, D = 1.5 in/hr

b) Mean jump frequency, f = 0.0833 Hz

Explanation:

a) The relationship between the diffusion coefficient, time and mean displacement and can be given by the expression:

^{2} = 2Dt..........(1)

Where <r> = mean displacement

D = Diffusion coefficient

t = time = 12 hrs

sum of the squares of the distance divided by 100 is 36 in2.

<r>²= 36 in²

Substituting these values into equation (1) above

36 = 2 * D *12\\36 = 24 D\\D = 36/24\\D = 1.5 in/hr

b) Mean jumping distance, <r> = 0.1 inches

Applying equation (1) again

Where D = 1.5 in/hr

^{2} = 2Dt

0.1^{2}  = 2 * 1.5t\\0.01 = 3t\\t = 0.01/3\\t = 0.0033 hrs\\t = 0.0033 * 3600\\t = 12 seconds

The mean jump frequency, f = 1/t

f = 1/12

f = 0.0833 Hz

8 0
2 years ago
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