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strojnjashka [21]
3 years ago
5

High-speed stroboscopic photographs show that the head of a 180 g golf club is traveling at 47 m/s just before it strikes a 46 g

golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 35 m/s. Find the speed of the golf ball just after impact.
Physics
2 answers:
inn [45]3 years ago
7 0

Answer:

47 m/s

Explanation:

golf club mass, mc = 180 g

golf ball mass, mb = 46 g

initial golf club speed, vc1 = 47 m/s

final golf club speed, vc2 = 35 m/s

initial golf ball speed, vb1 = 0 m/s

final golf ball speed, vb2 = ? m/s

The total momentum is conserved, then:

mc*vc1 + mb*vb1 = mc*vc2 + mb*vb2

Replacing with data and solving (dimension are omitted):

180*47 + 46*0 = 180*35 + 46*vb2

vb2 = (180*47 - 180*35)/46

vb2 = 47 m/s

lidiya [134]3 years ago
4 0

Answer:

47.17 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision.

mu+m'u' = mv+m'v'......................... Equation 1

Where m = mass of the golf club, m' = mass of the gulf ball, u = initial velocity of the gulf club, u' = initial velocity of the gulf ball, v = final velocity of the gulf club, v' = final velocity of the gulf ball.

Note: The gulf ball was at rest before impact.

Therefore,

mu = mv+m'v'

Make v' the subject of the equation

v' = (mu-mv)/m'....................... Equation 2

Given: m = 180 g = 0.180 kg, m' = 46 g = 0.046 kg, u = 47 m/s, v = 35 m/s

Substitute into equation 2

v' = [(0.18×47)-(0.18×35)]/0.046

v' = (8.47-6.3)/0.046

v' = 2.17/0.046

v' = 47.17 m/s

Hence the speed of the gulf ball just after impact = 47.17 m/s

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daser333 [38]

Answer:

<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b) The bullet travels horizontally 110.6 m</h2>

Explanation:

a)  Consider the vertical motion of bullet

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

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     Substituting

                      s = ut + 0.5 at²

                      1.5 = 0 x t + 0.5 x 9.81 xt²

                      t = 0.553 s

      Time elapsed before the bullet hits the ground is 0.553 seconds.

b) Consider the horizontal motion of bullet

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 200 m/s

        Acceleration, a = 0 m/s²  

        Time, t = 0.553 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 200 x 0.553 + 0.5 x 0 x 0.553²

                      s = 110.6 m

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6 0
2 years ago
A freight train rolls along a track with considerable momentum. If it were to roll at the same speed but had twice as much mass,
fomenos

Answer:

The momentum would be doubled

Explanation:

The magnitude of the momentum of the freight train is given by:

p=mv

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v is its speed

In this problem, we have that the speed of the train is unchanged, while the mass of the train is doubled:

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A container of gas is at a pressure of 1.3x10^5 Pa and a volume of 6 m^3. How much work is done by the gas if it expands at a co
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In thermodynamics, work of a system at constant pressure conditions is equal to the product of the pressure and the change in volume. It is expressed as follows:

W = P(V2 - V1)
W = 1.3x10^5 (2x6 - 6 )
<span>W = 780000 J
</span>
Hope this answers the question. Have a nice day.
5 0
2 years ago
An object is 15 cm in front of a diverging lens with a
Rainbow [258]

A) See ray diagram in attachment (-6.0 cm)

By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f = -10 cm is the focal length (negative for a diverging lens)

p = 15 cm is the distance of the object from the lens

Solving for q,

\frac{1}{q}=\frac{1}{-10 cm}-\frac{1}{15 cm}=-0.167 cm^{-1}

q=\frac{1}{-0.167 cm^{-1}}=-6.0 cm

B) The image is upright

As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:

h_i = - h_o \frac{q}{p}

where h_i, h_o are the size of the image and of the object, respectively.

Since q < 0 and p > o, we have that h_i >0, which means that the image is upright.

C) The image is virtual

As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.

This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual

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Answer:

Distance between two adjacent wave crests = 24m

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distance = (360 meters) / 15 = 24 meters (between two adyacent waves)

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