What type of electromagnetic wave can be used to transmit information?

a 2.35 water bucket is swung in a full cirlce of radius 0.824 m just fast enough so that the water doesn't fall out the top mean

tiny-mole [99]

Answer:

2.84 m/s

Explanation:

At the top position of the circular trajectory, the normal reaction is zero:

N = 0

So it means that the only force that is providing the centripetal force is the gravitational force (the weight of the bucket). Therefore we have:

$mg = m \frac{v^2}{r}$

where

m is the mass of the water bucket

g = 9.8 m/s^2 is the acceleration of gravity

v is the speed of the bucket

r = 0.824 m is the radius of the circle

Solving for v,

$v=\sqrt{gr}=\sqrt{(9.8 m/s^2)(0.824 m)}=2.84 m/s$

4 0

2 years ago

A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision

Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is $a=2,800m/s^{2}$

Explanation:

<u>Known Data</u>

We will asume initial speed has a negative direction, $v_{i}=-30m/s$ , final speed has a positive direction, $v_{f}=26m/s$ , $\Delta t=20ms=0.020s$ and mass $m_{b}$ .

<u>Initial momentum</u>

$p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s$

<u>final momentum</u>

$p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s$

<u>Impulse</u>

$I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s$

<u>Average Force</u>

$F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}$

<u>Average acceleration</u>

$F=ma$ , so $a=\frac{F}{m_{b}}$ .

Therefore, $a=\frac{2800m_{b} \ m/s^{2}}{m_{b}} =2800m/s^{2}$

8 0

3 years ago

A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound

NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

$f_n=\frac{nv_s}{4L}$

n: number of the harmonic = 3

vs: speed of sound = 340 m/s

L: length of the pipe = 2.5 m

You replace the values of n, L and vs in order to calculate the frequency:

$f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz$

hence, the frequency of the third overtone is 102 Hz

8 0

3 years ago

Find the speed of light in each of the following materials. (a) gallium phosphide m/s (b) carbon disulfide m/s (c) benzene

Oksanka [162]

Explanation:

We need to calculate the speed of light in each materials

(I). Gallium phosphide,

The index of refraction of Gallium phosphide is 3.50

Using formula of speed of light

$v=\dfrac{c}{\mu}$ ....(I)

Where, $\mu$ = index of refraction

c = speed of light

Put the value into the formula

$v=\dfrac{3\times10^{8}}{3.50}$

$v=8.6\times10^{7}\ m/s$

(II) Carbon disulfide,

The index of refraction of Gallium phosphide is 1.63

Put the value in the equation (I)

$v=\dfrac{3\times10^{8}}{1.63}$

$v=1.8\times10^{8}\ m/s$

(III). Benzene,

The index of refraction of Gallium phosphide is 1.50

Put the value in the equation (I)

$v=\dfrac{3\times10^{8}}{1.50}$

$v=2\times10^{8}\ m/s$

Hence, This is the required solution.

7 0

3 years ago

What is the potential energy of a 2 kg ball 15 m in the air?

oee [108]

Answer:

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 2 × 10 × 15

We have the final answer as

Hope this helps you

7 0

2 years ago