The answer to this is <span>KE = 8100 J</span>
Gradpoint? The answer is B
Answer:
i) Telescopes can be used to view far distant objects but the human eye can't view far distant objects.
ii) Telescopes uses two convex lenses producing a magnified image while the human eye only possesses one convex lens (image seen are smaller than that viewed under telescopes)
Explanation:
The telescopes can be used to view far distant objects due to their presence of two convex lenses. The two convex lenses are the objective lens (lens closer to object) and the eye piece lens (lens closer to eye). The object to be viewed forms an intermediate image first before the final image is seen using the eye piece lens.
The human eye only possess one convex lens and as such cannot view far ranged objects.
Answer:
V = 0.32 m /s
Explanation:
Momentum of the man on the sledge along with sledge
= (80+8) x 10 = 880 kg. m/s When the man jumps off the sledge , the velocity of the sledge will remain intact at 10 m/s
When the sledge hits the boulder and bounces back the momentum of the sledge - boulder system will remain conserved .
change in momentum of sledge = 8 x (10 +6 )
= 128 kg m/s
Change in the momentum of boulder
= 400 V -0
= 400V
400V = 128
V = 0.32 m /s
For rectilinear motions, derived formulas all based on Newton's laws of motion are formulated. The equation for acceleration is
a = (v2-v1)/t, where v2 and v1 is the final and initial velocity of the rocket. We know that at the end of 1.41 s, the rocket comes to a stop. So, v2=0. Then, we can determine v1.
-52.7 = (0-v1)/1.41
v1 = 74.31 m/s
We can use v1 for the formula of the maximum height attained by an object thrown upwards:
Hmax = v1^2/2g = (74.31^2)/(2*9.81) = 281.42 m
The maximum height attained by the model rocket is 281.42 m.
For the amount of time for the whole flight of the model rocket, there are 3 sections to this: time at constant acceleration, time when it lost fuel and reached its maximum height and the time for the free fall.
Time at constant acceleration is given to be 1.41 s. Time when it lost fuel covers the difference of the maximum height and the distance travelled at constant acceleration.
2ax=v2^2-v1^2
2(-52.7)(x) = 0^2-74.31^2
x =52.4 m (distance it covered at constant acceleration)
Then. when it travels upwards only by a force of gravity,
d = v1(t) + 1/2*a*t^2
281.42-52.386 = (0)^2+1/2*(9.81)(t^2)
t = 6.83 s (time when it lost fuel and reached its maximum height)
Lastly, for free falling objects, the equation is
t = √2y/g = √2(281.42)/9.81 = 7.57 s
Therefore, the total time= 1.41+6.83+7.57 = 15.81 s
