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3 years ago

What is the difference between light energy and heat energy

Physics

1 answer:

MrRissso [65]3 years ago

4 0

ANSWER:

What is the difference between heat and light? - Physics Stack ... Heat and light are different but they are both forms of energy. Heat is a form of kinetic energy contained in the random motion of the particles of a material. Light is a form of electromagnetic energy. As with other forms of energy, heat energy can be transformed into light energy and vice versa.

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Find the magnitude of the angular momentum of the second hand on a clock about an axis through the center of the clock face. Ass

Naddika [18.5K]

Answer:

L = 4.711 *10^{-6} kg m2/s

Explanation:

$i =\frac{ml^3}{3}$

$= \frac{0.00*.15^2}{3}$

=4.5*10^-5

angular velocity

$\omega = \frac{2\pi}{60}$

= 0.1047 rad/s

the angular momentum,

$L = I\omega$

$L = 4.5*10^{-5}* 0.1047 rad/s$

$L = 4.711 *10^{-6} kg m2/s$

3 0

2 years ago

Read 2 more answers

A bar on a hinge starts from rest and rotates with an angular acceleration α = 12 + 5t, where α is in rad/s2 and t is in seconds

Rama09 [41]

Answer:

Ф = 239.73 rad

Explanation:

α = 12 + 15×t

W = ∫α×dt

= ∫(12 + 5×t)×dt

= 12×t + 2.5×t^2

then:

Ф = ∫W×dt

= ∫(12×t + 2.5×t^2)dt

= 6×t^2 + 5/6×t^3

therefore the angle at t = 4.88s is:

Ф = 6×(4.88)^2 + 5/6×(4.88)^3

= 239.73 rad

5 0

3 years ago

Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh

frosja888 [35]

Answer:

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

Explanation:

Physically speaking, the resulting velocity of the helicopter ( $\vec v_{H}$ ), measured in meters per second, is equal to the absolute velocity of the wind ( $\vec v_{W}$ ), measured in meters per second, plus the velocity of the helicopter relative to wind ( $\vec v_{H/W}$ ), also call velocity at still air, measured in meters per second. That is:

$\vec v_{H} = \vec v_{W}+\vec v_{H/W}$ (1)

In addition, vectors in rectangular form are defined by the following expression:

$\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha)$ (2)

Where:

$\|\vec v\|$ - Magnitude, measured in meters per second.

$\alpha$ - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

$\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W})$ (3)

$\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)$

If we know that $\|\vec v_{W}\| = 25\,\frac{m}{s}$ , $\|\vec v_{H/W}\| = 125\,\frac{m}{s}$ , $\alpha_{W} = 240^{\circ}$ and $\alpha_{H/W} = 325^{\circ}$ , then the resulting velocity of the helicopter is:

$\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)$ $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$