All categories

2 years ago

Would it be true to say that in a standing wave there is no energy transmission?

Physics

1 answer:

cestrela7 [59]2 years ago

4 0

Answer: Yes, it is true.

Explanation:

The given statement is true, as in the standing wave there is no transmission of energy because the total or overall energy of the wave is basically given by the amplitude and frequency of the standing wave which is constant.

In the standing wave, the amount of the energy are equal at each point so that is why the energy is not being transferred. The standing wave are basically composed of the two travelling wave that carry energy in the opposite direction and it is also associate with the resonance.

You might be interested in

A projectile is launched horizontally from a 20-m tall edifice with a vox of 25 m/s. How long will it take for the projectile to

NISA [10]

Answer:

a) First let's analyze the vertical problem:

When the projectile is on the air, the only vertical force acting on it is the gravitational force, then the acceleration of the projectile is the gravitational acceleration, and we can write this as:

a(t) = -9.8m/s^2

To get the vertical velocity we need to integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

where v0 is the initial vertical velocity because the object is thrown horizontally, we do not have any initial vertical velocity, then v0 = 0m/s

v(t) = (-9.8m/s^2)*t

To get the vertical position equation we need to integrate over time again, to get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + p0

where p0 is the initial position, in this case is the height of the edifice, 20m

then:

p(t) = (-4.9m/s^2)*t^2+ 20m

The projectile will hit the ground when p(t) = 0m, then we need to solve:

(-4.9m/s^2)*t^2+ 20m = 0m

20m = (4.9m/s^2)*t^2

√(20m/ (4.9m/s^2)) = t = 2.02 seconds

The correct option is a.

b) The range will be the total horizontal distance traveled by the projectile, as we do not have any horizontal force, we know that the horizontal velocity is 25 m/s constant.

Now we can use the relationship:

distance = speed*time

We know that the projectile travels for 2.02 seconds, then the total distance that it travels is:

distance = 2.02s*25m/s = 50.5m

Here the correct option is a.

c) Again, the horizontal velocity never changes, is 25m/s constantly, then here the correct option is option b. 25m/s

d) Here we need to evaluate the velocity equation in t = 2.02 seconds, this is the velocity of the projectile when it hits the ground.

v(2.02s) = (-9.8m/s^2)*2.02s = -19.796 m/s

The velocity is negative because it goes down, and it matches with option d, so I suppose that the correct option here is option d (because the sign depends on how you think the problem)

4 0

2 years ago

How much time does it take a rhino running at 40 m/s to run 250 m?

lions [1.4K]

Answer:

Rhinoceroses are odd-toed ungulates native to sub-Saharan Africa and southern Asia, though all five living species have hugely contracted in range and number due to the influence of humans. Despite their titanic, tank-like bulk, rhinos can be amazingly swift: The fastest may reach at least 50 kilometers per hour (31 mph).

Explanation:

sana makatulong

8 0

3 years ago

In a jump spike, a volleyball player slams the ball from overhead and toward the opposite floor. Controlling the angle of the sp

MAVERICK [17]

Answer:

The ball would have landed 3.31m farther if the downward angle were 6.0° instead.

Explanation:

In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).

We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.

So, first we need to determine the components of the velocity of the ball, like this:

$V_{Ax}=V_{A}cos\theta$

$V_{Ax}=(21m/s)cos(-14^{o})$

$V_{Ax}=20.38 m/s$

$V_{Ay}=V_{A}sin\theta$

$V_{Ay}=(21m/s)sin(-14^{o})$

$V_{Ay}=-5.08 m/s$

we pick the positive one, so it takes 0.317s for the ball to hit on point A.

so now we can find the distance from the net to point A with this time. We can find it like this:

$x_{A}=V_{Ax}t$

$x_{A}=(20.38m/s)(0.317s)$

$x_{A}=6.46m$

Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:

$V_{Bx}=20.88 m/s$

$V_{By}=-2.195 m/s$

$y_{Bf}=y_{B0}+V_{0}t-\frac{1}{2}at^{2}$

$0=2.1m+(-2.195m/s)t-\frac{1}{2}(-9.8m/s^{2})t^{2}$

$-4.9t^{2}-2.195t+2.1=0$

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$t=\frac{-(-2.195)\pm\sqrt{(-2.195)^{2}-4(-4.9)(2.1)}}{2(-4.9)}$

t= -0.9159s or t=0.468s

we pick the positive one, so it takes 0.468s for the ball to hit on point B.

so now we can find the distance from the net to point B with this time. We can find it like this:

$x_{B}=V_{Bx}t$

$x_{B}=(20.88m/s)(0.468s)$

$x_{B}=9.77m$

So once we got the two distances we can now find the difference between them:

$x_{B}-x_{A}=9.77m-6.46m=3.31m$

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.

7 0

2 years ago

A passenger in a helicopter traveling upwards at 15 m/s accidentally drops a package out the window. If it takes 15 seconds to r

Alexeev081 [22]

Answer:

The helicopter was 1103.63 meters high when the package was dropped.

Explanation:

We consider positive speed as a downward movement

y: height (m)

t: time (s)

v₀: initial speed (m/s)

Δy = v₀t + $\frac{1}{2}$ gt²

Δy= 15 $\frac{m}{s}$ ×15 s + $\frac{1}{2}$ ×9.81 $\frac{m}{s^{2} }$ ×(15 s)²

Δy= 1103.63 m

8 0

2 years ago

What is the difference between a mineral and a rock

Sedbober [7]

A mineral is a naturally occurring, inorganic solid with a definite chemical composition and a crystalline structure formed by geological processes.

A rock is an aggregate of one or more minerals. It may also contain organic remains and mineraloids apart from regular mineral formations. Since rocks vary in their mineral and chemical composition, they are classified according to the process of their formation.

5 0

2 years ago

Read 2 more answers