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Sergeeva-Olga [200]

2 years ago

15

Trafegando por uma avenida com velocidade constante de 54km/h, num dado instante, o motorista percebe o sinal vermelho à sua fre

nte e pisa no freio até que, depois de 5 s, o automóvel para. Determine a aceleração escalar média do carro nesse intervalo de tempo em Km/h dividido por s e em m/s²

1 answer:

Furkat [3]2 years ago

5 0

Answer:

The acceleration is - 3 m/s2.

Explanation:

initial speed, u = 54 km/h = 15 m/s

final speed, v = 0

time, t = 5 s

Let the acceleration is a.

use first equation of motion

v = u + at

0 = 15 + a x 5

a = - 3 m/s2

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Addition of a metal slab of thickness "a" between the plates of a parallel plate capacitor of plate separation "d" is equivalent

Crazy boy [7]

Answer:

$K = \frac{d}{d+a}$

Explanation:

The capacitance of a capacitor in terms of the dielectric constant, area of the plate and the distance separating the plate is given by:

$C = \frac{\epsilon A}{d}$

Where A = Area of the plate

d = distance between the plates

$\epsilon =$ dielectric constant

Case 1:

When a meta slab of thickness, a, is added between the plates of the parallel plate capacitor , the effective separation between the plates becomes d+a

Therefore the capacitance of the capacitor becomes:

$C = \frac{\epsilon A}{d + a}$ .......................(1)

Case 2:

Introducing a dielectric with dielectric constant K between the plates, the capacitance of the capacitor becomes:

$C = \frac{K\epsilon A}{d}$ .........................(2)

Equating (1) and (2)

$\frac{K\epsilon A}{d} = \frac{\epsilon A}{d+a}\\\frac{K}{d} = \frac{1}{d+a} \\K = \frac{d}{d+a}$

3 0

3 years ago

PLEASE HELP ASAP!!

kirza4 [7]

Answer:

A. Distance over which the force is applied

Explanation:

As we know that in pulley system the mass of the car is balanced by the tension in the string

so here we will have

$T = r \times F$

so here in order to decrease the force needed to lift the car we have to increase Distance over which the force is applied

So here if we increase the distance over which force is applied then it will reduce the effort applied by us in this pulley system as the torque will be more if the distance is more.

7 0

3 years ago

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.

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Answer:

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Explanation:

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Answer:

a) α=7.9x10^-4 rad

b) θ=1.12x10^-4 rad

c) The Earth and the Moon cannot be seen without a telescope.

Explanation:

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4 0

2 years ago