What happens when a sound wave passes from water to air?
2 answers:
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Answer:
The friction force is 250 N
Explanation:
The desk is moving at constant velocity. This means that its acceleration is zero: a = 0. Newton's second law states that the resultant of the forces acting on the desk is equal to the product between mass (m) and acceleration (a):
In this case, we know that the acceleration is zero: a = 0, so also the resultant of the forces must be zero:
(1)
We are only interested in the forces acting along the horizontal direction, since it is the direction of motion. There are two forces acting in this direction:
- the pull, forward, F = 250 N
- the friction force, backward,
Given (1), we have
So the force of friction must be equal to the pull:
Answer:
Explanation:
Considering that this is parabolic motion, we know that the time the ball is in the air begins the instant it leaves the ground, reaches up to its max height, and then begins falling until it reaches the ground. Duh, right? Some important things happen during this trip. There are a few things we need to know in order to even begin the problem. Parabolic motion has x and y coordinates because it is 2-dimmensional; the acceleration in the x dimension is not the same as the acceleration in the y dimension; the velocity of an object at its max height is always 0; the time it takes to reach its max height (where the max height is half the distance the object travels) is half the time it takes to make the whole trip. Yikes. That's a lot to know and much to remember! Don't you just LOVE physics!?
For a. the hang time is the time the ball was in the air. Some of that stuff we talked about above is pertinent to solving this problem. We know that the velocity of the ball is 0 at its max height, and we also know that if we find the time it takes to reach its max height, we can double that number to find how long it was in the air for the whole trip. Use the one-dimensional equation
to find out how long it took to reach the max height. Even though we don't yet know the max height, we DO know that the velocity at that point is 0. BUT before we do that, since we are working in the y-dimension only, it would behoove us (benefit us) to find the velocity particular to this dimension. We are going to answer c. first, then backtrack.
c. wants the initial vertical velocity. That is found in the magnitude of the "blanket" or generic velocity times the sin of the angle, namely:
so
18 m/s Now we can use that as the initial upwards velocity in part a:
and filling in:
0 = 18 + (-9.8)t and
-18 = -9.8t so
t = 1.8 seconds. But remember, this is only half the time it was in the air. The whole trip, then, takes 2(1.8) which is
t = 3.6 seconds
That's a and c. Now for b:
b. asks for the x component of the velocity:
which works out to be the same as the vertical velocity, since the sin and cos of 45 degrees is the same:
and
18 m/s
Onto d:
d. wants the max height. Remember, it took 1.8 seconds to get to the max height, so using yet another one-dimensional equation:
Δx = v₀t + where Δx is the displacement, v₀ is the initial upwards velocity, a is the pull of gravity, and t is the time it takes to reach that max height (Δx, our unknown). Filling in:
Δx = and if you do the rounding correctly, you'll end up with this:
Δx = 32 - 16 so
the max height, Δx, is 16 meters.
e. wants the range. That translates to the distance the ball traveled. This is found in a glorified version of d = rt, where d is displacement, r is velocity, and t is...well, time (that doesn't change):
Δx = vt so
Δx = 18(3.6) remember that the ball was in the air for a total of 3.6 seconds, so
Δx = 65 meters.
Phew!!!!! That's a lot! I suggest you learn your physics or this will make you insane by the end of the course!
Answer:
Explanation:
First of all we shall calculate electric field near charge 2Q .
electric field due to charge Q = K x Q / (5 x 10⁻² )²
E₁ = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis
E₁ = KQ x 10⁴ i / 25
Similarly electric field due to charge 3Q near 2Q
= 3KQ x 10⁴ i / 25 . It is acting along y-axis
E₂ = 3KQ x 10⁴ j / 25
Similarly electric field due to charge 4Q near 2Q
= 4KQ x 10⁴ j / (25 x 2 )
= 2 KQ x 10⁴ / 25 . It is acting acting along north east direction
unit vector in north east direction = ( i + j )/ √2
So E₃ can be represented by
E₃ = 2 KQ x 10⁴ ( i + j ) / 25 x √2
Total field = KQ x 10⁴ i / 25 + 3KQ x 10⁴ j / 25 + 2 KQ x 10⁴ ( i + j ) / ( 25 x √2 )
= KQ x 10⁴ [ i + 3 j + √2 i + √2 j ) / 25
= 400 KQ ( 2.414 i + 4.414 j ) N / C
Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j ) x 2Q N
= 800 KQ² ( 2.414 i + 4.414 j ) N
= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N
= 108.63 ( i + 2 j ) N .
Magnitude of this force
= 108.63 x √5
= 243 N approx .
