Advection fog occurs when humid air move over cold ground

acceleration is defined as the rate of change for which characteristic? a. displacement b. position c. velocity d. time

fredd [130]

The answer is velocity.

7 0

3 years ago

A bicycle rider pushes a 13kg bicycle up a steep hill. the incline is 24 degree and the road is 275m long. the rider pushes the

Digiron [165]

Answer:

A. W = 6875.0 J.

B. W = -14264.6 J.

Explanation:

A. The work done by the rider can be calculated by using the following equation:

$W_{r} = |F_{r}|*|d|*cos(\theta_{1})$

Where:

$F_{r}$ : is the force done by the rider = 25 N

d: is the distance = 275 m

θ: is the angle between the applied force and the distance

Since the applied force is in the same direction of the motion, the angle is zero.

$W_{r} = |F_{r}|*|d|*cos(0) = 25 N*275 m = 6875.0 J$

Hence, the rider does a work of 6875.0 J on the bike.

B. The work done by the force of gravity on the bike is the following:

$W_{g} = |F_{g}|*|d|*cos(\theta_{2})$

The force of gravity is given by the weight of the bike.

$F_{g} = -mgsin(24)$

And the angle between the force of gravity and the direction of motion is 180°.

$W_{g} = |mgsin(24)|*|d|*cos(\theta_{2})$

$W_{g} = 13 kg*9.81 m/s^{2}*sin(24)*275 m*cos(180) = -14264.6 J$

The minus sign is because the force of gravity is in the opposite direction to the motion direction.

Therefore, the magnitude of the work done by the force of gravity on the bike is 14264.6 J.

I hope it helps you!

3 0

2 years ago

A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertica

Radda [10]

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed $\omega =40rad/s$

mass of $m_2=0.8 kg$ dropped at r=0.3 m from center

let $\omega _2$ be the final angular velocity of cylinder

Conserving Angular momentum

$L_1=L_2$

$\left ( \frac{m_1R^2}{2}\right )\omega =\left ( \frac{m_1R^2}{2}+m_2r^2\right )\omega _2$

$\left ( \frac{18\cdot 1.7^2}{2}\right )\cdot 40=\left ( \frac{18\cdot 1.7^2}{2}+0.8\cdot 0.3^2\right )\omega _2$

$26.01\times 40=26.082\times \omega _2$

$\omega _2=39.88 rad/s$

3 0

3 years ago

Select all of the statements that are true.

Strike441 [17]

I think its all four of them could be wrong but try all four !!!!!!

7 0

3 years ago

PLEASE HELP!! DUE SOON!!!

max2010maxim [7]

Answer:

See attached file :)

Hope this helps!

All the love, Ya boi Fraser :)

4 0

2 years ago

Read 2 more answers