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Dmitriy789 [7]
3 years ago
11

A golf ball with m = 46 g is struck a blow which makes an angle of 45o with the horizontal. The drive lands 200 m away on a flat

fairway. If the golf club and ball are in contact for a time of 7 ms, what is the average force of impact?
Physics
1 answer:
AfilCa [17]3 years ago
7 0

Answer:

Average force will be equal to 2908.57 N  

Explanation:

We have given mass of the ball m = 46 gram = 0.046 kg

Let velocity at which ball is projected is u m/sec

Angle at which ball is projected \Theta =45^{\circ}

Range of the ball is given R = 200 m

Range is equal to R=\frac{u^2sin2\Theta }{g}

200=\frac{u^2sin(90^{\circ})}{9.8}

u^2=1960

u = 44.27 m/sec

Change in momentum of the ball is equal to P=mu=0.46\times 44.27=20.36kgm/sec

Time of impact is given dt=7ms=0.007sec

Force is equal to rate of change of momentum

So force F=\frac{dP}{dt}=\frac{20.36}{0.007}=2908.57N

Force will be equal to 2908.57 N

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If the velocity of a long jumper during take-offat an angle of 30 degrees is 42 f/s, how fast is he moving forward (Vx) and how
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Answer:

Vx=  11.0865(m/s)

Vy=  6.4008(m/s)

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V= 42(ft/s) × 0.3048(m/ft) = 12.8016(m/s)

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Vx= 12.8016 (m/s) × cos(30°)= 11.0865(m/s)

And for the same theorem the speed on the Y axis will be:

Vy= 12.8016 (m/s) × sen(30°)= 6.4008(m/s)

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A battery charger is connected to a dead battery and delivers a current of 3.5 a for 4 hours, keeping the voltage across the bat
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Calculate the electric field at one corner of a square 50 cm on a side if the other corners are occupied by 250x10-7C (charges)
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The electric field at one corner of a square is 1614217 N/C.

Explanation:

The distance between x and y direction diagonals.

As per the given details the distance between diagonals is calculated as

0.5² + 0.5² = c²  =>  c = 0.707 m

Charge to the right:  In x direction

In order to find the electric charge towards x direction

we use e = kq/r² formula

As 'k' is coulomb's constant it's value is 9 x 10^{9} N m²/C²

e = (9 x 10^{9})(250 x 10^{-7}) / (0.5)²

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Charge diagonal:

e = kq/r²

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Similarly as shown in x direction the charge is same for y direction also

Charge below:  For y direction

e = kq/r²

e = (9 x 10^{9})(250 x 10^{-7}) / (0.5)²

e = 9 x 10^{5} N/C

Charge diagonal:

e = kq/r²

e = [(9 x 10^{9})(250 x 10^{-7}) / (0.5)²] sin 45

e = 159099 N/C

Y direction sum = 1059099 N/C

Resultant electric field strength:

1218198 ² + 1059099² = e²

e = 1614217 N/C [45 degrees below the horizontal]

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