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ollegr [7]
4 years ago
14

The process of heat radiation is the ONLY method of heat transfer that can occur

Physics
2 answers:
AysviL [449]4 years ago
6 0

Answer:

Through space

Explanation:

The radiation from the sun heats the outer atmosphere

lina2011 [118]4 years ago
5 0
Radiation emitted by a body is a consequence of thermal agitation of its composing molecules. so...<span> electromagnetic waves ?</span>
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A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen
PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})

Where:

\omega_{o}, \omega - Initial and final angular velocities, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

\theta_{o}, \theta - Initial and final angular position, measured in radians.

Then,

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}

Given that \omega_{o} = 0\,\frac{rad}{s}, \omega = 0.70\,\frac{rad}{s} and \theta-\theta_{o} = 4.9\,rad, the angular acceleration is:

\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}

\alpha = 0.05\,\frac{rad}{s^{2}}

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

\omega = \omega_{o} + \alpha \cdot t

Where t is the time measured in seconds.

The time is cleared and obtain after replacing every value:

t = \frac{\omega-\omega_{o}}{\alpha}

If \omega_{o} = 0\,\frac{rad}{s},  \omega = 0.70\,\frac{rad}{s} and \alpha = 0.05\,\frac{rad}{s^{2}}, the required time is:

t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }

t = 14\,s

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

\bar \alpha = \frac{\omega-\omega_{o}}{t}

If \omega_{o} = 0\,\frac{rad}{s},  \omega = 0.70\,\frac{rad}{s} and t = 14\,s, the average angular acceleration is:

\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}

\bar \alpha = 0.05\,\frac{rad}{s^{2}}

The average angular acceleration is 0.05 radians per square second.

4 0
3 years ago
Jose gets up from his seat on the bus to move closer to the front. Just as he begins to walk forward, the bus stops at a light.
aleksklad [387]

Answer:

She falls forward

Explanation:

Dunno, just factss

5 0
3 years ago
Read 2 more answers
A car driver traveling at a speed of 108km per hour ,sees a traffic light and stopped after travelling for 20seconds .Find the a
navik [9.2K]

Answer:

– 2.5 m/s²

Explanation:

We have,

• Initial velocity, u = 180 km/h = 50 m/s

• Final velocity, v = 0 m/s (it stops)

• Time taken, t = 20 seconds

We have to find acceleration, a.

\longrightarrow a = (v ― u)/t

\longrightarrow a = (0 – 50)/20 m/s²

\longrightarrow a = –50/20 m/s²

\longrightarrow a = – 5/2 m/s²

\longrightarrow a = – 2.5 m/s² (Velocity is decreasing) [Answer]

6 0
3 years ago
2. Tomas is hanging from a tree limb, that is inclined at a 65° angle. The force
LUCKY_DIMON [66]

Answer:

57 N

Explanation:

Were are told that the force

of gravity on Tomas is 57 N.

And it acts at an inclined angle of 65°

Thus;

The vertical component of the velocity is; F_y = 57 sin 65

While the horizontal component is;

F_x = 57 cos 65

Thus;

F_y = 51.66 N

F_x = 24.09 N

The net force will be;

F_net = √((F_y)² + (F_x)²)

F_net = √(51.66² + 24.09²)

F_net = √3249.0837

F_net = 57 N

4 0
3 years ago
You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi
Mama L [17]

Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

6 0
3 years ago
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