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il63 [147K]
4 years ago
7

A 0.0450-kg golf ball initially at rest is given a speed of 25.0 m/s when a club strikes. If the club and ball are in contact fo

r 2.00 ms, what average force acts on the ball? *
Physics
1 answer:
LuckyWell [14K]4 years ago
5 0

Answer:

Average force, F = 562.5 N

Explanation:

Mass of the golf ball, m = 0.045 kg

Initially, it is at rest, u = 0

Final speed of the ball, v = 25 m/s

The club and the ball are in contact for, t=2\ ms=2\times 10^{-3}\ s

We need to find the average force acting on the ball. It can be calculated using the formula as :

F\times t=m(v-u)

F=m\dfrac{v-u}{t}

F=0.045\times \dfrac{25}{2\times 10^{-3}}  

F = 562.5 N

So, the average force acting on the ball is 562.5 N. Hence, this is the required solution.

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A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an ang
koban [17]

Answer:

magnitude of the magnetic field 0.692 T

Explanation:

given data

rectangular dimensions = 2.80 cm by 3.20 cm

angle of 30.0°

produce a flux Ф = 3.10 × 10^{-4}  Wb

solution

we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm

and here angle between magnitude field and area will be ∅ = 90 - 30

∅ = 60°

and flux  is express as

flux Ф = \int \vec{B}.d\vec{A}   .................1

and Ф = BA cos∅    ............2

so B = \frac{\phi }{Acos\theta }    

and we know

A = ab

so

B = \frac{\phi }{abcos\theta }    ..............3

put here value

B =  \frac{3.10\times 10^{-4} }{2.80 \times 10^{-2}\times 3.20 \times 10^{-2}\times cos60}  

solve we get

B = 0.692 T

8 0
3 years ago
Across:
vladimir2022 [97]

Answer:

it would be c i just had it

Explanation:

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6 0
2 years ago
A weightlifter presses a 200 N weight 0.5 m over his head in 2 s. What is the power of the weightlifter
icang [17]

Answer:

50 watts

Explanation:

Applying,

Power (P) = Workdone (W)/Time(t)

But,

Work done (W) = Force (F)×distance(d)

Therefore,

P = Fd/t..................... Equation 1

Where P =  power of the weightlifter, F = Force applied, d = distance, t = time.

From the question,

Given: F = 200 N, d = 0.5 m, t = 2 s

Substitute these values into equation 1

P = (200×0.5)/2

P = 100/2

P = 50 watts

7 0
3 years ago
Consider a horizontal layer of the dam wall of thickness dx located a distance x above the reservoir floor. what is the magnitud
sergejj [24]
The force on the layer will be equivalent to the weight of water on it. This is given by:
F = mg; m is the mass of water and g is the acceleration due to gravity.
5 0
3 years ago
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Hello please help i’ll give brainliest
balu736 [363]
D, I hope this is correct!
6 0
3 years ago
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