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Nezavi [6.7K]
4 years ago
8

In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.0800 s, during which

time it experiences an acceleration of 229 m/s2. The ball is launched at an angle of 52.3 ° above the ground. Determine the (a) horizontal and (b) vertical components of the launch velocity.
Physics
1 answer:
Elena-2011 [213]4 years ago
5 0
The initial velocity of the ball is 0. Applying:
v = u + at
v = 0 + 229 x 0.08
v = 18.3 m/s

a)
Vx = Vcos(∅)
Vx = 18.3cos(52.3)
Vx = 11.2 m/s

b)
Vy = Vsin(∅)
Vy = 18.3sin(52.3)
Vy = 14.5 m/s
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taurus [48]

Answer:

D. Cooler

Explanation:

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How much of the Moon is always illuminated one time? Explain your answer.
Natalija [7]

Answer:

50% of it .

Explanation:

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A body with a mass of 5 Kg has the same kinetic energy as a second body. The second body has a mass of 10 kg and is moving at a
vitfil [10]
First body:

E_{k_1}=\frac{1}{2}.m.v^2\\
\\
E_{k_1}=\frac{1}{2}.5.v_1^2=\frac{5}{2}.v_1^2

Second body:

E_{k_2}=\frac{1}{2}.m.v_2^2\\
\\
E_{k_2}=\frac{1}{2}.10.(20)^2=2.000 \ J

From description of the task we have:

E_{k_1}=E_{k_2}\\
\\
\frac{5}{2}.v_1^2=2.000 \ J\\
\\
5.v_1^2=4.000 \ J\\
\\
v_{1}^2=800 \ J\\
\\
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Which instrument is used to measure liquid volume?
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Volumetric cylinders and volumetric flasks

3 0
3 years ago
Read 2 more answers
A tire is filled with air at 10 ∘C to a gauge pressure of 250 kPa. Part A If the tire reaches a temperature of 45 ∘C, what fract
Alex_Xolod [135]

Answer:

The air fraction to be removed is 0.11

Given:

Initial temperature, T = 10^{\circ} = 283 K

Pressure, P = 250 kPa

Finally its temperature increases, T' = 45^{\circ} = 318 K

Solution:

Using the ideal gas equation:

PV = mRT

where

P = Pressure

V = Volume

m = no. of moles of gas

R = Rydberg's Constant

T = Temperature

Now,

Considering the eqn at constant volume and pressure, we get:

mT = m'T'

Thus

\frac{m}{m'} = \frac{T'}{T}                      (1)

Now, the fraction of the air to be removed for the maintenance of pressure at 250 kPa:

y = \frac{m - m'}{m} = 1 - \frac{m'}{m}

From eqn (1):

y = 1 - \frac{T}{T'}

y = 1 - \frac{283}{318} = 0.11

6 0
4 years ago
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