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muminat
3 years ago
9

This part of the electromagnetic spectrum can be used to generate power. (2 points) A: gamma rays B: radio waves C: microwaves D

: infrared rays
Physics
1 answer:
ANTONII [103]3 years ago
6 0

A) gamma rays

good luck

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How much energy was released if rocket hydrogen fuel was burnt ?​
suter [353]
When ignited, the gas mixture converts to water vapor and releases energy, which sustains the reaction: 241.8 kJ of energy (LHV) for every mole of H2 burned.” A mole of hydrogen weighs 2 grams. So, this is a LHV (lower heating value) of 120.9 kJ/gram of hydrogen when heat of vaporization is subtracted.
6 0
2 years ago
Read 2 more answers
Maximum voltage produced in an AC generator completing 60 cycles in 30 sec is 250V. (a) What is period of armature? (b) How many
Vlada [557]

Answer:

a. 2 Hz b. 0.5 cycles c . 0 V

Explanation:

a. What is period of armature?

Since it takes the armature 30 seconds to complete 60 cycles, and frequency f = number of cycles/ time = 60 cycles/ 30 s = 2 cycles/ s = 2 Hz

b. How many cycles are completed in T/2 sec?

The period, T = 1/f = 1/2 Hz = 0.5 s.

So, it takes 0.5 s to complete 1 cycles. At t = T/2 = 0.5/2 = 0.25 s,

Since it takes 0.5 s to complete 1 cycle, then the number of cycles it completes in 0.25 s is 0.25/0.5 = 0.5 cycles.

c. What is the maximum emf produced when the armature completes 180° rotation?

Since the emf E = E₀sinθ and when θ = 180°, sinθ = sin180° = 0

E = E₀ × 0 = 0

E = 0

So, at 180° rotation, the maximum emf produced is 0 V.

8 0
3 years ago
a force of 50 newtons pulls a rope attached to a 150 newton sled across a horizontal surface at a constant velocity of 5 meters
lakkis [162]

Answer:

I don't know the answer I hope you find it tho good luck##

3 0
3 years ago
According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about
NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, r=4\ AU=1.496\times 10^{11}\ m

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

T^2=\dfrac{4\pi^2}{GM}r^3

M is the mass of the sun

T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3    

T^2=\sqrt{9.96\times 10^{14}}\ s

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0
3 years ago
A train is travelling towards the station on a straight track. It is a certain distance from the station when the engineer appli
AleksandrR [38]

Answer:

500 m

Explanation:

t = Time taken

u = Initial velocity = 50 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -2.5 m/s²

Equation of motion

v=u+at\\\Rightarrow 0=50-2.5\times t\\\Rightarrow \frac{-50}{-2.5}=t\\\Rightarrow t=20\ s

Time taken by the train to stop is 20 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow s=50\times 20+\frac{1}{2}\times -2.5\times 20^2\\\Rightarrow s=500\ m

∴ The engineer applied the brakes 500 m from the station

4 0
3 years ago
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