Ask question

Ask question

All categories

3 years ago

5

A wheel accelerates from rest for 2.0s with a constant acceleration of 3.0 rad/s^2. if the wheel has a radius of 0.50m, through

what distance in m has a point on the rim moved?

1 answer:

romanna [79]3 years ago

4 0

Ф = 1/2αt²
s = rФ

s = 1/2αrt²

s distance
α = 3 rad/s²
t = 2 s
r = 0.5 m

You might be interested in

Can a measurement be exact

Ulleksa [173]

Can a measurement be exact? No

3 0

3 years ago

Sliding friction is _ than the static friction.

Alona [7]

Answer:

less

Explanation:

Sliding friction is always less than static friction. This is because in sliding friction, the bodies slide with each other and thus the effect of friction is not more. However, it does not happen in the case of static friction.

4 0

3 years ago

A frame hanging on a wall is held by two cables. The tension in each cable is 30 N, and the cables make an angle of 45° with the

Dimas [21]

Answer: option D) 42.4 N

The weight of the frame is balanced by the vertical component of tension.

W = T sin θ + T sin θ = 2 T sin θ

The tension in each cable is T = 30 N

Angle made by the cables with the horizontal, θ = 45°

⇒ W = 2×30 N × sin 45° = 2 × 30 N × 0.707 = 42.4 N

Hence, the weight of the frame is 42.4 N. Correct option is D.

6 0

4 years ago

When Casey looks at a CD or DVD she notices that there is a rainbow pattern that is created by the light. She became curious and

ipn [44]

Answer:

Explanation:

diffraction

6 0

3 years ago

Read 2 more answers

A Carnot engine operates with an efficiency of 26.0% when the temperature of its cold reservoir is 281 K. Assuming that the temp

VLD [36.1K]

Answer:

The temperature of cold reservoir should be 246.818 K for efficiency of 35%

Explanation:

In first case we have given efficiency of Carnot engine = 26 % = 0.26

Temperature of cold reservoir $T_L=281K$

We know that efficiency of Carnot engine is given by

$\eta =1-\frac{T_L}{T_H}$

$0.26 =1-\frac{281}{T_H}$

$T_H=379.72K$

For second Carnot engine efficiency is given as 35% = 0.35

And temperature of hot reservoir is same so $T_H=379.72K$

So $0.35=1-\frac{T_L}{379.72}$

$T_L=246.818K$

So the temperature of cold reservoir should be 246.818 K for efficiency of 35%

4 0

4 years ago