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Dimas [21]
2 years ago
10

PLEASE USE KINEMATIC EQUATION!

Physics
1 answer:
valkas [14]2 years ago
5 0

At its maximum height h, the football has zero vertical velocity, so if it was kicked with initial upward speed v, then

0² - v² = -2gh

Solve this for v :

v² = 2gh

v = √(2gh)

The height y of the football t seconds after being kicked is

y = vt - 1/2 gt²

Substitute v = √(2gh), replace y = h, and solve for h when t = 3.8 s :

h = √(2gh) t - 1/2 gt²

h = √(2gh) (3.8 s) - 1/2 g (3.8 s)²

h ≈ (16.8233 √m) √h - 70.756 m

(By √m, I mean "square root meters"; on its own this quantity doesn't make much physical sense, but we need this to be consistent with √h. h is measured in meters, so √h is measured in √m, too.)

h - (16.8233 √m) √h + 70.756 m = 0

Use the quadratic formula to solve for √h :

√h = ((16.8233 √m) ± √((16.8233 √m)² - 4 (70.756 m))) / 2

Both the positive and negative square roots result in the same solution,

√h ≈ 8.411 √m

Take the square of both sides to solve for h itself:

(√h)² ≈ (8.411 √m)²

⇒   h ≈ 70.756 m ≈ 71 m

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makkiz [27]
Some guidance notes which may help.To calculate the current flow, Ohm's law can be used. This can be written as current=voltage/resistance, or I=V/R. V is 1.5V.R for the copper wire quoted would be calculated as R = resistivity x length/cross sectional area. The area would be calculated from the formula area = pi x diameter squared/4So, R=resistivity x length divided by (pi x diameter squared/4)Until is the resistivity of copper is known, that's about as far as can be gone.Any further questions, please ask.
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3 years ago
When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6
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Answer:

it is separated by 80 cm distance

Explanation:

As per Coulombs law we know that force between two point charges is given by

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here we know that

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q_2 = +5.6 \mu C

force between two charges is given as

F = 0.66 N

now we have

F = \frac{kq_1q_2}{r^2}

0.66 = \frac{(9\times 10^9)(8.4 \mu C)(5.6 \mu C)}{r^2}

r = 0.8 m

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3 0
3 years ago
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When might it be harder to stop a vehicle moving at 30 km/h than one moving at 60 km/h?
rodikova [14]

Answer:

when the momentum of the vehicle moving at 30 km/h is higher than the one from the vehicle moving at 60 km/h

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It's much harder to stop a freight truck moving at 30 km/h than a hot wheels car moving at 60 km/h.

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Calculate a rate of cooling down of air from 80 C to 5C Show calculation. Give an answer in cubic meters per minute and cfm.
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Explanation:

Given that,

Rate of cooling of air

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\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})

Where, dT=T_{1}-T_{2}

Here, T =\dfrac{T_{1}+T_{2}}{2}

T_{0} = 25°C  (surrounding temperature)

dt = 1 minute

\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})

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\dfrac{80-5}{1}=c(\dfrac{85}{2}-25)

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Are worm holes a real phenomenon?
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So to cut the story short, it is not a proven phenomenon only theoretical.
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