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3 years ago

5

As a car accelerates its tires increase their angular velocity from 114 rad/s to 131 rad/s in 0.43 seconds. What is the angular

acceleration of the tires?

2 answers:

e-lub [12.9K]3 years ago

5 0

I think it's 39.53

(please do calculate it. I am not completely sure)

levacccp [35]3 years ago

5 0

Angular acceleration = (change in angular speed) / (time for the change)

Change in angular speed = (speed at the end) - (speed at the start)

Change in angular speed = (131 rad/s) - (114 rad/s) = 17 rad/s

Angular acceleration = (17 rad/s) / (0.43 sec)

Angular acceleration = (17 / 0.43) rad/s²

<em>Angular acceleration = 39.5 rad/s²</em>

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Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is th

katrin [286]

Complete Question

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long.What Average force is excreted on the Nail

Answer:

$F=2*10^{4}N$

Explanation:

From the question we are told that:

Mass $m=0.500kg$

Initial Velocity $V=15.0m/s$

Distance $x=2.80mm=>0.00280m$

Diameter $d=2.50mm=>0.00250m$

Length $l=6.00cm=>0.6m$

Generally the equation for Force is mathematically given by

$F=\frac{mv^2}{2d}$

$F=\frac{0.500*15^2}{2.80*10^{-3}}$

$F=2*10^{4}N$

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2 years ago

While eating a peanut butter and jelly sandwich peanut butter drips on Dana shirt she tries to remove the peanut butter from her

MakcuM [25]

Answer:

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3 years ago

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.3 c

Mashcka [7]

Answer with Explanation:

We are given that

Diameter of pipe, $d_1=0.573 m$

$v_1=13.5 m/s$

$v_2=5.83 m/s$

Volume flow rate of the petroleum along the pipe= $Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})$

$Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s$

By equation of continuity

$A_1v_1=A_2v_2$

$\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2$

$d^2_2=\frac{v_1}{v_2}d^2_1$

$d_2=\sqrt{\frac{v_1}{v_2}}d_1$

$d_2=0.573\sqrt{\frac{13.5}{5.83}}$

$d_2=0.87 m$

$d_2=0.87\time 100=87 cm$

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4 0

3 years ago

what is the name of the area around a charged object where the object can exert a force on other charged objects?

shusha [124]

Answer:

An electric field is a region around a charged object where the object's electric force is exerted on other charged objects. Electric fields get weaker the farther away they are from the charge. An electric field is invisible. You can use the field line to represent it.

Explanation:

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2 years ago

A piece of wood is floating in a tub of water. A second piece of wood rests on top of the first piece; it does NOT touch the wat

andreev551 [17]

Answer:

C)

Explanation:

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3 years ago