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Maru [420]
3 years ago
8

Using a rope that will snap if the tension in it exceeds 379 N, you need to lower a bundle of old roofing material weighing 473

N from a point 6.50 m above the ground. (a) What magnitude of the bundle's acceleration will put the rope on the verge of snapping
Physics
1 answer:
Tema [17]3 years ago
8 0

Answer:

Acceleration = a = -1.94 m/s^2

Explanation:

Weight of the bundle = W = mg  

                      m = W/g = 473/9.8  

                      m = 48.26 kg

There are two forces acting on the bundle such as tension in the upward direction and weight in the downward direction. So, according to newton’s second law of motion:  

Sum of forces = ma  

                           T - W = ma  

                           a = (T – W)/m

                           a = (379 - 473)/48.26

                           a = -1.94 m/s^2

Negative sign shows that the acceleration is in the downward direction.

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A golf ball (m = 59.2 g) is struck a blow that makes an angle of 50.5 ◦ with the horizontal. The drive lands 130 m away on a fla
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Answer:

The force is calculated as 338.66 N

Explanation:

We know that force is given by

\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}=\frac{m(v_{f}-v_{o})}{\Delta t}

We know that range of a projectile is given by

R=\frac{v_{o}^{2}sin(2\theta )}{g}

it is given that R=130 m applying values in the above equation we get

R=\frac{v_{o}^{2}sin(2\theta )}{g}\\\\v_{0}=\sqrt{\frac{Rg}{sin(2\theta )}}\\\\\therefore v_{o}=\sqrt{\frac{130\times 9.81}{sin(2\times 50.5_{o}} )}\\\\v_{o}=36.04m/s

Thus the force is obtained as

\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}=\frac{59.2\times 10^{}-3(36.04-0)}{6.3\times 10^{-3}}

Thus force equals F=338.66N

4 0
3 years ago
Read 2 more answers
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