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alexandr1967 [171]
3 years ago
13

During a very quick stop, a car decelerates at 6.8 m/s^2. Assume the forward motion of the car corresponds to a positive directi

on for the rotation of the tires (and that they do not slip on the pavement).(a) What is the angular acceleration of its 0.280-m-radius tires, assuming they do not slip on the pavement? (b) How many revolutions do the tires make before coming to rest, given their initial angular velocity is 95.0 rad/s ? (c) How long does the car take to stop completely? (d) What distance does the car travel in this time?
Physics
1 answer:
geniusboy [140]3 years ago
8 0

Answer:

-24.28571 rad/s²

29.57239 revolutions

3.91176 seconds

52.026478 m

Explanation:

a_t = Tangential acceleration = -6.8 m/s²

r = Radius of wheel = 0.28

\omega_i = Initial angular velocity = 95 rad/s

\theta = Angle of rotation

\omega_f = Final angular velocity

t = Time taken

Angular acceleration is given by

\alpha=\frac{a_t}{t}\\\Rightarrow \alpha=\frac{-6.8}{0.28}\\\Rightarrow \alpha=-24.28571\ rad/s^2

The angular acceleration is -24.28571 rad/s²

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-95^2}{2\times -24.28571}\\\Rightarrow \theta=185.80885\ rad=185.80885\times \frac{1}{2\pi}\\\Rightarrow \theta=29.57239\ rev

The number of revolutions is 29.57239

\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-95}{-24.28571}\\\Rightarrow t=3.91176\ s

The time it takes for the car to stop is 3.91176 seconds

Linear distance

s=r\theta\\\Rightarrow s=0.28\times 185.80885\\\Rightarrow s=52.026478\ m

The distance the car travels is 52.026478 m

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Answer:

Orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

Explanation:

The gravitational force is responsible for the orbital motion of the planet, satellite, artificial satellite, and other heavenly bodies in outer space.

When an object is applied with a velocity that is equal to the velocity of the orbit at that location, the body continues to move forward. And, this motion is balanced by the gravitational pull of the second object.

The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.

The velocity of the orbit is given by the relation,

                                    V = \sqrt{\frac{GM}{R + h} }

Where

                   V - velocity of the orbit at a height h from the surface

                    R - Radius of the second object

                    G - Gravitational constant

                    h - height from the surface

The body will be in orbital motion when its kinetic motion is balanced by gravitational force.

                         1/2 mV^{2} = GMm/R

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3 years ago
Which of the following is not true of a form?
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Answer:

It can only display one record at a time

Explanation:

Form ;

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3.It may contain images in the background.

4.This can be sorted data regardless of its source of information.

Only option C is wrong.

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Do objects tend to stay moving because of a force called Interia
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the net force applied to an object is directly proportional to the acceleration undergone by that object

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How much time will have passed by the time a sample of an isotope has decayed to 6.25% of its original mass if the half-life is
aleksandrvk [35]

Answer:

60 days.

Explanation:

Let the original mass (N₀) = 1 g

Amount remaining (N) = 6.25% of its original mass

= 6.25% × 1

= 6.25/100 × 1

= 0.0625 g

Half life (t½) = 15 days

Time (t) =?

Next, we shall determine the rate of decay. This can be obtained as follow:

Decay constant (K) = 0.693/ half life

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Half life (t½) = 15 days

Decay constant (K) =?

K = 0.693 / t½

K = 0.693 / 15

K = 0.0462 / day

Finally, we shall determine the time taken for the sample of the isotope to decay to 6.25% of its original mass.

This can be obtained as follow:

Original amount (N₀) = 1 g

Amount remaining (N) = 0.0625 g

Decay constant (K) = 0.0462 / day

Time (t) =?

Log (N₀/N) = Kt/2.3

Log (1/0.0625) = 0.0462 × t / 2.3

Log 16 = 0.0462 × t / 2.3

1.2041 = 0.0462 × t /2.3

Cross multiply

0.0462 × t = 1.2041 × 2.3

Divide both side by 0.0462

t = (1.2041 × 2.3)/0.0462

t = 59.9 ≈ 60 days

Therefore, the time taken for the sample of the isotope to decay to 6.25% of its original mass is 60 days

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