Think about the motion of a roller coaster car, after completing a trip around the track

How does the speed of a sound wave change as the medium through which it traveks changes?

uranmaximum [27]

Increase as density increase and vise versa.
<span>The wavelength increases when a sound wave travels from a less dense to a more dense medium, the speed increases, and the frequency stays the same.</span>

8 0

3 years ago

4. There is a flashing yellow light at the intersection you are approaching. What does the flashing yellow light indicate, and w

OleMash [197]

It indicates that you should proceed with caution. You should slow down and be prepared to stop to safely navigate this intersection.

4 0

4 years ago

Please help with this one

olasank [31]

Answer:

OPTION A is the correct answer

6 0

3 years ago

The 1000-lb elevator is hoisted by the pulley system and motor M. If the motor exerts a constant force of 500 lb on the cable, d

Nezavi [6.7K]

Answer:

the power that must be supplied to the motor is 101.45 hp

Explanation:

Given that,

The weight of elevator = 1000lb

The motor exerts a constant force of 500 lb on the cable

The load has been hoisted s = 15 ft starting from rest

The motor has an efficiency of e = 0.65

According to the equation of motion:

F = ma

$3(500) - 1000 = \frac{1000}{32.2} * aa = 16.1 ft/s^2$

$v^2 - u^2 = 2a (S - So)\\\\v^2 - (0)^2 = 2 * 16.1 (15-0)\\\\v = 21.98m/s$

To calculate the output power:

P(out) = F. v

P(out) = 3 (500) * 21.98

P(out) = 32970 lb.ft/s

As efficiency is given and output power is known, we can calculate the input power.

ε = P(out) / P(in)

0.65 = 32970 / P(in )

P(in) = 50,723 lb.ft/s

P(in) = 50,723 / 500 hp

= 101.45 hp

the power that must be supplied to the motor is 101.45 hp

5 0

4 years ago

An electric motor spins at 1000 rpm and is slowing down at a rate of 10 t rad/s2 ; where t is measured in seconds. (a) If the mo

REY [17]

Answer:

a) The tangential component of acceleration at the edge of the motor at $t = 1.5\,s$ is -1.075 meters per square second.

b) The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

Explanation:

The angular aceleration of the electric motor ( $\alpha$ ), measured in radians per square second, as a function of time ( $t$ ), measured in seconds, is determined by the following formula:

$\alpha = -10\cdot t\,\left[\frac{rad}{s^{2}} \right]$ (1)

The function for the angular velocity of the electric motor ( $\omega$ ), measured in radians per second, is found by integration:

$\omega = \omega_{o} - 5\cdot t^{2}\,\left[\frac{rad}{s} \right]$ (2)

Where $\omega_{o}$ is the initial angular velocity, measured in radians per second.

a) The tangential component of aceleration ( $a_{t}$ ), measured in meters per square second, is defined by the following formula:

$a_{t} = R\cdot \alpha$ (3)

Where $R$ is the radius of the electric motor, measured in meters.

If we know that $R = 7.165\times 10^{-2}\,m$ , $\alpha = 10\cdot t$ and $t = 1.5\,s$ , then the tangential component of the acceleration at the edge of the motor is: