A spotlight on the ground shines on a wall 12 meters away. If a man 2 meters high walks away from the spotlight towards the wall
at a speed of 1.6 meters per second, how fast is the length of his shadow on the wall decreasing when he is 4 meters away from the wall?
1 answer:
Answer: The length of the shadow on the wall is decreasing by 0.6m/s
Explanation:
the specified moment in the problem, the man is standing at point D with his head at point E.
At that moment, his shadow on the wall is y=BC.
The two right triangles ΔABC and ΔADE are similar triangles. As such, their corresponding sides have equal ratios:
ADAB=DEBC
8/12=2/y,∴y=3 meters
If we consider the distance of the man from the building as x then the distance from the spotlight to the man is 12−x.
(12−x) /12=2/y
1− (1 /12x )=2 × 1/y
Let's take derivatives of both sides:
−1 / 12dx = −2 × 1 / y^2 dy
Let's divide both sides by dt:
−1/12⋅dx/dt=−2/y^2⋅dy/dt
At the specified moment:
dxdt=1.6 m/s
y=3
Let's plug them in:
−1/121.6) = - 2/9 × dy/dt
dy/dt = 1.6/12 ÷ 2/9
dy/dt = 1.6/12 × 9/2
dy/dt = 14.4/24 = 0.6m/s
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Answer:
The final velocity of the second car is 57 m/s south.
Explanation:
This is an elastic collision between two train cars. In this case, the total kinetic energy between the two bodies will remain the same.
The formula to apply is :
where ;
Given in the question that;
Apply the formula as;
{14650*18}+{3825*11} = {14650 *6} + {3825 * v₂f}
263700+42075=87900 + 3825v₂f
305775 =87900 + 3825v₂f
305775-87900 = 3825v₂f
217875=3825v₂f
217875/3825 =v₂f
56.96 = v₂f
<u>57 m/s = v₂f { nearest whole number}</u>
For a photographer that wishes to determine the color of light that he can use in a dark room that will not expose the films he is processing, having used a Blue Incandescent bulb, he should proceed to use a Red Incandescent bulb for the next trial.
The photographer in question is performing an experiment. For these kinds of experiments it is important to identify the variables present, which can be of three kinds:
- Control variables
- Dependent variables
- Independent variables
For this experiment, the dependent variable is the exposure of the light onto the films, given that this is what we wish to measure. The independent variable will be the color of the light being used which is what will affect the dependent variable.
The remaining variable must be the control variable. Unlike the previous variables, we can have more than one of these. The control variable is there to make sure that only the dependent variable is affecting the outcome. We do this by keeping the control variable the same through each trial, which is why the photographer should not change the type of bulb in the second experiment, changing only the color of the light.
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