A basketball rolls off a deck that is 3.2 m from the pavement. The basketball lands 0.75 m from the edge of the deck. How fast w
1 answer:
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Answer:
x = 0.0685 m
Explanation:
In this exercise we can use the relationship between work and energy conservation
W = ΔEm
Where the work is
W = F x
The energy can be found in two points
Initial. Just when the block with its spring spring touches the other spring
Em₀ = K = ½ m v²
Final. When the system is at rest
=
= ½ k₁ x² + ½ k₂ x²
We can find strength with Newton's second law
∑ F = F - fr
Axis y
N- W = 0
N = W
The friction force has the equation
fr = μ N
fr = μ W
The job
W = (F – μ W) x
We substitute in the equation
(F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)
½ x² (k₁ + k₂) + (F - μ W) x - ½ m v² = 0
We substitute values and solve
½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0
x² 30 + 14.4 x - 1,125 = 0
x² + 0.48 x - 0.0375 = 0
We solve the second degree equation
x = [-0.48 ±√(0.48 2 + 4 0.0375)] / 2
x = [-0.48 ± 0.617] / 2
x₁ = 0.0685 m
x₂ = -0.549 m
The first result results from compression of the spring and the second torque elongation.
The result of the problem is x = 0.0685 m
Answer:
F = 2 × 10⁻³ N
Explanation:
Given:
frequency, f = 2400 MHz
Height, h = 25cm = 0.25 m
Area of the base, A = 30 cm x 30 cm = 900 cm² = 0.09 m²
Energy of the microwave, E = 0.50 mJ = 0.5 x 10⁻³ J
Now, the time taken by the wave from top to the base, t = h/c
here, c is the speed of the light
thus,
t = 0.25/(3 x 10⁸) = 8.33 x 10⁻¹⁰ s
The radiation pressure = Intensity/c
now, the intensity is given as:
I = Power/ area
also,
Power = Energy/ time = 0.5 x 10⁻³ J/8.33 x 10⁻¹⁰ s = 600000 W
thus,
I = 600000 W/ 0.09 m² = 6666666.6 W/m²
substituting the value in the formula for pressure due to radiation, we have
= 6666666.6 W/m²/(3 x 10⁸)
also
pressure = Force/ area
thus,
Force/ area = 6666666.6 W/m²/(3 x 10⁸)
or
Force (F) = (6666666.6 W/m² × 0.09 m²)/(3 x 10⁸)
or
F = 2 × 10⁻³ N
Answer:
a) the mug hits the floor 0.7425m away from the end of the bar. b) |V|=5.08m/s θ= -72.82°
Explanation:
In order to solve this problem, we must first start by doing a drawing of the situation. (see attached picture).
a)
From the drawing we can see that we are dealing with a two dimensions movement problem. So in order to find out how far away from the bar the mug will fall, we need to start by finding how long it will take the mug to be in the air, so we analyze the vertical movement of the mug.
In order to find the time we need to use the following formula, which contains the data we know:
we know that and that
as well, so the formula is simplified to:
we can now solve this for t, so we get:
we know that and that
the acceleration of gravity is negative because the mug is moving downwards. So we substitute them into the given formula:
which yields:
t=0.495s
we can now use this to find the horizontal distance the mug travels. We know that:
so we can solve this for x, so we get:
and we can now substitute the values we know:
x=(1.5m/s)(0.495s)
which yields:
x=0.7425m
b) Now that we know the time it takes the mug to hit the floor, we can use it to find the final velocity in the y-direction by using the following formula:
we know the initial velocity in the vertical direction is zero, so we can simplify the formula:
so we can solve this for the final velocity:
in this case the acceleration is the same as the acceleration of gravity (which is negative) so we can substitute that and the time we found on the previous part to get:
which yields:
so now we know the components of the final velocity, which are:
and
so now we can find the speed by determining the magnitude of the vector, like this:
so we get:
which yields:
|V|=5.08m/s
now, to find the direction of the impact, we can use the following equation:
so we get:
which yields:
