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Arte-miy333 [17]
3 years ago
9

Which of the following are Electromagnetic Waves?

Physics
1 answer:
earnstyle [38]3 years ago
8 0

Answer:

b

Explanation:

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A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force cons
shepuryov [24]

Answer:

v=2.556m/s

Explanation:

From the conservation of mechanical energy

K_{E1}+U_1=K_{E2}+U_2

\frac{1}{2}m*v_1^2+\frac{1}{2}*K*x_1^2=\frac{1}{2}m*v_2^2+\frac{1}{2}*K*x_2^2

x_2=0.08m

v_1=0 m/s

Solve to velocity v2

m*v_2^2=k*x_1^2-k*x_2^2

v^2=\frac{k}{m}*(x_1^2-x_2^2)

v^2=\frac{5.5N/m}{0.24kg}*(0.54m^2-0.080^2)

v=\sqrt{6.54m^2/s^2}=2.556m/s

4 0
3 years ago
If the primary coil of a transformer has 200 turns and is supplied with 120 v ac power, how many turns must the secondary coil h
Oduvanchick [21]
The ratio of the turns to the voltage should be equal
i.e: 200/120 = t/12
so the secondary coil should have 20 turns
5 0
3 years ago
The thermal expansion of a solid is caused by:
Shkiper50 [21]

Answer:

C

Explanation:

The increase in the distance between equilibrium positions for the vibrating atoms.

6 0
3 years ago
A person travels by car from one city to an-
Mademuasel [1]
<h2>Answer:</h2>

<h2>Explanation:</h2>

First, let's refer to the distance formula:

d=v*t, where d is distance, v is velocity or speed and t is time.

Now, let's find the distance covered by each individual speed that the car had:

<h3>1. Speed 1.</h3>

In order to use the formula, we need to convert minutes into hours since the speed is given in km/h.

21.1 min/60= 0.35 h.

Now, apply the distance formula.

d=(0.35h)*(86.8km/h)= 30.38 km.

<h3>2. Speed 2.</h3>

Convert minutes to hours again and do the same calculations.

10.6min/60=0.18h

d=(0.18h)*(106km/h)= 19.08 km.

<h3>3. Speed 3.</h3>

36.5min/60= 0.61h

d=(0.61h)*(30.9km/h)= 18.85 km.

<h3>4. Obtain the total distance.</h3>

The total distance must be given by the addition of all individual distances traveled by the car on each speed:

Total distance= 30.38 km + 19.08 km + 18.85 km= 68.31 km.

5 0
1 year ago
Early LCD backlights use _______________ technology, popular for its low power use, even brightness, and long life.
Gala2k [10]

Answer:

cold cathode fluorescent lamp

Explanation:

A cold cathode fluorescent lamp (CCFL) is a lighting system that uses two phenomena: electron discharge and fluorescence.

CCFLs are mainly used as light sources for backlights, since they are compact and durable than ordinary fluorescent lamps. They exhibit a wide range of brightness and color (color temperature and chromaticity) that can be achieved by varying the pressure and type of the material injected into the glass tube. The thickness and type of the phosphor used to coat the inner wall of the tube also plays a role in altering the color and brightness.

8 0
3 years ago
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