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mel-nik [20]
2 years ago
6

During soccer practice, Maya kicked a soccer ball 37° off the ground at 25 m/s. What was the ball's speed 2.2 s after she kicked

it?
Physics
1 answer:
ki77a [65]2 years ago
5 0

<u>Answer</u>

3.44 m/s


<u>Explanation</u>

The motion apply the equations of Newton's law of motion. The ball is acceleration is -9.8 m/s² (acceleration due to gravity. It is negative because the ball is going against gravity, so it is decelerating).

The first equation of Newton's law of motion is;

V = U + at

Where V is the final velocity, U is the initial velocity, a is acceleration and t the time taken.

V = 25 + (-9.8 × 2.2)

    = 25 - 21.56

     = 3.44 m/s

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Acceleration is measured in_________<br><br><br> m<br><br><br> g<br><br><br> m/s<br><br><br> m/s2
Nana76 [90]
Acceleration is measured in m/s².

Answer: m/s²
4 0
3 years ago
If a 75 W lightbulb is 15% efficient, how many joules of light energy does the bulb produce every minute?
stiks02 [169]

Answer:

1 W = 1 J / sec       Definition of watt is 1 joule / sec

So if a bulb uses 75 J / sec it must use

75 J/s * 60 sec / min = 4500 J/min    energy used by bulb

If bulb is 15% efficient then the light delivered is

P = 4500 J / min * .15 = 675 J / min

4 0
2 years ago
A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di
siniylev [52]

Answer:

1408.685 KN/C

Explanation:

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = 8.99*10^9

  • The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

                                  E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\

part a)

Electric Field strength at point P: a = 0.75 m

E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2  

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

 

6 0
3 years ago
As a stands, her entire weight is momentarily placed on the heels of her high-heeled shoes. Calculate the pressure exerted on th
Lena [83]

Answer:

Pressure will be 6072449.952Pa

Explanation:

We have given mass of the women m = 65 kg

Radius of the heels r = 0.578 cm = 0.00578 m

We have to find the pressure

We know that pressure is given by

P=\frac{F}{A}=\frac{mg}{A}

So force F = mg = 65×9.8 = 637 N

Area A=\pi r^2=3.14\times 0.00578^2=1.049\times 10^{-4}m^2

So pressure p=\frac{637}{1.049\times 10^{-4}}=6072449.952Pa

5 0
3 years ago
Average mass of a toy car(small, hot wheels)?
Tresset [83]
2.4 ounces is the average weight of a hot wheels car.
6 0
3 years ago
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