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Crazy boy [7]
3 years ago
10

Electromagnetic waves are produced when _____.

Physics
2 answers:
GuDViN [60]3 years ago
8 0

Answer: A

Explanation: when an electric charge accelerates it gains kinetic energy which is then expelled as radiation due to its vibration or acceleration

Tju [1.3M]3 years ago
6 0

Answer: The correct answer is a.

Explanation:

Electromagnetic wave is a wave in which the magnetic field and the electric field are perpendicular to each other.

Both fields are directed in the direction of the propagation of the electromagnetic wave. It travels with the speed of the light.

The motion of the electric field produces both electric field and the magnetic field.

The electromagnetic waves are produced when an electric charge vibrates or accelerates.

Therefore, the correct option is (a).

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A car goes from 0 to 26.8 m/s in 6.2 s. What is the average acceleration of the car?
Nutka1998 [239]

Answer:

0 - 60 mph = 0 - 26.8 m/s = 0 - 96.6 km/h; 0 - 100 km/h = 0 - 27.8 m/s = 0 - 62.1 mph.

Explanation:

5 0
3 years ago
Two buses are driving along parallel freeways that are 5mi apart, one heading east and the other heading west. Assuming that eac
Oksanka [162]

Answer:

101.54m/h

Explanation:

Given that the buses are 5mi apart, and that they are both driving at the same speed of 55m/h, rate of change of distance can be determined using differentiation as;

Let l be the be the distance further away at which they will meet from the current points;

l=\sqrt{13^2-5^2}=12m\\\\\frac{dl}{dt}=-(55m/h+55m/h})\\\\=-110m/h#The speed toward each other.

\frac{dh}{dt}=0, \ \ \ \ h=constant\\\\h^2+l^2=b^2\\\\2h\frac{dh}{dt}+2l\frac{dl}{dt}=2b\frac{db}{dt}\\\\2\times5\times0+2\times12\times(-110)=2\times13\frac{db}{dt}\\\\\frac{db}{dt}=-101.54m/h

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h

4 0
3 years ago
Jaime lifts a package weighing 75N. if she lifts it 1.2 m, what work has she done
9966 [12]
Answer:
90 J
Explanation:
W=fd
W=(75)(1.2)
W= 90 J
6 0
3 years ago
A wave has a wavelength of 3.0 m, a frequency of 25.0 Hz, and an amplitude of 14.0 cm. The wave travels in the positive x-direct
Mrac [35]

Answer:

The total number of oscillations made by the wave during the time of travel is 1.4 Oscillations. Strictly speaking, the number of complete oscillations is 1.

Explanation:

The required quantity is the number of complete oscillations made by the traveling wave. The amplitude time and frequency are not needed to calculate the number of oscillations as it is the ratio of the distance traveled to the wavelength( minimum distance that must be traveled to complete one oscillation) of the wave. So the total number of oscillations is 1.4 while the number of complete oscillations is 1 (strictly speaking). The detailed solution to this question can be found in the attachment below. Thank you!

4 0
3 years ago
The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters p
beks73 [17]

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

v(t)=ate^{-6t}   (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]  (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a

hence, the maximum speed is v_max = ((1/6)e^-1)a

5 0
3 years ago
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