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3 years ago

Which single force acts on an object in freefall?

Physics

2 answers:

nirvana33 [79]3 years ago

8 0

Gravity i think but if uts not gravity then it will be air resistance

balandron [24]3 years ago

3 0

It's A. Gravity...I know cuz u took test

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A force of 30.0 N is applied to a 3.00 kg object for 3.00 seconds. Calculate the velocity experienced by the object.

olganol [36]

Answer:

Explanation:

F = ma and

$a=\frac{v}{t}$

We have F, we have m, but in order to solve for v, we need a.

30.0 = 3.00a so

a = 10.0 m/s/s. Plug that in for a in the second equation and solve for v:

$10.0=\frac{v}{3.00}$ so

v = 10.0(3.00) so

v = 30.0 m/s

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2 years ago

Jana is observing a young autistic boy in a preschool classroom. Her job is to note each time the boy gets up from his chair wit

svp [43]

The answer is tallying

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3 years ago

Choose all the answers that apply.

Serhud [2]

According to Newton's first law of motion:

-- objects at rest will remain at rest unless acted upon by an outside force

-- objects in motion will remain in motion unless acted upon by an outside force

8 0

3 years ago

Which of the following is not used in calculating acceleration?

Leno4ka [110]

Answer:

l think C am not pretty show

4 0

2 years ago

You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

Answer: The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

Explanation:

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

3 0

3 years ago