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Levart [38]
3 years ago
15

A soccer player takes a free kick from a spot that is 20 m from the goal. The ball leaves his foot at an angle of 38 ∘, and it e

ventually hits the crossbar of the goal, which is 2.4 m from the ground. At what speed did the ball leave his foot?
Physics
1 answer:
Sergio039 [100]3 years ago
7 0
This dude can kick the ball
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The total amount of energy in a closed system must stay constant.

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An RC circuit takes t = 2.5 ms to charge to 35% of its full charge after it weas connected to a battery, What is the total time
marin [14]

Answer: time taken to charge to 95%

t = -5.80[ln(1-0.95)]

t = 17.38ms

Explanation:

For an RC Charging circuit

Where Vs

Vc = Vs (1 - e^(-t/RC))

Vc/Vs = 1 - e^(-t/RC)

-t/RC = ln(1 - Vc/Vs)

t = -RC[ln(1 - Vc/Vs)] and RC = k = -t/ln(1 - Vc/Vs)

Where ;

Vc = voltage across the capacitor

Vs = voltage supply

t = charging time = 2.5ms

k = RC = time constant.

Vc/Vs = 0.35

To calculate the time constant k;

k = -t/ln(1- Vc/Vs)

k = -2.5/ln(1-0.35)

k = 5.80ms

time taken to charge to 95%

t = -5.80[ln(1-0.95)]

t = 17.38ms

4 0
3 years ago
The body falls in a free fall 11s.Calculate: a) from what height the body of the paddle) what path did it fall during the last s
JulsSmile [24]

Answer:

a) h = 593.50 m

b) h₁₁ = 103 m

c) vf = 107.91 m/s

Explanation:

a)

We will use second equation of motion to find the height:

h = v_{i}t + (\frac{1}{2})gt^2

where,

h = height = ?

vi = initial speed = 0 m/s

t = time taken = 11 s

g = 9.81 /s²

Therefore,

h = (0\ m/s)(11\ s) + (\frac{1}{2})(9.8\ m/s^2)(11\ s)^2\\\\

<u>h = 593.50 m</u>

b)

For the distance travelled in last second, we first need to find velocity at 10th second by using first equation of motion:

v_{f} = v_{i} + gt

where,

vf = final velocity at tenth second = v₁₀ = ?

t = 10 s

vi = 0 m/s

Therefore,

v_{10} = 0\ m/s + (9.81\ m/s^2)(10\ s)\\\\v_{10} = 98.1\ m/s

Now, we use the 2nd equation of motion between 10 and 11 seconds to find the height covered during last second:

h = v_{i}t + (\frac{1}{2})gt^2

where,

h = height covered during last second = h₁₁ =  ?

vi = v₁₀ = 98.1 m/s

t = 1 s

Therefore,

h_{11} = (98.1\ m/s)(1\ s) + (\frac{1}{2})(9.8\ m/s^2)(1\ s)^2\\\\

<u>h₁₁ = 103 m</u>

c)

Now, we use first equation of motion for complete motion:

v_{f} = v_{i} + gt

where,

vf = final velocity at tenth second = ?

t = 11 s

vi = 0 m/s

Therefore,

v_{f} = 0\ m/s + (9.81\ m/s^2)(11\ s)

<u>vf = 107.91 m/s</u>

8 0
3 years ago
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