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3 years ago

The medium of a wave is _____.

1 answer:

7 0

B. A medium of the earth’s atmosphere is why we can hear sound. In space, you wouldn’t have an atmosphere or or any kind of medium for sound waves to travel through, which is why you wouldn’t be able to hear sound in space

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The electric field strength is 5.50×10^4 N/C inside a parallel-plate capacitor with a 2.50 mm spacing. A proton is released from

valentina_108 [34]

Answer:

$v=1.6\times10^{5}m/s$

Explanation:

The equation that relates electric field strength and force that a charge experiments by it is F=qE.

Newton's 2nd Law states F=ma, which for our case will mean:

$a=\frac{F}{m}=\frac{qE}{m}$

We know from accelerated motion that $v^2=v_0^2+2ad$

In our case the proton is released from rest, so $v_0=0m/s$ and we get $v=\sqrt{2ad}$

Substituting, we get our final velocity equation:

$v=\sqrt{\frac{2qEd}{m}}$

For the proton we know that $q=1.6\times10^{-19}C$ and $m=1.67\times10^{-27}kg$ . Writing in S.I. d=0.0025m, we obtain:

$v=\sqrt{\frac{2(1.6\times10^{-19}C)(5.5\times10^{4}N/C)(0.0025m)}{(1.67\times10^{-27}kg)}}=162318.5m/s=1.6\times10^{5}m/s$

6 0

3 years ago

CHEGG 42 mT magnetic field points due west. If a proton of kinetic energy 9 x 10-12 J enters this field in an upward direction,

alexdok [17]

Answer:

The magnitude of the Force is $F = 697 *10^{-15}N$ and the direction is South

Explanation:

From the question we are told that

The magnetic field point due west and since East point toward the positive x -axis $(i)$ then this magnetic field would be mathematically represented as

$\= B = 42(-i)mT = 42*10^{-3} (-i) T$

Now from the question we are told that the kinetic energy is

$KE = 9*10^{-12}J$

Now this kinetic energy can be mathematically represented as

$KE = \frac{1}{2}mv^2$

Where m is the mass of proton which has a general value of

$m = 1.67*10^{-27}kg$

Now making the subject of the formula

$v = \sqrt{\frac{KE}{0.5 * m} }$

Substituting values we have

$v = \sqrt{\frac{9*10^{-12}}{0.5 * 1.67*10^{-27}} }$

$= 10.37*10^7m/s$

Now from the question we are told that proton is moving upward which is in the positive z direction so the velocity of the proton would be in the positive

So the velocity would be

$\= v = 10.37*10^{7} \r k \ m/s$

Now the magnetic Force can be mathematically represented as

$\= F = q \= v * \= B$

Where q is the charge on the proton which has a general value of $q =1.6*10^{-19}C$

Now substituting the value

$\= F = 1.6*10^{-19 } * (10.37 *10^7) \r k * (42 *10^{-3})(-i)$

$= 697*10^{-15} J$

Now according to Fleming's left hand rule the direction of the magnetic force is south toward the negative Y - direction $(-j)$

So the force can be denoted as

$\= F = 697*10^{-15}(-j) N$

6 0

3 years ago

At time t=0 a grinding wheel has an angular velocity of 30.0 rad/s . It has a constant angular acceleration of 35.0 rad/s2 until

Rama09 [41]

Answer:

(A) 570 rad

(B) 10 s

Explanation:

The equations of motion for circular motions are used.

Initial angular velocity, $\omega_0 = 30.0 \text{ rad/s}$

Angular acceleration, $\alpha =35.0 \text{ rad/s}^2$

(A)

At t = 2.00 s, the angular displacement, θ, is given by

$\theta = \omega_0t+\frac{1}{2}\alpha t^2 = (30\times 2) + \frac{1}{2}\times35\times2^2=60+70 = 130\text{ rad}$

After this time, it decelerates through an angular displacement of 440 rad.

Total angular displacement = 130 + 440 rad = 570 rad

(B)

At the time the circuit breaker tips, the angular velocity is given by

$\omega = \omega_0+\alpha t = 30.0+(35.0\times 2) = 30.0+70.0 =100.0\ \text{rad/s}$

This becomes the initial angular velocity for the decelerating motion. Because it stops, the final angular velocity is 0 rad/s. The time for this part of the motion is calculated thus:

$\theta_2 = \left(\dfrac{\omega_i+\omega_f}{2}\right)t$