The work done onto the car is 506,250 J
The work done on a system implies an increase in the internal energy of the system as a result of some forces acting on the system from the outside.
From the parameters given:
- The mass of the car = 1500 kg
- The initial speed = 30 m/s
- The final speed = 15 m/s
The work done onto the car refers to the change in the kinetic energy (i.e. ΔK.E)
= 506,250 J
Therefore, we can conclude that the work done on the car is 506,250 J
Learn more about work done here:
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A car acting as an object in front of a biconvex lens between F and 2 F on the object side of the lens. There is a light ray parallel to the principal axis that is bent through F on the image side of the lens. There is a ray straight through the center of the lens. The rays intersect below the x axis further than 2 F away from the lens and farther from the principal axis than the object is tall.
<u> The image produced by the lens is (b) inverted and real</u>
Explanation:
A real image occurs where the rays converge.
Real images can be produced both by the concave mirrors or converging lenses, but the condition is that the object of consideration is always placed far away from the mirror or the lens than the focal point, and thus the real image produced is inverted.
A car acting as an object in front of a biconvex lens between F and 2 F on the object side of the lens. There is a light ray parallel to the principal axis that is bent through F on the image side of the lens. There is a ray straight through the center of the lens. The rays intersect below the x axis further than 2 F away from the lens and farther from the principal axis than the object is tall.
<u> The image produced by the lens is (b) inverted and real</u>
Answer:
Circular wave
Explanation:
Circular waves are special types of mechanical waves. They all travel through a material medium or some times a vacuum.
An example of such wave is a ripple caused by dropping a stone in a tank of water.
A wave that propagates in circular form on the surface of water falls into this category.
Answer:
4.8 m/s
Explanation:
When she catches the train,
- They will have travelled the same distance.and
- Their speeds will be equal
The formula for the distance covered by the train is
d = ½at² = ½ × 0.40t² = 0.20t²
The passenger starts running at a constant speed 6 s later, so her formula is
d = v(t - 6.0)
The passenger and the train will have covered the same distance when she has caught it, so
(1) 0.20t² = v(t - 6.0)
The speed of the train is
v = at = 0.40t
The speed of the passenger is v.
(2) 0.40t = v
Substitute (2) into (1)
0.20t² = 0.40t(t - 6.0) = 0.40t² - 2.4 t
Subtract 0.20t² from each side
0.20t² - 2.4t = 0
Factor the quadratic
t(0.20t - 2.4) = 0
Apply the zero-product rule
t =0 0.20t - 2.4 = 0
0.20t = 2.4
(3) t = 12
We reject t = 0 s.
Substitute (3) into (2)
0.40 × 12 = v
v = 4.8 m/s
The slowest constant speed at which she can run and catch the train is 4.8 m/s.
A plot of distance vs time shows that she will catch the train 6 s after starting. Both she and the train will have travelled 28.8 m. Her average speed is 28.8 m/6 s = 4.8 m/s.
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