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3 years ago

A 15,000-watt radio station transmits at 1700 kHz.

Physics

2 answers:

viva [34]3 years ago

6 0

the answer is a

i had this on a quiz

notsponge [240]3 years ago

3 0

$E_{\gamma}=hf= (6.62\times10^-^3^4 Js)(1700\times10^3Hz)=1.12\times10^-^2^7J/photon$
Answer A

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A proton traveling at 17.6° with respect to the direction of a magnetic field of strength 3.28 mT experiences a magnetic force o

umka2103 [35]

Answer:

a) The proton's speed is 5.75x10⁵ m/s.

b) The kinetic energy of the proton is 1723 eV.

Explanation:

a) The proton's speed can be calculated with the Lorentz force equation:

$F = qv \times B = qvBsin(\theta)$ (1)

Where:

F: is the force = 9.14x10⁻¹⁷ N

q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C

v: is the proton's speed =?

B: is the magnetic field = 3.28 mT

θ: is the angle between the proton's speed and the magnetic field = 17.6°

By solving equation (1) for v we have:

$v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s$

Hence, the proton's speed is 5.75x10⁵ m/s.

b) Its kinetic energy (K) is given by:

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

Where:

m: is the mass of the proton = 1.67x10⁻²⁷ kg

$K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV$

Therefore, the kinetic energy of the proton is 1723 eV.

I hope it helps you!

3 0

3 years ago

Which changes in an electric motor will make the motor stronger? select 3 options. using a stronger permanent magnet using a wea

Masja [62]

1. Using Strong Permanent. 2. increasing the current. 3. Decreasing the space between Magnets

Explanation:

Brainiest

4 0

2 years ago

When a court has "___ ___" they can hear a case once a lower court has ruled on it.

notka56 [123]

Your answer is going to be Appellate jurisdiction.

3 0

3 years ago

A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.

SVEN [57.7K]

Answer:

a) Knowns, initial speed $v_{i}=13.0 m/s$ , final speed $v_{f}=0 m/s$ and gravity due it is a constant $g=9.8m/s^{2}$

b) The maximum high reached by the dolphin is $y_{max}=8.62 m$

c) Total time is $t=2.65s$

Explanation:

a) First of all the initial speed is given at the start of the problem, gravity is constant and final speed is known any object thrown straight up reaches its max high at $0m/s$ speed.

b) Second, now that we know final speed we use $v_{f} =v_{i}-gt$ , as we clear for $t=\frac{v_{i}-v_{f} }{g}=\frac{20.0m/s}{9.8m/s^{2} }=1.32 s$ .

Then we use $y=v_{i}t-\frac{1}{2} gt^{2}=(20.0m/s)(1.32s)-\frac{1}{2} (9.8m/s^{2} ) (1.32s)^{2} =8.62m$

c)Third, finally we can use $y=v_{i}t-\frac{1}{2} gt^{2}$ , as we know $y=0m$ when the dolphin fall into the water again and $v_{i} =13.0m/s$ , then we have $0=(13m/s)t-\frac{1}{2} (9.8m/s^{2} )t^{2}$ is a quadratic form $0=t(13.0-4.9t)$ so we have $t_{i}=0s$ and $t_{f}=\frac{13}{4.90} =2.65s$