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3 years ago

A plane startling from rest accelerates at 3 m/s Square. Calculate the increase in velocity after 25 sec

Physics

1 answer:

Tatiana [17]3 years ago

3 0

Answer:

75 m/s

Explanation:

We can apply motion equations here

V = U + a * t

V = velocity @ t time

U = initial velocity

a = acceleration

t = time taken

V = U + a * t

V = 0+ 3 * 25

V = 75 m/s

After 25 seconds , subjected to the given acceleration velocity is increased from 0 to 75 m/s

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3 years ago

In each of two coils the rate of change of the magnetic flux in a single loop is the same. The emf induced in coil 1, which has

irinina [24]

Answer:

Coil 2 have 235 loops

Explanation:

Given

The number of loops in coil 1 is n ₁= 159

The emf induced in coil 1 is ε ₁ = 2.78 V

The emf induced in coil 2 is ε ₂ = 4.11 V

Let

n ₂ is the number of loops in coil 2.

Given, the emf in a single loop in two coils are same. That is,

ϕ ₁/n ₁= ϕ ₂ n ₂⟹ 2.78/159 = 4.11/ n ₂

n₂= $\frac{159 * 4.11}{2.78}$

n₂=235

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4 0

3 years ago

Read 2 more answers

A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen

PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})$

Where:

$\omega_{o}$ , $\omega$ - Initial and final angular velocities, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

Then,

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\theta-\theta_{o} = 4.9\,rad$ , the angular acceleration is:

$\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}$

$\alpha = 0.05\,\frac{rad}{s^{2}}$

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

$\omega = \omega_{o} + \alpha \cdot t$

Where $t$ is the time measured in seconds.

The time is cleared and obtain after replacing every value:

$t = \frac{\omega-\omega_{o}}{\alpha}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $\alpha = 0.05\,\frac{rad}{s^{2}}$ , the required time is:

$t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }$

$t = 14\,s$

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

$\bar \alpha = \frac{\omega-\omega_{o}}{t}$

If $\omega_{o} = 0\,\frac{rad}{s}$ , $\omega = 0.70\,\frac{rad}{s}$ and $t = 14\,s$ , the average angular acceleration is:

$\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}$

$\bar \alpha = 0.05\,\frac{rad}{s^{2}}$

The average angular acceleration is 0.05 radians per square second.

4 0

3 years ago

Help me quick!!! please!!

Rudik [331]

Answer:

39.240 W

Explanation:

Let's start by calculating the work done by the engine. We can assume that it is the same work done by the weight of the object to bring it from 40m to the surface: as much energy it takes to bring it up, the same ammount it takes to bring it down. Said work is $w= \vec F\cdot \vec{h} = mg h = 1000 \times 9.81\times 40 = 392.400 J$

At this point we can simply apply the definition of power, that is $P = \frac wt$ , to get the power of the engine is $39.240 W$

4 0

3 years ago

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