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3 years ago

What is the radius of an automobile tire that turns with a period of 0.091 s and has a linear

Physics

1 answer:

faust18 [17]3 years ago

3 0

Answer:

1) The radius of the tire is approximately 0.28966 meters

2) The centripetal force is the force that keeps a body moving on a circular path

Explanation:

1) The linear speed of the automobile tire = 20.0 m/s

The period with which the tire turns = 0.091 s

The period = The time it takes to make a complete turn

Therefore;

The number of turns in 1 second = 1/0.091 ≈ 10.989 turns

The distance covered with 10.989 turns, assuming no friction = 10.989 × The circumference of the tire

∴ The distance covered with 10.989 turns, assuming no friction = 10.989 × 2 × π × Radius of the tire

From the speed of the car, 20.0 m/s, we have;

The distance covered in 1 second = 20.0 meters

Therefore;

10.989 × 2 × π × Radius of the tire = 20.0 meters

Radius of the tire = (20.0 meters)/(10.289 × 2 × π) ≈ 0.28966 meters

The radius of the tire ≈ 0.28966 meters

2) The centripetal force is the force required to maintain the curved motion of an object, and having a direction towards the center of the rotary motion.

The centripetal force is given by the formula, $F = \dfrac{m \cdot v^2}{r}$

Where;

F = The centripetal force

m = The mass of the object

v = The linear velocity of the object

r = The radius of the rotational motion.

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If for every action force an equal and opposite reaction force exlsts, how can anything ever be accelerated?

klasskru [66]

Answer:

Explanation:

The net force is the force between action and reaction and when this forces are not the same an acceleration is spurred.

3 0

3 years ago

Assume that block A which has a mass of 30 kg is being pushed to the left with a force of 75 N along a frictionless surface. Wha

Veronika [31]

Answer:

The force of friction acting on block B is approximately 26.7N. Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.

Explanation:

The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.

To solve this problem, start with setting up the net force equations for both block A and B:

$F_{Anet} = m_A\cdot a_A = F - F_{fr}\\F_{Bnet} = m_B\cdot a_B = F_{fr}$

where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:

$a_B = a_A-0.5 \frac{m}{s^2}$

Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:

$30kg\cdot a_A = 75N - F_{fr}\\24kg\cdot a_B = F_{fr}\\a_B= a_A-0.5 \frac{m}{s^2}\\\implies \\a_A=\frac{87}{54}\frac{m}{s^2},\,\,\,a_B=\frac{10}{9}\frac{m}{s^2}\\F_{fr} = 24kg \cdot \frac{10}{9}\frac{m}{s^2}=\frac{80}{3}kg\frac{m}{s^2}\approx 26.7N$

The force of friction acting on block B is approximately 26.7N.

This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.

6 0

3 years ago

Read 2 more answers

A spherically-spreading EM wave comes from a 104.0 W source. At a distance of 9.6 m, what is the intensity of the wave?

Nookie1986 [14]

Answer:

Approximately 0.0898 W/m².

Explanation:

The intensity of light measures the power that the light delivers per unit area.

The source in this question delivers a constant power of $\rm 104.0\; W$ . If the source here is a point source, that $\rm 104.0\; W$ of power will be spread out evenly over a spherical surface that is centered at the point source. In this case, the radius of the surface will be 9.6 meters.