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3 years ago

Wave energy can only be transmitted through a material medium. wave energy can only be transmitted through a material medium.

1 answer:

3 0

The correct answer is False.
In fact, electromagnetic waves (which carry energy), do not need a material medium to travel, because they can travel in vacuum as well. So, wave energy can be transmitted also through vacuum.

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A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's

blondinia [14]

Answer:

a) $z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

b) $v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

Explanation:

The particle position is given by:

$z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0$

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

$z'(t) =v(t) = -13t$

Now we can replace the velocity for t=1.75 s

$v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}$

For t = 3.0 s we have:

$v(3.0s) = -13*3.0 =-39 \frac{m}{s}$

Part b

For this case we can find the average velocity with the following formula:

$v_{avg}= \frac{z_f - z_i}{t_f -t_i}$

And we can find the positions for the two times required like this:

$z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m$

$z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m$

And now we can replace and we got:

$V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}$

8 0

3 years ago

What is the mass of 294N weight on Earth

STALIN [3.7K]

Answer:

m=30kg

Explanation:

F=mg

294=m(9.8)

m=294/9.8

m=30kg

Hope that helps :)

7 0

4 years ago

The maximum speed with which you can throw a stone is about 25 m/s. Can you hit a window 50 m away and 13 m up from the point wh

Nookie1986 [14]

Answer:

no, the maximum height = 12.3m

Explanation:

The equation for the trajectory of the ball for a given angle can be derived from:

(1) $x=cos\phi v_0t$

and

(2) $y=-\frac{1}{2}gt^2 +sin\phi v_0t$

Combining equation 1 and 2:

(3) $y=xtan\phi -\frac{gx^2}{2cos^2\phi v_0^2}$

Taking the derivative:

(4) $\frac{dy}{d\phi}=\frac{xsec^2\phi}{v_0^2}(v_0^2-gxtan\phi)$

Setting equation 4 equal to zero to calculate the maximum angle:

(5) $\phi=tan^{-1}(\frac{v_0^2}{gx})$

Plugging equation 5 back into equation2 confirms, the trajectory can't reach the target. The maximum height is 12.3m.

3 0

3 years ago

Read 2 more answers

Which objects can be electrically polarized?

Vera_Pavlovna [14]

Answer:

the answer is C: neutral objects

4 0

3 years ago

Calculate the time in which a tuning fork of frequency 234 hz completes 26 vibrations. (Plz write formula also)

Leokris [45]

Frequency is inversely promotional to time: f=1/t So, 234=26/t 234/26=t
the answer 9seconds

4 0

4 years ago