2 years ago

I NEED HELP WITH THE QUESTION ABOVEEE

Physics

1 answer:

Sergio039 [100]2 years ago

3 0

I think its b if not then I'm sorry buddy:(

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A sprinter starts from rest, and reaches their top speed of 12 m/s in 3 seconds.

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A 6.2 kg ladder, 1.97 m long, rests on two sawhorses. Sawhorse A is 0.64 m from one end of the ladder, and sawhorse B is 0.17 m

Keith_Richards [23]

Answer:

42.69 N and 18.07 N

Explanation:

We are given that

Mass of ladder=6.2 kg

Length of ladder=1.97 m

Distance of Sawhorse A from one end=0.64 m

Distance of sawhorse B from other end=0.17 m

Let center of Ladder= $\frac{1.97}{2}=0.985 m$

Now, the distance of sawhorse A from center=r=0.985-0.64=0.345 m

Distance of sawhorse B from center of ladder=0.985-0.17=0.815 m

Force one ladder due to gravity=mg= $6.2\times 9.8=60.76N$

Where $g=9.8 m/s^2$

Torque applied on Sawhorse A= $0.345F_a$

Torque applied on Sawhorse B= $0.815F_b$

In equilibrium

$0.345F_a=0.815F_b$

$F_b=\frac{0.345}{0.815}F_a$

Total force= $F_a+F_b$

$F_a+\frac{0.345}{0.815}F_a=60.76$

$\frac{0.815F_a+0.345F_a}{0.815}=60.76$

$\frac{1.16}{0.815}F_a=60.76$

$F_a=\frac{60.76\times 0.815}{1.16}=42.69 N$

$F_b=\frac{0.345}{0.815}\times 42.69=18.07 N$

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3 years ago

In a carnival ride, passengers stand with their backs against the wall of a cylinder. The cylinder is set into rotation and the

Lerok [7]

Answer:

Minimum speed will be equal to 2.213 m/sec

Explanation:

We have given radius of the r = 2 m

Coefficient of friction $\mu =0.25$

At minimum speed frictional force will be equal to centripetal force

So $\mu mg=\frac{mv^2}{r}$

$\frac{v^2}{r}=\mu g$

$v=\sqrt{\mu rg}=\sqrt{0.25\times 2\times 9.8}=2.213m/sec$

So the minimum speed will be equal to 2.213 m/sec

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2 years ago