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3 years ago

Weight is measured in units called

Physics

1 answer:

Aleksandr-060686 [28]3 years ago

8 0

Pounds

If you are talking about the unit of measurement for weight is that of force it would be Newtons.

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A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 6.0s

pashok25 [27]

Answer:

109656.25 Nm

Explanation:

$\omega_f$ = Final angular velocity = 1.5 rad/s

$\omega_i$ = Initial angular velocity = 0

$\alpha$ = Angular acceleration

t = Time taken = 6 s

m = Mass of disk = 29000 kg

r = Radius = 5.5 m

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{1.5-0}{6}\\\Rightarrow \alpha=0.25\ rad/s^2$

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=\dfrac{1}{2}mr^2\alpha\\\Rightarrow \tau=\dfrac{1}{2}29000\times 5.5^2\times 0.25\\\Rightarrow \tau=109656.25\ Nm$

The torque specifications must be 109656.25 Nm

5 0

3 years ago

13. A transformer has a primary coil with 600 turns and a secondary coil with 300 turns. If the output voltage is 320 volts, wha

vivado [14]

His is a step down transformer since n(primary) is greater than n(seconcary). You relate the input voltage with the ouput voltage with the following equation:

<span>Vout = n2/n1*Vin (n2/n1 is essentially your 'transfer function' that dictates what a specified input would produce) </span>

<span>Solving the equation: </span>

<span>Vin = Vout*n1/n2 = (320V)*(600/300) = 640 V </span>

<span>This is checked by seeing if Vin is greater than Vout, which it is for a step down transformer.</span>

5 0

3 years ago

2.)

Oduvanchick [21]

Answer:

$-22.7 m/s^2$

Explanation:

This is a uniformly accelerated motion, so we can determine the deceleration of the car by using a suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For the car in this problem,

u = 27.8 m/s

v = 0

s = 17 m

Solving for a, we find the acceleration:

$a=\frac{v^2-u^2}{2s}=\frac{0-27.8^2}{2(17)}=-22.7 m/s^2$

4 0

3 years ago

It takes a minimum distance of 76.50 m to stop a car moving at 15.0 m/s by applying the brakes (without locking the wheels). Ass

Vinvika [58]

The minimum stopping distance when the car is moving at 32.0 m/s is 348.3 m.

<h3>Acceleration of the car </h3>

The acceleration of the car before stopping at the given distance is calculated as follows;

v² = u² + 2as

when the car stops, v = 0

0 = u² + 2as

0 = 15² + 2(76.5)a

0 = 225 + 153a

-a = 225/153

a = - 1.47 m/s²

<h3>Distance traveled when the speed is 32 m/s</h3>

If the same force is applied, then acceleration is constant.

v² = u² + 2as

0 = 32² + 2(-1.47)s

2.94s = 1024

s = 348.3 m

Learn more about distance here: brainly.com/question/4931057

#SPJ1

5 0

2 years ago

Select the correct answer. The motion of a car on a position-time graph is represented with a horizontal line. What does this in

enyata [817]

If the car's motion appears as a horizontal line on a <u><em>position-time </em></u>graph, it shows that as time changes, the car's position doesn't change.

This is just a complicated way to say that the car is <em>not moving</em>.<em> (A)</em>

7 0

3 years ago