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3 years ago

How are heat and light waves produced on and in the sun?

2 answers:

8 0

All of the electromagnetic energy radiated from the sun (and from
other stars) is the product of nuclear fusion in its core.

lesya [120]3 years ago

3 0

Heat waves are the radiation released from the Sun, the energy and the light is what is visible of this emission of energy.

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Which is faster a bowling ball or a golf ball

timofeeve [1]

Golf ball because it weighs less so it has more power to yeah

8 0

3 years ago

Read 2 more answers

A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.

Katen [24]

Answer:

9.4 m/s

Explanation:

The work-energy theorem states that the work done on an object is equal to the change in kinetic energy of the object.

So we can write:

$W=K_f - K_i$

where in this problem:

W = -36.733 J is the work performed on the car (negative because its direction is opposite to the motion of the car)

$K_i = 66,120 J$ is the initial kinetic energy of the car

$K_f$ is the final kinetic energy

Solving for Kf,

$K_f = W+K_i = -36,733+66,120=29,387 J$

The kinetic energy of the car can be also written as

$K_f = \frac{1}{2}mv^2$

where:

m = 661 kg is the mass of the car

v is its final speed

Solving, we find

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.4 m/s$

8 0

3 years ago

PHYSICS HELP PLEASE!! MUST SHOW MATH WORK!!

SVETLANKA909090 [29]

<u>Answer:</u>

1) Distance traveled by bird = 403 meter

2)Average speed = 1.66 km /hour

3) Zcceleration = 2 $m/s^2$

<u>Explanation:</u>

1) Distance traveled = Speed * Time taken = 31 * 13 = 403 meter.

2) Average speed = Total distance covered / Time taken for that distance to cover.

Total distance covered = 2+0.5+2.5 = 5 km

Time taken = 3 hours

Average speed = 5/3 = 1.66 km /hour

3) Acceleration is defined as the rate of change of velocity, so acceleration a = change in velocity/time.

Change in velocity = 14 - 6 = 8 m/s

Time = 4 seconds

So acceleration = 8 / 4 = 2 $m/s^2$

6 0

3 years ago

Photoelectrons with a maximum speed of 8.00 • 106 m/sec are ejected froma surface in the presence of light with a frequency of 6

Andru [333]

The kinetic energy is given by:

$K=\frac{1}{2}mv^2$

We know the mass and the maximum speed, plugging their values in the expression above we have:

$\begin{gathered} K=\frac{1}{2}(9.1\times10^{-31})(8\times10^6)^2 \\ K=2.91\times10^{-17}\text{ J} \end{gathered}$

Therefore, the answer is d.

5 0

1 year ago

FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine

Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

$\tau=Fd=mg\cdot R$

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

$\tau=I\alpha$

therefore, equating this to the above equation gives

$mg\cdot R=I\alpha$

solving for alpha gives

$\alpha=\frac{mgR}{I}$

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

$\theta=\frac{1}{2}\alpha t^2$

Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

$\omega^2=0+2(1.47)(12.97)$ $w^2=38.12$ $\boxed{\omega=6.17.}$

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

3 0

1 year ago