Please help me with this question

How does a television signal differ from a standard radio wave

Llana [10]

Signal of television as radio waves are of same origin: electromagnetic waves
The difference between them is only in the frequency in which they are issued.

8 0

4 years ago

Read 2 more answers

A child drops a ball. The ball hits the ground and bounces. The graph to the left shows the velocity-time graph for the ball fro

WINSTONCH [101]

Answer:

Consider the velocity-time graph attached below.

The velocity-time graph represents the acceleration of a body under a force.

We can see that is the graph that if a child release the ball above the ground at A, it hits the ground at B. Bounces back with a reaches the top again at C, and hits the ground again at D.

The slope of velocity time graph represents acceleration. From A to B, velocity in increasing constantly with respect to time, which means constant acceleration from A to B. AS velocity increase, momentum of the ball also increases, which results in the increase of Kinetic energy.

At B, the ball hits the ground, the velocity decreases, momentum decrease s, because kinetic energy is transferred from the ball to the ground, due to which the ball would not attain the same height after the bounce.

Then the velocity remains negative at C, which means that now the ball is moving in opposite direction till C. It reaches its new at height at C, which is not the same as that of A because of lost in Kinetic Energy, and fall again.

4 0

3 years ago

A horizontal disk with a radius of 10 cm rotates about a vertical axis through its center. The disk starts from rest at t = 0 an

Tom [10]

Answer:

0.69s

Explanation:

10 cm = 0.1 m

Let t be the time that radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude. At that time we have the angular velocity would be

$\omega = \alpha t = 2.1 t$

And so the radial acceleration is

$a_r = \omega^2 r = (2.1t)^2 r = 2.1^2 t^2 * 0.1= 0.441 t^2 m/s^2$

The tangential acceleration is always the same since angular acceleration is constant:

$a_t = \alpha * r = 2.1 * 0.1 = 0.21 m/s^2$

For these 2 quantities to be the same

$a_r = a_t$

$0.441 t^2 = 0.21$

$t^2 = 0.21/0.441 = 0.4762$

$t = \sqrt{0.4762} = 0.69 s$

6 0

4 years ago

The speed of a certain proton is 350 km/s. If the uncertainty in its momentum is 0.100%, what is the necessary uncertainty in it

Evgesh-ka [11]

Answer:

$\Delta x = 1.807 \times 10^{-10}m$

Explanation:

mass of proton, m = 1.67 x 10^-27 kg

speed of proton, v = 350 km/s = 350,000 m/s

Momentum of proton, p = mass x speed

p = 1.67 x 10^-27 x 350000 = 5.845 x 10^-22 kg m /s

uncertainty in momentum, Δp = 0.1 % of p

Δp = $\frac{0.1\times 5.845 \times 10^{-22}}{100}=5.845 \times 10^{-25}$

According to the principle

$\Delta x\times \Delta p \geq \frac{h}{2\pi }$

where, Δx be the uncertainty in position

$\Delta x\times 5.845 \times 10^{-25}= \frac{6.634 \times 10^{-34}}{2\times 3.14}$

$\Delta x = 1.807 \times 10^{-10}m$

5 0

3 years ago

I need help with question two

zaharov [31]

I can't see numbers here, so here are all the answers:
1) the frequency is c/λ = 3e8/556e-9 = 5.39e14Hz
2) light travels at <span>299,792,458 m/s. So in nanoseconds it's 0.299792458m. This is about 1/3 of a meter which is about one foot.
3) length is L = ct = (</span>299,792,458 m/s)(6e-15) = 1.799e-6m or 1.799μm

6 0

3 years ago