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2 years ago

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A car is moving in uniform circular motion. If the cars speed were to double to keep the car moving with the same radius the acc

eleration would?
A- Increase by a factor of 2
B- Increase by a factor of 4
C- Decrease by a factor of 2
D- Decrease by a factor of 4

1 answer:

Stells [14]2 years ago

5 0

Answer:

<em>The centripetal acceleration would increase by a factor of 4</em>

<em>Correct choice: B.</em>

Explanation:

<u>Circular Motion</u>

The circular motion is described when an object rotates about a fixed point called center. The distance from the object to the center is the radius. There are other magnitudes in the circular motion like the angular speed, tangent speed, and centripetal acceleration. The formulas are:

$v_t=w\ r$

$\displaystyle a_c=\frac{v_t^2}{r}$

If the speed is doubled and the radius is the same, then

$\displaystyle a_c=\frac{(2v_t)^2}{r}$

$\displaystyle a_c=4\frac{v_t^2}{r}$

The centripetal acceleration would increase by a factor of 4

Correct choice: B.

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List three (3) specific examples of each biological<br> Homeostasis:<br> Metabolism:<br> Growth:

Andre45 [30]

Answer:

Homeostasis: When bacteria or viruses that can make you ill get into your body, your lymphatic system kicks in to help maintain homeostasis.

Metabolism: The processes of making and breaking down glucose molecules are example of metabolism.(respiration and photosynthesis)

Growth:The liver continues to form new cells to replace senescent and dying ones.

Hope these examples help you.

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2 years ago

The angular velocity of a flywheel obeys the equa tion w(1) A Br2, where t is in seconds and A and B are con stants having numer

makkiz [27]

Answer:

$A \to rad/s$

$B \to rad/s^3$

Explanation:

$\omega_z(t)=A + Bt^2$

Required

The units of A and B

From the question, we understand that:

$\omega_z(t) \to rad/s$

This implies that each of $A$ and $Bt^2$ will have the same unit as $\omega_z(t)$

So, we have:

$A \to rad/s$

$Bt^2 \to rad/s$

The unit of t is (s); So, the expression becomes

$B * s^2 \to rad/s$

Divide both sides by $s^2$

$B \to \frac{rad/s}{s^2}$

$B \to rad/s^3$

5 0

2 years ago

A sinusoidal voltage is given by the expression ????(????)=20cos(5π×103 ????+60°) V. Determine its (a) frequency in hertz, (b) p

MA_775_DIABLO [31]

<em>There are some placeholders in the expression, but they can be safely assumed</em>

Answer:

(a) $f=1617.9\ Hz$

(b) $T=0.618\ ms$

(c) $A=20 \ Volts$

(d) $\varphi=60^o$

Explanation:

<u>Sinusoidal Waves </u>

An oscillating wave can be expressed as a sinusoidal function as follows

$V(t)&=A\cdot \sin(2\pi ft+\varphi )$

Where

$A=Amplitude$

$f=frequency$

$\varphi=Phase\ angle$

The voltage of the question is the sinusoid expression

$V(t)=20cos(5\pi\times 103t+60^o)$

(a) By comparing with the general formula we have

$f=5\pi\times 103=1617.9\ Hz$

$\boxed{f=1617.9\ Hz}$

(b) The period is the reciprocal of the frequency:

$\displaystyle T=\frac{1}{f}$

$\displaystyle T=\frac{1}{1617.9\ Hz}=0.000618\ sec$

Converting to milliseconds

$\boxed{T=0.618\ ms}$

(c) The amplitude is

$\boxed{A=20 \ Volts}$

(d) Phase angle:

$\boxed{\varphi=60^o}$

4 0

2 years ago

A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist

Paha777 [63]

Answer:

Part a)

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

Explanation:

As we know by energy conservation the total energy at the bottom of the bowl is given as

$\frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy

now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

$\frac{1}{2}mv^2 = mgh'$

now we know that

$I = \frac{2}{5}mr^2$

$\omega = \frac{v}{r}$

so we have

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh$

$\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh$

$\frac{7}{10}mv^2 = mgh$

$\frac{1}{2}mv^2 = \frac{10}{14}mgh$

so the height on the smooth side is given as

$h' = \frac{10}{14} h$

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

$h' = \frac{10}{14} h$

so on smooth it will go to lower height

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3 years ago

Read 2 more answers

An electromagnet is a solenoid with a piece of ferromagnetic material within it.

Vlada [557]

ANSWER:
The answer will be OT

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2 years ago