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3 years ago

12

A child attaches a rubber ball to a string and whirls it around in a circle overhead. If the string is 0.2m long and the ball's

speed is 16 m/s, what is the ball's centripetal acceleration?

1 answer:

AlladinOne [14]3 years ago

4 0

Answer:

The centripetal acceleration, $a_c = 1280 m/s^2$

Explanation:

Centripetal acceleration is given by the formula:

$a_c = \frac{v^2}{r}$

Where the speed of the ball, v = 16 m/s

Length of the string, r = 0.2 m

Substituting these parameters into the centripetal equation above:

$a_c = \frac{16^2}{0.2}\\a_c = \frac{256}{0.2}\\a_c = 1280 m/s^2$

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Answer:

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Explanation:

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2 years ago

A block is pulled across a flat surface at a constant speed using a force of 50 newtons at an angle of 60 degrees above the hori

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The magnitude of the friction force is 25 N

Explanation:

To solve this problem, we just have to analyze the forces acting on the block along the horizontal direction. We have:

The horizontal component of the pulling force, $F cos \theta$ , where F = 50 N is the magnitude and $\theta=60^{\circ}$ is the angle between the direction of the force and the horizontal; this force acts in the forward direction
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According to Newton's second law, the net force acting on the block in the horizontal direction must be equal to the product between the mass of the block and its acceleration:

$\sum F_x = ma_x$

where

m is the mass of the block

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However, the block is moving at constant speed, so the acceleration is zero:

$a_x = 0$

So the equation becomes

$\sum F_x = 0$ (1)

The net force here is given by

$\sum F_x = F cos \theta - F_f$ (2)

And so, by combining (1) and (2), we find the magnitude of the friction force:

$F cos \theta - F_f = 0\\F_f = F cos \theta = (50)(cos 60^{\circ})=25 N$

Learn more about force of friction:

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The climate of a town on the coast will have less dramatic changes than a town further inland. Is this statement true or false?

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2 years ago

The information on a can of soda indicates that the can contains 355 mL. The mass of a full can of soda is 0.369 kg, while an em

Vilka [71]

Answer:

$\rho=995.50\ kg.m^{-3}$

$\bar w=9765.887\ N.m^{-3}$

$s=0.9955$

Explanation:

Given:

volume of liquid content in the can, $v_l=0.355\ L=3.55\times 10^{-4}\ L$

mass of filled can, $m_f=0.369\ kg$

weight of empty can, $w_c=0.153\ N$

<u>So, mass of the empty can:</u>

$m_c=\frac{w_c}{g}$

$m_c=\frac{0.153}{9.81}$

$m_c=0.015596\ kg$

<u>Hence the mass of liquid(soda):</u>

$m_l=m_f-m_c$

$m_l=0.369-0.015596$

$m_l=0.3534\ kg$

<u>Therefore the density of liquid soda:</u>

$\rho=\frac{m_l}{v_l}$ (as density is given as mass per unit volume of the substance)

$\rho=\frac{0.3534}{3.55\times 10^{-4}}$

$\rho=995.50\ kg.m^{-3}$

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$\bar w=\frac{m_l.g}{v_l}=\rho.g$

$\bar w=995.5\times 9.81$

$\bar w=9765.887\ N.m^{-3}$

Specific gravity is the density of the substance to the density of water:

$s=\frac{\rho}{\rho_w}$

where:

$\rho_w=$ density of water

$s=\frac{995.5}{1000}$

$s=0.9955$

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3 years ago

Read 2 more answers