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bekas [8.4K]
3 years ago
9

What is the acceleration of a 32-kg object if a 6.75-N force is applied to it?

Physics
1 answer:
professor190 [17]3 years ago
5 0
F=MA
*newton 2nd law*
F=6.75N
M=32kg

A= F/M
A=6.75/32
equal
a.0.21m/s2
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If you quadruple the temperature of a black body, by what factor will the total energy radiated per second per square meter incr
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then we can make a ratio between the state of before quadruple (with subscript 1) and after (with subscript 2) as:

\frac{P_{1} }{P_{2} } =\frac{T_{1}^4 }{T_{2}^4}

As

T_{2}=4T_{1}

Then

\frac{P_{1} }{P_{2} } =\frac{T_{1}^4 }{(4T_{1})^4}

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pshichka [43]

Answer:

v_average = (d₂-d₁) / Δt

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Find the work w1 done on the block by the force of magnitude f1 = 95.0 n as the block moves from xi = -5.00 cm to xf = 4.00 cm .
Vlad [161]
<h3><u>Answer;</u></h3>

= 8.55 Joules

<h3><u>Explanation;</u></h3>

Work done is the product of force and the distance moved by an object.

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Force = 95 Newtons

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Thus;

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