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2 years ago

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A coil with 55 loops has its flux change from 0 Wb to 0.0266 Wb in a certain amount of time, generating 5.22 V of EMF. How much

time elapsed?

1 answer:

Eduardwww [97]2 years ago

4 0

Answer:

0.280 s

Explanation:

I set it up as 5.22=(55)(0.0266)/x and then solved for x to be 2.80.

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A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2

Ahat [919]

Answer:

Approximately $0.608$ (assuming that $g = 9.81\; \rm N\cdot kg^{-1}$ .)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

Let $m$ represent the mass of this car.
Let $r$ represent the radius of the circular track.

This answer will approach this question in two steps:

Step one: determine the centripetal force when the car is about to skid.
Step two: calculate the coefficient of static friction.

For simplicity, let $a_{T}$ represent the tangential acceleration ( $1.90\; \rm m \cdot s^{-2}$ ) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to $90^\circ$ or $\displaystyle \frac{\pi}{2}$ radians.

The angular acceleration of this car can be found as $\displaystyle \alpha = \frac{a_{T}}{r}$ . ( $a_T$ is the tangential acceleration of the car, and $r$ is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ( $u$ and $v$ ) to (tangential) acceleration $a_{T}$ and displacement $x$ :

$v^2 - u^2 = 2\, a_{T}\cdot x$ .

The idea is to solve for the final angular velocity using the angular analogy of that equation:

$\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta$ .

In this equation, $\theta$ represents angular displacement. For this motion in particular:

$\omega(\text{initial}) = 0$ since the car was initially not moving.
$\theta = \displaystyle \frac{\pi}{2}$ since the car travelled one-quarter of the circle.

Solve this equation for $\omega(\text{final})$ in terms of $a_T$ and $r$ :

$\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}$ .

Let $m$ represent the mass of this car. The centripetal force at this moment would be:

$\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}$ .

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, $m\, g$ .

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let $\mu_s$ denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

$F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g$ .

The size of this force should be equal to that of the centripetal force when the car is about to skid:

$\mu_s\, m\, g = \pi\, m\, a_{T}$ .

Solve this equation for $\mu_s$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g}$ .

Indeed, the expression for $\mu_s$ does not include any unknown letter. Let $g = 9.81\; \rm N\cdot kg^{-1}$ . Evaluate this expression for $a_T = 1.90\;\rm m \cdot s^{-2}$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608$ .

(Three significant figures.)

7 0

3 years ago

it is determined that a certain light wave has a wavelength of 3.012x10^-12 m. the light travels at 2.99x10^8 m/s. what is the f

Dahasolnce [82]

Answer:

$9.93\times 10^{19}Hz$

Explanation:

Speed of light is the product of its wavelength and frequency, expressed as

S=fw

Where s represent speed, f is frequency while w is wavelength

Making f the subject of the formula then

f=s/w

Substituting 2.99x10^8 m/s for s and 3.012x10^-12 m for w then

$f=\frac {2.99\times 10^{8}}{3.012\times 10^{-12}}=9.926958831341\times 10^{19}\\f\approx 9.93\times 10^{19}Hz$

Therefore, the frequency equals to $9.93\times 10^{19}Hz$

4 0

2 years ago

If a 50 kg student is standing on the edge of a cliff. Find the student’s gravitational potential energy if the cliff is 40 m hi

Oduvanchick [21]

you can find it using the equation: potential energy=mass*gravitational acceleration*height.

energy=50kg*9.8N/kg*40m=19600Nm=19600J or 19.6kJ

Sometimes they use 10 instead of 9.8 for the g constant.

Rember to make me Brainliest!!!

3 0

3 years ago

Why does a lorry travelling at 30mph have more kinetic energy than a car travelling at the same velocity?

rjkz [21]

Answer:

Because it has more mass

Explanation:

To understand this, think about the equation of kinetic energy

KE = $\frac{1}{2}$ m $v^{2}$

Kinetic energy depends on both the velocity (v) as well as the mass (m).

Because a lorry is bigger and heavier than a car, it will have more mass. With more mass, at the same velocity the lorry with have more kinetic energy.

5 0

3 years ago

How does The centripetal fotce change if the car has less Mass?

vaieri [72.5K]

Answer:

The centripetal force acting on the car is proportional to the mass of the car.

Explanation:

Let,

The mass of the car be 'm'

The velocity of the car moving in the curved path be 'v'

The radius of the curved path be 'r'

According to physics, a body moving ion circular path experience a force directed along the radius of the path. This force is called centripetal force.

The formula for centripetal force is,

<em>F = mv²/r</em>

Where,

a = v²/r

So, if the mass of the car changes, the centripetal force also changes proportionally according to the above equation.

5 0

3 years ago