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Nezavi [6.7K]
3 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]3 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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What kind of employment is regarded as national level employment?
yanalaym [24]

Answer:

In most cases, the sources of information for both levels are same as follows:

Verbal or personal contacts : This information is transferred through verbal method. ...

Different media: Media is one of the best sources of information in these days.

Explanation:

7 0
2 years ago
A car is traveling north at 17.7 m/s . After 6 it’s velocity is 141 in the same direction. Find the magnitude and direction of t
Furkat [3]

By equation of motion we have   v = u + at

Where u = Initial velocity, v = final velocity, t = time taken and a = acceleration

Here v = 141 m/s, u = 17.7 m/s and t = 6 s

On substitution we will get

        141 = 17.7+ 6a

       So, a = (141-17.7)/6 = 20. 55 m/s^{2}

       Aceeleration = 20. 55 m/s^{2} along north direction.


3 0
3 years ago
A scene in a movie has a stuntman falling through a floor onto a bed in the room below. The plan is to have the actor fall on hi
IgorC [24]

Answer:

m=17.79Kg

Explanation:

In this process energy must be conserved. On the initial stage, there will be only gravitational potential energy, while on the final stage there will be only elastic potential energy, so they will be equal. We write this as:

U_g=U_e

Which is the same as:

mgh=\frac{k \Delta x^2}{2}

So we can obtain our mass from there, and for our values:

m=\frac{k \Delta x^2}{2gh}=\frac{(65144 N/m)(0.1333m)^2}{2(9.8m/s^2)(3.32m)}=17.79Kg

4 0
3 years ago
Practice: The speed of sound at sea level is normally about 340 m/s. A car honks its horn as it drives toward an observer. The f
stepan [7]

Answer:

25.5 m/s

Explanation:

The Doppler effect occurs when there is relative motion between a source of a wave and an observer. In such situation, there is a shift in the apparent frequency of the wave perceived by the observer.

The formula that gives the apparent frequency perceived by the observer is:

f'=\frac{v\pm v_o}{v\pm v_s}f

where

f is the real frequency of the wave

f' is the apparent frequency of the wave

v is the speed of the wave

v_s is the velocity of the source (negative if the source is moving towards the observer, positive otherwise)

v_o is the velocity of the observer (positive if the observer is moving towards the source, negative otherwise)

In this problem:

v = 340 m/s is the speed of sound

f = 800 Hz is the frequency of the horn

f' = 860 Hz is the apparent frequency

v_o=0 (the observer is at rest)

Re-arranging the equation for v_s, we can find the velocity of the horn and the driver:

f'=\frac{v}{v-v_s}f\\(v-v_s)f'=vf\\vf'-v_sf'=vf\\v_s=v\frac{f'-f}{f'}=(340)\frac{860-800}{860}=25.5 m/s

So, 25.5 m/s towards the observer.

3 0
3 years ago
Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling witho
8090 [49]

Answer:

Hoop will reach the maximum height

Explanation:

let the mass and radius of solid ball, solid disk and hoop be m and r  (all have same radius and mass)

They all  are rolled with similar initial speed v

by the law of conservation of energy we can write

K_{trans}+K_{rot}= P

for solid ball

[tex]\frac{1}{2}mv^2+\frac{1}{2}I_{ball}\omega^2= mgh_{ball}

putting I_{ball}=\frac{2}{5}mr^2 and \omega=\frac{v}{r} in the above equation and solving we get

h_{ball}= 0.071v^2

now for solid disk

[tex]\frac{1}{2}mv^2+\frac{1}{2}I_{disk}\omega^2= mgh_{disk}

putting I_{ball}=\frac{1}{2}mr^2 and \omega=\frac{v}{r} in the above equation and solving we get

h_{disk}= 0.076v^2

for hoop

[tex]\frac{1}{2}mv^2+\frac{1}{2}I_{hoop}\omega^2= mgh_{hoop}

putting I_{hoop}=mr^2 and \omega=\frac{v}{r} in the above equation and solving we get

h_{hook}= 0.10v^2

clearly from the above calculation we can say that the Hoop will reach the maximum height

5 0
3 years ago
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