Ask question

Ask question

All categories

Dahasolnce [82]

3 years ago

12

a fan is rotating clockwise and its acceleration has a positive sign. is the angular velocity of the fan speeding up, slowing do

wn, or constant?

1 answer:

balu736 [363]3 years ago

5 0

Answer:

The angular velocity is slowing down.

Explanation:

By convention, if a rigid body is rotating clockwise, the angular velocity is negative.
If the angular acceleration has a positive sign, since the angular acceleration and the angular velocity have opposite signs, this means that the angular velocity is slowing down.

You might be interested in

Im so confused can someone help. I’ll give you BRAINLIETS

Hatshy [7]

I found the answer sheet online for you

3 0

3 years ago

A 2.0-kg object is attached to a spring (k = 55.6 N/m) that hangs vertically from the ceiling. The object is displaced 0.045 m v

rusak2 [61]

Answer:

Maximum acceleration will be $1.251m/sec^2$

Explanation:

We have given mass of the object m = 2 kg

Spring constant k = 55.6 N/m

Amplitude is given as A = 0.045 m

We know that maximum acceleration in SHM is given by

Maximum acceleration $=A\omega ^2$

We know that $\omega ^2=\frac{k}{m}=\frac{55.6}{2}=27.8$

So maximum acceleration = $27.8\times 0.045=1.251m/sec^2$

8 0

3 years ago

Read 2 more answers

Explain the relationship of charles law to absolute zero

vivado [14]

Chalres law and absolute zero . in 1780s, by experiments

8 0

3 years ago

Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg

vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

$\omega = \frac{v}{R}$

Where,

$\omega =$ Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

$\alpha = \frac{\omega}{t}$

Where

$\alpha =$ Angular acceleration

$\omega =$ Angular velocity

t = Time

Our values are

$v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})$

$v = 16.67m/s$

$r = 0.25m$

$t=6s$

Replacing at the previous equation we have that the angular velocity is

$\omega = \frac{v}{R}$

$\omega = \frac{ 16.67}{0.25}$

$\omega = 66.67rad/s$

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

$\alpha = \frac{\omega}{t}$

$\alpha = \frac{66.67}{6}$

$\alpha = 11.11rad/s^2$

Therefore the angular acceleration of a point on the outer edge of the tires is $11.11rad/s^2$

5 0

3 years ago

A 2,000 kg car is moving at 15m/s when it collides with a 1200kg car sitting still. Briefly compare the impulse imparted on the

I am Lyosha [343]

impulse = F × t

The greater the impulse exerted on something, the greater will be the change in momentum.

impulse = change in momentum

Ft = ∆(mv)

4 0

3 years ago

Read 2 more answers