All categories

3 years ago

A mirage is created when light is refracted ___.

Physics

2 answers:

omeli [17]3 years ago

7 0

Answer:

through layers of hot air just above a surface,causing it to follow a curved path.

elena-14-01-66 [18.8K]3 years ago

5 0

. . . downward (through the air).

You might be interested in

A uniform rod is set up so that it can rotate about a perpendicular axis at one of its ends. The length and mass of the rod are

igor_vitrenko [27]

Answer:

0.231 N

Explanation:

To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be

$\alpha = \frac{\theta}{t} = \frac{6.37}{9.87} = 0.6454 rad/s^2$

If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:

$I = \frac{mL^2}{3} = \frac{1.27*0.847^2}{3} = 0.303kgm^2$

According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is

$T = I\alpha = 0.303*0.6454 = 0.196 Nm$

So the force acting on the other end to generate this torque mush be:

$F = \frac{T}{L} = \frac{0.196}{0.847} = 0.231 N$

4 0

4 years ago

What is the acceleration of a 32-kg object is a 6.75-N force is being applied to it?

zimovet [89]

<span>0.21 m/s2

A.)

hope this helps best of luck :)
</span>

6 0

3 years ago

If a force of 32000N exerted pressure of 160N/m² , find the area on which the force acts.

Alex17521 [72]

Answer: 200m^2

Explanation:

160N=32000N/x

x*160N=32000

x=200m^2

4 0

2 years ago

A force of 15 newtons is applied to both Object A with a mass of 25 kilograms and Object B with a mass of 50 kilograms. What is

love history [14]

acceleration of object ais help the acceleration of an object b OD

3 0

3 years ago

A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mas

algol [13]

Answer:

6.3 rev/s

Explanation:

The new rotation rate of the satellite can be found by conservation of the angular momentum (L):

$L_{i} = L_{f}$

$I_{i}*\omega_{i} = I_{f}*\omega_{f}$

The initial moment of inertia of the satellite (a solid sphere) is given by:

$I_{i} = \frac{2}{5}m_{s}r^{2}$

Where $m_{s}$ : is the satellite mass and r: is the satellite's radium

$I_{i} = \frac{2}{5}m_{s}r^{2} = \frac{2}{5}1900 kg*(4.6 m)^{2} = 1.61 \cdot 10^{4} kg*m^{2}$

Now, the final moment of inertia is given by the satellite and the antennas (rod):

$I_{f} = I_{i} + 2*I_{a} = 1.61 \cdot 10^{4} kg*m^{2} + 2*\frac{1}{3}m_{a}l^{2}$

Where $m_{a}$ : is the antenna's mass and l: is the lenght of the antenna

$I_{f} = 1.61 \cdot 10^{4} kg*m^{2} + 2*\frac{1}{3}150.0 kg*(6.6 m)^{2} = 2.05 \cdot 10^{4} kg*m^{2}$

So, the new rotation rate of the satellite is:

$I_{i}*\omega_{i} = I_{f}*\omega_{f}$

$\omega_{f} = \frac{I_{i}*\omega_{i}}{I_{f}} = \frac{1.61 \cdot 10^{4} kg*m^{2}*8.0 \frac{rev}{s}}{2.05 \cdot 10^{4} kg*m^{2}} = 6.3 rev/s$

Therefore, the new rotation rate of the satellite is 6.3 rev/s.

I hope it helps you!

5 0

4 years ago