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3 years ago

2. The weight of a basketball player is 6 ft and 1 12 inches. Change his height to:

Physics

1 answer:

Elza [17]3 years ago

4 0

Answer:

a feat

Explanation:

cause the mf already 7ft on the dot there is no such thing as 6 ft 12

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Which are ways to improve the design of this experiment? Check all that apply.

forsale [732]

-Reduce the sample size so the experiment can be done faster.-Increase the sample size from 6 cups to 12 cups of sand and water.-Use more legible handwriting when recording data.-Use more precise digital thermometers.-Use more precise scales that measure to the hundredth of a gram.

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4 years ago

Read 2 more answers

Speed of 26.7 m/s in 3.06 s. How far had the car traveled by the time the final speed was achieved?

Alik [6]

Answer:

Below

Explanation:

You can use this equation to find the distance :

distance = velocity x time

distance = (26.7)(3.06)

= 81.702 m

Rounding to 3 sig figs

= 81.7 m

Hope this helps

6 0

3 years ago

Where does the energy released in a nuclear decay reaction come from

erastovalidia [21]

The answer is electrons

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2 years ago

You find a measurement or 34 kg written in a laboratory notebook. What physical characteristic is this a measurement of?

Elis [28]

Planck's constant. A physical constant adopted in 2011 by the CGPM.

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3 years ago

a body of a mass 1kg suspended from a spring is found to stretch the spring by 10cm. (A) what is the spring constant? what is pe

Nikitich [7]

Answer:

(A) The spring constant is 98 N/m.

(B) The period of oscillation is 0.635 s.

Explanation:

Given:

Mass of the body is, $m=1\ kg$

Extension length of the spring is, $x=10\ cm=0.1\ m$

Now, let 'k' be the spring constant.

The force acting on the body is due to gravity only and is equal to its weight.

So, weight of the body is given as:

$F_g=mg\\F_g=1\times 9.8\\F_g=9.8\ N$

Now, we know that for a spring-mass system, the net force acting on the body is equal to the product of the spring constant and extension length. Therefore,

$F_g=kx\\9.8=k(0.1)\\k=\frac{9.8}{0.1}=98\ N/m$

Hence, the spring constant is 98 N/m.

(B)

Period of oscillation of a body of spring-mass system in SHM is given as:

$T=2\pi\sqrt{\frac{m}{k}}$

Plug in the given values and solve for period 'T'. This gives,

$T=2\pi\sqrt{\frac{1}{98}}\\T=2\pi\times 0.101\\T=0.635\ s$

Therefore, the period of oscillation is 0.635 s.

5 0

3 years ago