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UNO [17]
2 years ago
6

When a golfer tees off, the head of her golf club which has a mass of 158 g is traveling 48.2 m/s just before it strikes a 46.0

g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s. Neglect the mass of the club handle and determine the speed of the golf ball just after impact.
Physics
1 answer:
adoni [48]2 years ago
3 0

Answer:

v₂ = 53.23 m/s

Explanation:

Given that,

The mass of a golf club, m₁ = 158 g = 0.158 kg

The initial speed of a golf club, u₁  =  48.2 m/s

The mass of a golf ball, m₂ = 46 g = 0.046 kg

It was at rest, u₂ = 0

Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s, v₁ = 32.7 m/s

We use the conservation of energy to find the speed of the golf ball just after impact as follows :

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.158(48.2)-0.158(32.7)}{0.046}\\\\=53.23\ m/s

So, the speed of the golf ball just after the impact is equal to 53.23 m/s.

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a 2kg marble is moving at 3 m/s when it strikes another 4kg marble moving in the opposite direction at -3 m/s. What will be the
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Momentum is conserved after the collision 

Momentum of 2 Kg before collision = 2 * 3 = 6
Momentum of 4 kg  before collision = 4 * -3 = -12

so 6 + -12 = 2 * -4 + 4 *x   where x is velocity of 4kg marble.

4x - 8 = -6
4x = 2
x = 0.5 

Velocity of 4 kg marble is 0.5 m/s after collision

The 2 kg marble will move in the opposite direction to which it was moving before the collision.
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Why is a spectrum of colors produced when white light passes through a prism?
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4 0
3 years ago
A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught
meriva

Answer:

The initial velocity of the softball is 14.711 meters per second.

Explanation:

This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.

From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:

y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2} (Eq. 1)

Where:

y_{o} - Initial height of the softball, measured in meters.

y - Final height of the softball, measured in meters.

v_{o} - Initial velocity of the softball, measured in meters per second.

t - Time, measured in seconds.

g - Gravitational acceleration, measured in meters per square second.

If we know that y = y_{o}, t = 3.56\,s and g = -9.807\,\frac{m}{s^{2}}, the initial velocity of the softball is:

v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0

3\cdot v_{o} -44.132\,m= 0

v_{o} = 14.711\,\frac{m}{s}

The initial velocity of the softball is 14.711 meters per second.

8 0
3 years ago
What water pressure must a pump that is located on the first floor supply to have water on the thirteenth of a building with a p
irga5000 [103]

The water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

The given parameters;

  • <em>Pressure on the 13 th floor, P₁ = 35 PSI</em>
  • <em>Distance between each floor, d = 10 ft</em>

The vertical pressure of the water is calculated as follows;

P = \rho gh\\\\\frac{P}{h} = \rho g\\\\\frac{P}{h} = k\\\\\frac{P_1}{h_1} = \frac{P_2}{h_2} \\\\

The vertical height of the first floor from the 13th floor = 130 ft

The vertical height of the 13 ft floor = 10  ft

P_1 = \frac{P_2 h_1}{h_2} \\\\P_1 = \frac{35 \times 130}{10} \\\\P_1 = 455 \ PSI

Thus, the water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

Learn more about vertical height and pressure here: brainly.com/question/15691554

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