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Murrr4er [49]
3 years ago
7

How do reflection and refraction of waves affect communication?​

Physics
2 answers:
romanna [79]3 years ago
5 0

Answer:

Reflection involves a change in direction of waves when they bounce off a barrier. Refraction of waves involves a change in the direction of waves as they pass from one medium to another.

defon3 years ago
5 0

refraction is the change in direction of a wave passing from one medium to another

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If a force of 19 N is applied to a crate to push it 23 m, how much work is done on the crate by the force?
swat32

Answer:

437 Joules

Explanation:

Use the formula for work directly

(work) = (force) x (displacement)

to get

(work) = (19 N) x (23 m) = 437 Joules

4 0
3 years ago
How do iquantum tunel through a wall and how long will it take me?
Anon25 [30]

Answer:

The phenomenon known as "tunneling" is one of the best-known predictions of quantum physics, because it so dramatically confounds our classical intuition for how objects ought to behave. If you create a narrow region of space that a particle would have to have a relatively high energy to enter, classical reasoning tells us that low-energy particles heading toward that region should reflect off the boundary with 100% probability. Instead, there is a tiny chance of finding those particles on the far side of the region, with no loss of energy. It's as if they simply evaded the "barrier" region by making a "tunnel" through it.

Explanation:

5 0
3 years ago
Describe the difference between potencial and kinetic energy
avanturin [10]
Potential energy is energy stored in an object. kinetic energy is energy of motion
8 0
3 years ago
An airplane starts from A and flies to B at a constant speed. After reaching B it returns to A at the same speed. There was no w
Dafna1 [17]

Answer:

When there is wind it takes longer

Explanation:

With no wind, the round trip time is

t_1=\frac{d}{v}+\frac{d}{v}=\frac{2d}{v}

When we have a constant wind speed w

t_2=\frac{d}{v-w} +\frac{d}{v+w} =\frac{2vd}{v^{2} -w^{2}}

comparing the reciprocal times;

\frac{1}{t_2}=\frac{v^{2}-w^{2} }{2vd}=\frac{v}{2d}-\frac{w^{2}}{2vd}   \leq \frac{v}{2d}=\frac{1}{t_1}

This means that t1 is smaller than t2, ergo, it takes longer with wind

4 0
3 years ago
A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa
Nataly_w [17]

Answer:

V_p = 267.258 m/s

V_n = 38.375 m/s      

Explanation:

using the law of the conservation of the linear momentum:

P_i = P_f

where P_i is the inicial momemtum and P_f is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

P_i = m(270 m/s)

P_f = mV_P + M_nV_n

where m is the mass of the proton and V_p is the velocity of the proton after the collision, M_n is the mass of the nucleus and V_n is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = mV_p + 14mV_n

then, m is cancelated and we have:

270 = V_p + 14V_n

This is a elastic collision, so the kinetic energy K is conservated. Then:

K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2

and

Kf = \frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2

then,

\frac{1}{2}m(270)^2 =  \frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2

here we can cancel the m and get:

\frac{1}{2}(270)^2 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2

now, we have two equations and two incognites:

270 = V_p + 14V_n  (eq. 1)

\frac{1}{2}(270)^2 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2

in the second equation, we have:

36450 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2  (eq. 2)

from this last equation we solve for V_n as:

V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }

and replace in the other equation as:

270 = V_p + 14\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }

so,

V_p = -267.258 m/s

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

V_n = \frac{270+267.258}{14}

V_n = 38.375 m/s  

3 0
3 years ago
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