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3 years ago

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Physics

1 answer:

eduard3 years ago

5 0

Answer:

The average force the golf club exerts on the ball is 600 N

Explanation:

Newton's second law of motion states that force, F, is directly proportional to the rate of change of momentum produced

F = m× (v₂ - v₁)/(Δt)

The given parameters of the motion of the ball are;

The mass of the ball, m = 45 g = 0.045 kg

The initial velocity of the ball, v₁ = 0 m/s

The speed with which the ball was hit by the golfer, v₂ = 40 m/s

The duration of contact between the golf club and the ball, Δt = 3 ms = 0.003 seconds (s)

By Newton's law of motion, the average force, 'F', which the golf club exerts on the ball is therefore, given as follows;

F = 0.045 kg × (40 m/s - 0 m/s)/(0.003 s) = 600 N

The average force the golf club exerts on the ball = F = 600 N.

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A professor, with dumbbells in his hands and holding his arms out, is spinning on a turntable with an angular velocity. What hap

Olenka [21]

Answer:

<em>His angular velocity will increase.</em>

Explanation:

According to the conservation of rotational momentum, the initial angular momentum of a system must be equal to the final angular momentum of the system.

The angular momentum of a system = $I$ 'ω'

where

$I$ ' is the initial rotational inertia

ω' is the initial angular velocity

the rotational inertia = $mr'^{2}$

where m is the mass of the system

and r' is the initial radius of rotation

Note that the professor does not change his position about the axis of rotation, so we are working relative to the dumbbells.

we can see that with the mass of the dumbbells remaining constant, if we reduce the radius of rotation of the dumbbells to r, the rotational inertia will reduce to $I$ .

From

$I$ 'ω' = $I$ ω

since $I$ is now reduced, ω will be greater than ω'

therefore, the angular velocity increases.

5 0

3 years ago

A positively charged particle is in the center of a parallel-plate capacitor that has charge ±Q on its plates. SUppose the dista

slamgirl [31]

Answer:

Stay the same

Explanation:

First of all, let's find how the capacitance of the capacitor changes.

Initially, it is given by

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

From the formula, we see that the capacitance is inversely proportional to the separation between the plates. In this problem, the distance between the plates is doubled, so the capacitance will be halved:

$C' = \frac{1}{2}C$

The potential difference across the capacitor is given by

$V= \frac{Q}{C}$

where

Q is the charge on the plates

C is the capacitance

We see that the voltage is inversely proportional to the capacitance. We said that the capacitance has halved: therefore, the potential difference across the two plates will double:

$V' = 2 V$

Now we can analyze the electric field between the plates of the capacitor, which is given by

$E=\frac{V}{d}$

we said that:

- The voltage has doubled: V' = 2 V

- The distance between the plates has doubled: d' = 2 d

therefore, the new electric field will be

$E'=\frac{2V}{2d}=\frac{V}{d}=E$

So, the electric field is unchanged. And since the force on the particle at the center is directly proportional to the electric field:

F = qE

Then the force on the particle will stay the same.

4 0

3 years ago

A cylindrical, 0.500-m rod has a diameter of 0.02 m. The rod is stretched to a length of 0.501 m by a force of 3000 N. What is t

Salsk061 [2.6K]

Answer:

Y = 4.775 x 10⁹ Pa = 4.775 GPa

Explanation:

First, we calculate the stress on the rod:

$stress = \frac{Force}{Area} = \frac{3000\ N}{\pi r^2} \\\\stress = \frac{3000\ N}{\pi (0.01\ m)^2}\\\\stress = 9.55\ x\ 10^6\ Pa = 9.55 MPa\\$

Now, we calculate the strain:

$strain = \frac{Change\ in Length}{Original\ Length}\\\\strain = \frac{0.501\ m - 0.5\ m}{0.5\ m}\\\\strain = 0.002\\$

Now, we will calculate the Young's Modulus (Y):

$Y = \frac{stress}{strain}\\\\Y = \frac{9.55\ x\ 10^6\ Pa}{0.002} \\$

6 0

3 years ago

A 900 kg vehicle moves around a curve with an incline of 20\circ∘ at a speed of 12.5 m/s. If the curve has a radius of 50 meters

valentina_108 [34]

Answer:

The normal force experienced by the car is approximately 8223.2 N

Explanation:

The question relates to banking of road where the centripetal force for the circular motion of the vehicle is provided by the horizontal component of the normal reaction

The mass of the vehicle that moves around the curve, m = 900 kg

The incline of the curve, θ = 20°

The speed with which the vehicle moves around the curve, v = 12.5 m/s

The radius of the curve, R = 50 meters

We have;

$N \cdot sin(\theta) = \dfrac{m \cdot v^2}{R}$

Where;

θ = The angle of inclination of the road = 20°

N = The normal force experienced by the car

m = The mass of the car = 900 kg

v = The velocity with which the car is moving = 12.5 m/s

R = The radius of the curve around which the vehicle moves = 50 m

$\therefore N = \dfrac{m \cdot v^2}{R \cdot sin(\theta)} = \dfrac{900 \times (12.5)^2}{50 \times sin(20^{\circ})} = 8223.1998754586828969046217875927$

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2 years ago

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Sever21 [200]

why does the baseball have to be autographed?

8 0

2 years ago

Read 2 more answers