750W 30mins...750x30x60 joules ... 75000x18 ... about 1,500,000j
D
The net force on the object is 5 N.
Explanation:
We can calculate the net force on the object by using Newton's second law of motion:
where
F is the net force on an object
m is the mass of the object
a is its acceleration
For the object in this problem, we have
m = 1 kg is its mass
is its acceleration
Substituting into the equation, we find the net force:
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The cost of developing thermonuclear power with plasmabe defended because D. It can provide an inexpensive power source.
<h3>How did the
cost of developing t
hermonuclear power defended?</h3>
The cost of developing thermonuclear power defended becvause we can see in the paragraph how it was told that the generation of ths power can be donee through the understanding of the occurrence of plasmain nature,
It should be noted that this thermonuclear power with plasmabe posses the characteristics which make it to exist in the ionosphere, and it can be felt in the flames as well; as in the chemical and nuclearexplosions.
In conclusion the power can be seen as an inexpensive source power because the p[roduction of this power cn be found in most of the thing that can be found around us as discused above.
Therefore, option D is correct.
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I’ll refer to electromagnetic radiation as EMR.
Visible light is a very small subset of EMR. Many other ranges like infrared, ultraviolet, or gamma must be detected by special equipment.
EMR is what makes up light, and as we know from any high school physics class, light exhibits both particle-like properties (photoelectric effect and Compton scattering) and wave-like properties (refraction, diffraction, double-slit & single-slit experiment).
EMR can travel without a medium, like the vast emptiness of space. It can also travel with a medium. It can transmit through various materials albeit at a slower speed, like water, earth’s atmosphere, glass etc.
The propagation speed of EMR in space is 3x10^8 m/s, which is a speed unattainable by any of our current means of transportation. I would say that’s quite fast.
Answer:
k = 1073.09 N/m
A = 0.05 m
Explanation:
Given:
- Time period T = 0.147 s
- maximum speed V_max = 2 m/s
- mass of the block m = 0.67 kg
Find:
- The spring constant k
- The amplitude of the motion A.
Solution:
- A general simple harmonic motion is modeled by:
x (t) = A*sin(w*t)
- The velocity of the above modeled SHM is:
v = dx / dt
v(t) = A*w*cos(w*t)
- Where A is the amplitude in meters, w is the angular speed rad/s and time t is in seconds.
- We can see that maximum velocity occurs when (cos(w*t)) maximizes i.e it is equal to 1 or -1. Hence,
- V_max = A*w
- Where w is related to mass of the object and spring constant k as follows,
w = sqrt ( k / m )
- The relationship between w angular speed and Time period T is:
w = 2*pi / T
- Equating the above two equations we have,
m*(2*pi / T)^2 = k
- Hence, k = 0.67*(2*pi / 0.157)^2
k = 1073.09 N / m
- So, amplitude A is:
A = V_max*sqrt ( m / k )
A = 2*sqrt ( 0.67 / 1073.09 )
A = 0.05 m