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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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4 0
2 years ago
Can someone help with my physics homework? please
Murrr4er [49]

Answer:

a) 19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options:  (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) 24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

Explanation:

a) If force applied in the cart is uniform, that is, constant in magnitude and direction and is parallel to distance travelled by the car, the work done on the cart is defined by the following equation:

W = F\cdot \Delta s (1)

Where:

F - Force applied by the cart, measured in newtons.

\Delta s - Distance travelled by the car, measured in meters.

W - Work done on the cart, measured in joules.

If we know that F = 888\,N and \Delta s = 22\,m, then the work done on the cart is:

W =(888\,N)\cdot (22\,m)

W = 19536\,J

19536 joules of work are done.

b) The work is done by the engine on the structure of the cart.

c) There are three options:  (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.

d) If we know that F = 1111\,N and \Delta s = 22\,m , then the work on the cart is:

W = (1111\,N)\cdot (22\,m)

W = 24442\,N

24442 joules of work are done.

e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.

f) The gas consumption is directly proportional to the square of velocity and mass of the cart and, hence, to the work done on the cart. In consequence, we conclude that the bigger engine uses more gas to go 22 meters.

g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.

3 0
3 years ago
Help please, i don't know what the answer is but i accidently clicked D .....
GaryK [48]

Answer:

You are right correct option is D.

4 0
3 years ago
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