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S_A_V [24]
4 years ago
8

16. An object has a mass of 13.5 kilograms. What force is required to accelerate it to a rate of 9.5 m/s2?

Physics
2 answers:
V125BC [204]4 years ago
8 0
D. 128.25 because force=mass x acceleration
gtnhenbr [62]4 years ago
8 0

Answer:

D. 128.25 N

Explanation:

The force exerted on an object and the acceleration of the object are related by the following equation (Newton's second law):

F=ma

where

F is the force exerted on the object

m is the mass of the object

a is the acceleration

In this problem, we know:

- The mass of the object: m = 13.5 kg

- The acceleration: a = 9.5 m/s^2

Therefore, we can find the force required to give the object this acceleration:

F=ma=(13.5 kg)(9.5 m/s^2)=128.25 N

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a sphere of diameter 6•0cm is moulded into a thin uniform wire of diameter 0•2mm.calculate the length of the wire in metres​
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<u>Explanation:</u>

Given, Diameter of sphere = 6 cm

We know that, radius can be found by taking the half in the diameter value. So,

       \text { sphere radius, } R=\frac{D}{2}=\frac{6}{2}=3 \mathrm{cm}=3 \times 10^{-2} \mathrm{m}

Similarly,

      \text { wire radius, } r=\frac{0.2}{2}=0.1 \mathrm{mm}=1 \times 10^{-3} \mathrm{m}

We know the below formulas,

          \text {volume of sphere}=\frac{4}{3} \times \pi \times R^{3}

          \text {volume of wire}=\pi \times r^{2} \times l

When equating both the equations, we can find length of wire as below, where \pi=\frac{22}{7}

          \frac{4}{3} \times \pi \times R^{3}=\pi \times r^{2} \times l

         \frac{4}{3} \times \frac{22}{7} \times\left(3 \times 10^{-2}\right)^{3}=\frac{22}{7} \times\left(1 \times 10^{-3}\right)^{2} \times l

The \pi value gets cancelled as common on both sides, we get

           \frac{4}{3} \times 27 \times 10^{-6}=10^{-6} \times l

The 10^{-6} value gets cancelled as common on both sides, we get

           l=4 \times 9=36 m

7 0
3 years ago
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