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Elenna [48]
3 years ago
14

A ball is tossed into the air from height of 8 feet with an initial velocity of 8 feet per second. find the time r(in seconds) i

t takes for the object to reach the ground by solving the equation -16t + 8t + 8 = 0 you trimmed a large strip of wallpaper from a scrap to fit into the corner of a wall you are wallpapering. you trimmed 15 inches from the length
Physics
1 answer:
-Dominant- [34]3 years ago
3 0
For this case we have the following equation:
 -16t + 8t + 8 = 0
 The first thing we must do is rewrite the equation correctly.
 We have then:
 -16t ^ 2 + 8t + 8 = 0
 Solving the polynomial we have:
 t1 = -1/2
 t2 = 1
 We discard the negative root because we want to find the time.
 Answer:
 it takes for the object to reach the ground about:
 t = 1 second
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\oint \vec{B}\vec{dl} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}

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B= Magnetic Field

l = length

\mu_0 = Vacuum permeability

\epsilon_0 = Vacuum permittivity

Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

B(2\pi r) = \mu_0 \epsilon_0 \frac{d(EA)}{dt}

Recall that the speed of light is equivalent to

c^2 = \frac{1}{\mu_0 \epsilon_0}

Then replacing,

B(2\pi r) = \frac{1}{C^2} (\pi r^2) \frac{d(E)}{dt}

B = \frac{r}{2C^2} \frac{dE}{dt}

Our values are given as

dE = 2150N/C

dt = 5s

C = 3*10^8m/s

D = 0.440m \rightarrow r = 0.220m

Replacing we have,

B = \frac{r}{2C^2} \frac{dE}{dt}

B = \frac{0.220}{2(3*10^8)^2} \frac{2150}{5}

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So, I = P/A and I₀ = P₀/A

Now, if I = 4I₀,

P/A = 4P₀/A

P = 4P₀

Now, energy E = Pt, where t = time. So, P = E/t and P₀ = E₀/t

Substituting P and P₀ into the equation, we have

P = 4P₀

E/t = 4E₀/t

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Since the energy is four times the initial energy, the energy output increases by a factor of 4.

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