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Ugo [173]
3 years ago
12

Consider a golf club hitting a golf ball that results in the following graph of force versus time on a 45 gram golf ball. If the

final velocity of the ball has a magnitude of 41 m/s, determine the value of Fmax

Physics
1 answer:
katen-ka-za [31]3 years ago
4 0

The question is missing the graph. So, the graph is attached below.

Answer:

The value of F_{max} is 7380 N.

Explanation:

Given:

Mass of the ball is, m=45\ g=0.045\ kg

Initial velocity of the ball is, u=0\ m/s(Assuming)

Final velocity of the ball is, v=41\ m/s

From the graph,

Time interval for which the force acts, \Delta t=0.5\ ms=0.5\times 10^{-3}\ s

Height of the triangle is equal to the maximum force acting on the ball = F_{max}

Now, we know that, impulse acting on the ball is equal to the area under the curve of force and time graph.

So, impulse is equal to the area of the triangle and is given as:

Impulse = \frac{1}{2}\times base\times height

Here, base is time interval \Delta t and height is F_{max}.

Impulse = \frac{1}{2}\times (0.5\times 10^{-3})\times F_{max}

Impulse = 0.25\times 10^{-3}F_{max}

Now, we also know that, impulse is equal to the change in momentum of a body.

Therefore, change in momentum (Δp) is given as:

\Delta p=m(v-u)\\\Delta p = 0.045(41-0)=1.845\ kgms^{-1}

Now, change in momentum is equal to impulse acting on the ball. Thus,

0.25\times 10^{-3}F_{max}=1.845\\\\F_{max}=\frac{1.845}{0.25\times 10^{-3}}\\\\F_{max}=7380\ N

Therefore, the value of F_{max} is 7380 N.

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Scilla [17]

Explanation:

Given that,

Mass of a body, m = 1 kg

Force constant, k = 16 N/m

We need to find the angular frequency and the frequency of oscillation.

(a) The angular frequency of a body is given by :

\omega=\sqrt{\dfrac{k}{m}} \\\\=\omega=\sqrt{\dfrac{16}{1}} \\\\=4\ rad/s

(b) The frequency of oscillation is given by :

f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{4}{2\pi}\\\\=\dfrac{2}{\pi}\ Hz

Hence, this is the required solution.

7 0
3 years ago
Light is incident on the left face of an isosceles prism; with an apex angle of 49o, such that the light exiting the right face
Sunny_sXe [5.5K]

Answer:

\mu = 1.645

Explanation:

By Snell's law we know at the left surface

\theta_i = 19^o

\theta_r = ?

\mu_1 = 1

\mu_2 = \mu

now we have

1 sin19 = \mu sin\theta_r

0.33 = \mu sin\theta_r

now on the other surface we know that

angle of incidence = \theta_r'

\theta_e = 90

so again we have

\mu sin\theta_r' = 1 sin90

so we have

\theta_r = sin^{-1}\frac{0.33}{\mu}

\theta_r' = sin^{-1}\frac{1}{\mu}

also we know that

\theta_r + \theta_r' = 49

sin^{-1}\frac{0.33}{\mu} + sin^{-1}\frac{1}{\mu} = 49

By solving above equation we have

\mu = 1.645

3 0
3 years ago
Read 2 more answers
The graph represents velocity over time.<br> What is the acceleration?
Semmy [17]

Answer:

I think the answer is 0.2 m/s2

Explanation:

7 0
2 years ago
a football player kicks a ball with a mass of 0.42kg. The average acceleration of the football was 14.8 m/s2. How much force did
Zarrin [17]

Answer:

6.216 N

Explanation:

As for Newton's second law of motion

F=ma

where F= the acting force

           m=subjected mass

           a= the acceleration

applying F=ma to the football

F=m*a

 =0.42*14.8

 =6.216 N

6.216 N of a force is supplied to the ball

5 0
3 years ago
A 10 kg mass rests on a table. What acceleration will be generated when a force of 5 N is applied? a 0.5 m/s2 b 3.5 m/s2 c 5 m/s
rosijanka [135]

Answer:

a= 0.5m/s^2

Explanation:

Force applied on an object is known as

F=m.a  (Newton's second law states it)

a=F/m

a=5/10=0.5m/s^2

3 0
3 years ago
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