The weight of a rock is 8.5 pounds. Does this statement include

An auto race takes place on a circular track. A car completes one lap in a time of 25.0 s, with an average tangential speed of 5

ludmilkaskok [199]

Answer:

(a) Angular speed will be 0.2512 rad/sec

(b) Radius will be 213.375 m

Explanation:

We have given time to complete 1 lap = 25 sec

We know that 1 lap = $2\pi$ radian

(a) So angular speed $=\frac{2\pi }{25}=\frac{2\times 3.14}{25}=0.2512rad/sec$

(B) Tangential velocity is given as v = 53.6 m/sec

We know that tangential velocity is given by

$v=\omega r$

So radius $r=\frac{v}{\omega }=\frac{53.6}{0.2512}=213.375m$

4 0

4 years ago

Why is it not advisable to touch a canvas tent from inside when it is raining?

Sauron [17]

Answer: When you touch wet canvas, surface tension will draw water to your finger. However, the drop left behind where you touched, like any irregular point on an overhead surface, will draw condensation from inside the tent if it is humid.

3 0

3 years ago

Explain the relationship between mass and acceleration

ycow [4]

Answer:

Explanation: Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.

3 0

3 years ago

Read 2 more answers

023 (part 1 of 2) 10.0 points

Annette [7]

Answer:

Part 1

The angular speed is approximately 1.31947 rad/s

Part 2

The change in kinetic energy due to the movement is approximately 675.65 J

Explanation:

The given parameters are;

The rotation rate of the merry-go-round, n = 0.21 rev/s

The mass of the man on the merry-go-round = 99 kg

The distance of the point the man stands from the axis of rotation = 2.8 m

Part 1

The angular speed, ω = 2·π·n = 2·π × 0.21 rev/s ≈ 1.31947 rad/s

The angular speed is constant through out the axis of rotation

Therefore, when the man walks to a point 0 m from the center, the angular speed ≈ 1.31947 rad/s

Part 2

Given that the kinetic energy of the merry-go-round is constant, the change in kinetic energy, for a change from a radius of of the man from 2.8 m to 0 m, is given as follows;

$\Delta KE_{rotational} = \dfrac{1}{2} \cdot I \cdot \omega ^2 = \dfrac{1}{2} \cdot m \cdot v ^2$

I = m·r²

Where;

m = The mass of the man alone = 99 kg

r = The distance of the point the man stands from the axis, r = 2.8 m

v = The tangential velocity = ω/r

ω ≈ 1.31947 rad/s

Therefore, we have;

I = 99 × 2.8² = 776.16 kg·m²

$\Delta KE_{rotational}$ = 1/2 × 776.16 kg·m² × (1.31947)² ≈ 675.65 J

3 0

3 years ago

Is aluminum foil reflecting onto something conduction, convection, or radiation?

mojhsa [17]

I had the SAME problem, put down Radiation and it’s thermal/light.

4 0

3 years ago