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Lapatulllka [165]
3 years ago
5

Cathode-ray tubes produce images on the principle of induced emf. true or false?

Physics
1 answer:
yan [13]3 years ago
3 0
The appropriate response is false. Cathode-beam ray does not deliver pictures on the guideline of instigated emf. The cathode-ray is a high-vacuum tube in which cathode beams deliver an iridescent picture on a fluorescent screen, utilized mostly in TVs and workstations.
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¿ qué es un motor de explosión?
ArbitrLikvidat [17]

Answer:

es un motor de combustión interna con encendido por chispa.

5 0
3 years ago
In order to transmit information via radio waves, the waves need to be changed somehow. For car radios this can happen in two wa
laila [671]

Answer: amplitude

Explanation: This describes the maximum amount of the displacement of a particle from it rest position. Usually, it is measured in metres

Since we are considering AM which is amplitude modulation, a technique used in electronic communication, most commonly for broadcasting information through a radio carrier wave. In amplitude modulation, the amplitude (signal strength) of the carrier wave is diversified in proportion to that of the message signal being broadcasted.

7 0
3 years ago
Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is
Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of c_p is determined via the expression:

c_p = \frac{KR}{K_1}

where ;

R = universal gas constant = \frac{8.314 \ J}{28.97 \ kg.K}

k = constant = 1.4

c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}

c_p= 1.004 \ kJ/kg.K

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}     ------ equation(1)

we can rewrite the above equation as :

0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}

T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}

where:

T_1  = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K

T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}

T_2 = 402.36 \ K

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation:\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}

P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}

P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}

P_2 = 17.79 \ bar

Therefore, the exit pressure is 17.79 bar

7 0
3 years ago
Describe three possible careers in physical science
Amanda [17]

Answer:

physical sicence boilagiy sxcience and earth science

Explanation:

6 0
3 years ago
You lift a 50 N object 2 meters off the ground what work did you do on the object
mixer [17]
Work equals force × displacement (distance between initial point and end point is displacement)
if u follow this it becomes
work = 50 × 2 which is equal to 100

comment if u have more questions
4 0
3 years ago
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