PROJECTILE MOTION FOR TWO Rocks-VELOCITY-TIME GRAPHS

stealth61 [152]

Answer:

If not because you know you just can't

3 0

2 years ago

In the Hydrogen atom, the energy spacing between the is 4.07 x 101 J (Joules). When an is the frequency of the photons emitted?

agasfer [191]

Answer:

The frequency of the photon is $3.069\times10^{14}\ Hz$ .

Explanation:

Given that,

Energy $E=4.07\times10^{-19}\ J$

We need to calculate the energy

Using relation of energy

$E_{4}-E_{2}=\Delta E$

Where, $\Delta E$ = energy spacing

$4h\nu-2h\nu=4.07\times10^{-19}$

$\nu=\dfrac{4.07\times10^{-19}}{2h}$

Put the value of h into the formula

$\nu=\dfrac{4.07\times10^{-19}}{2\times6.63\times10^{-34}}$

$\nu=3.069\times10^{14}\ Hz$

Hence, The frequency of the photon is $3.069\times10^{14}\ Hz$ .

4 0

2 years ago

Suppose the original segment of wire is stretched to 10 times its original length. How much charge must be added to the wire to

Debora [2.8K]

Here we want to study how the linear charge density changes as we change the measures of our body.

We will find that we need to add 9*Q of charge to keep the linear charge density unchanged.

I will take two assumptions:

The charge is homogeneous, so the density is constant all along the wire.

As we work with a linear charge density we work in one dimension, so the wire "has no radius"

Originally, the wire has a charge Q and a length L.

The linear charge density will be given by:

λ = Q/L

Now the length of the wire is stretched to 10 times the original length, so we have:

L' = 10*L

We want to find the value of Q' such that λ' (the linear density of the stretched wire) is still equal to λ.

Then we will have:

λ' = Q'/L' = Q'/(10*L) = λ = Q/L

Q'/(10*L) = Q/L

Q'/10 = Q

Q' = 10*Q

So the new charge must be 10 times the original charge, this means that we need to add 9*Q of charge to keep the linear charge density unchanged.

If you want to learn more, you can read:

brainly.com/question/14514975

6 0

2 years ago

Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour in 2.22 s

muminat

Answer:

Acceleration = 10.06 m/s²

Explanation:

1 mile = 1.6093km

1609.3m = 1 mile

1 m = $\frac{1}{1609}$ mile

50.0 miles/hour = $\frac{50 * 1609.3}{60 * 60}$ m/s

= 22.35m/s

from equation

S = Ut + 1/2 at²

v = U + at

22.35 = 0 + a * 2.22

a = 22.35 ÷ 2.22

= 10.06 m/s²

4 0

3 years ago

(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is . Find the time i

a_sh-v [17]

Answer:

a) It took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves.

b) The minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds.

Explanation:

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensitie. That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

The distribution of the radiation in the electromagnetic spectrum can also be given in wavelengths, but it is more frequent to work with it at frequencies:

Gamma rays

X-rays

Ultraviolet rays

Visible region

Infrared

Microwave

Radio waves.

Any radiation that belongs to electromagnetic spectrum has a speed in vacuum of $3x10^{8}m/s$ .

a) Find the time it took for his voice to reach the Earth via radio waves.

To know the time that took for Neil Armstrong's voice to reach the Earth via radio waves, the following equation can be used:

$c = \frac{d}{t}$ (1)

Where v is the speed of light, d is the distance and t is the time.

Notice that t can be isolated from equation 1.

$t = \frac{d}{c}$ (2)

The distance from the Earth to the Moon is $3.85x10^{8} m$ , therefore.

$t = \frac{3.85x10^{8} m}{3x10^{8}m/s}$

$t = 1.28s$

Hence, it took 1.28 seconds to Neil Armstrong's voice to reach the Earth via radio waves.

b) Determine the minimum time that will be required for a message from Mars to reach the Earth via radio waves.

The distance from the Earth to the Mars at its closest approach is $5.76x10^{10}m$ , therefore.

$t = \frac{5.76x10^{10}m}{3x10^{8}m/s}$

$t = 192s$

Hence, the minimum time that will be required for a message from Mars to reach the Earth via radio waves is 192 seconds.

3 0

3 years ago