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2 years ago

7

A race car travels in a circular track of radius 200m. If the car move with a constant speed of 80 m/s, find (a) its angular vel

ocity and (b) its tangential acceleration

1 answer:

Helen [10]2 years ago

3 0

Answer:

(a) 0.40 rad/s

(b) 0 m/s

Explanation:

Since the car is moving at a constant speed, the tangential acceleration of the car is 0 m/s.

To find its angular velocity, we use the equation ω = v/r.

Therefore, the angular velocity ω = 80/200 = 0.40 rad/s.

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2) [25 pts] The bob of mass m=5 kg shown in the figure is being held by a force F. If the angle is 0⃗

Colt1911 [192]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The tension is $T = 48.255 \ N$

b

The time taken is $t = 0.226 \ s$

c

The position for maximum velocity is

S = 0

d

The maximum velocity is $V_{max} =0.384 \ m/s$

Explanation:

The free body for this question is shown on the second uploaded image

From the question we are told that

The mass of the bob is $m_b = 5 \ kg$

The angle is $\theta = 10^o$

The length of the string is $L = 0.5 \ m$

The tension on the string is mathematically represented as

$T = mg cos \theta$

substituting values

$T = 5 * 9.8 cos(10)$

$T = 48.255 \ N$

The motion of the bob is mathematically represented as

$S = A sin (w t + \frac{\pi }{2} )$

=> $S = A sin (wt)$

Where $w$ is the angular speed

and $\frac{\pi}{2}$ is the phase change

At initial position S = 0

So $wt = cos^{-1} (0)$

$wt = 1$

Generally $w$ can be mathematically represented as

$w = \frac{2 \pi }{T}$

Where T is the period of oscillation which i mathematically represented as

$T = 2 \pi \sqrt{\frac{L}{g} }$

So

$t = \frac{1}{w}$

$t = \frac{T}{2 \pi}$

$t = \sqrt{\frac{L}{g} }$

substituting values

$t = \sqrt{\frac{0.5}{9.8} }$

$t = 0.226 \ s$

Looking at the equation

$wt = 1$

We see that maximum velocity of the bob will be at S = 0

i. e $w = \frac{1}{t}$

The maximum velocity is mathematically represented as

$V_{max} = w A$

Where A is the amplitude which is mathematically represented as

$A = L sin \theta$

So

$V_{max} = \frac{2 \pi }{T } L sin \theta$

$V_{max} = \frac{2 \pi }{2 \pi } \sqrt{\frac{g}{L} } L sin \theta$ Recall $T = 2 \pi \sqrt{\frac{L}{g} }$

$V_{max} = \sqrt{gL} sin \theta$

substituting values

$V_{max} = \sqrt{9.8 * 0.5 } sin (10)$

$V_{max} =0.384 \ m/s$

6 0

2 years ago

What is the acceleration of a 1000 kg car subject to a 550 N net force

Xelga [282]

Answer:

The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2

Explanation:

Given:

F = 550 N

m = 1000 kg

To Find:

a = ?

Solution:

So by the equation by Newton's 2nd Law of Motion,

F = m x a

550 N = 1000 kg x a

a = 550 N/ 1000 kg

a = 0.55 m/s^2

Therefore,

The acceleration of a 1000 kg car subject to a 550 N net force = 0.55 m/s^2

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4 0

2 years ago

could you help me with this question: Aspherical shell is formed by taking a solid sphere of radius 24.0 cm and hollowing out a

kumpel [21]

Answer:

La vdd tengo la misma duda, de q prepa eres?

Explanation:

Nada más para saber

5 0

2 years ago

What is the name of the line drawn perpendicular to the surface where a light ray strikes?

AnnyKZ [126]

It is the normal line

7 0

2 years ago

A ball bounced by a basketball player on the floor bounces back up at her. Newton's First Law Newton's Second Law Newton's Third

taurus [48]

Answer:

Newton's Third Law of Motion

Explanation:

Newton's Third Law of Motion which states that, for every action there is an equal but opposite reaction.

This ultimately implies that, in every interaction, there is a pair of forces acting on the two interacting objects.

In this scenario, a ball bounced by a basketball player on the floor bounces back up at her.

According to Newton's Third Law of Motion, the statement above simply means that in every interaction, there is a pair of forces acting on the two interacting objects i.e the ball and floor. The size of the force on the ball equals the size of the force on the floor. These two forces are called action and reaction forces and are the subject of Newton's third law of motion.

Hence, the ball bounced by the basketball player on the floor would bounce back in equal magnitude.

4 0

2 years ago