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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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LenaWriter [7]

Answer:

485520 m

Explanation:

v_{o} = initial velocity of the projectile = 1360 m/s

v_{f} = final velocity of the projectile = \left ( \frac{2}{5} \right )v_{_{o}} = \left ( \frac{2}{5} \right )(1360) = 544 m/s

a = acceleraton due to gravity on moon = - 1.6 m/s²

h = Altitude of the projectile

Using the kinematics equation

v_{f}^{2} = v_{o}^{2} + 2 a h

Inserting the values

544^{2} = 1360^{2} + 2 (-1.6) h

h = 485520 m

3 0
3 years ago
A car speeds up from rest to +16 m/ s in 4s. calculate the acceleration
Pani-rosa [81]

The magnitude of acceleration is (change in speed) / (time for the change).

Change in speed = (speed at the end) - (speed at the beginning) =

                                   (16 m/s)  -  (0)  =  16 m/s .

Time for the change  =  4 s .

Magnitude of acceleration = (16 m/s) / (4 s) = 4 m/s per sec = 4 m/s² .


6 0
3 years ago
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You are tossing a ball directly up into the air. Before the ball leaves your hand, you are exerting a force directed against the
DaniilM [7]

Answer:

F n = 0.2 N

Explanation:

given,                                      

you are exerting force of 10 N on the ball.

mass of the ball = 1 kg              

acceleration due to gravity  = 9.8 m/s²

normal force on the ball = ?          

normal force is force exerted by the object to counteract the force from other object.                

normal force acting on the ball will be

F n = F - mg                          

F n = 10 - 1 × 9.8                        

F n = 10 -9.8                    

F n = 0.2 N            

Hence, normal force acting on the ball is equal to 0.2 N

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3 years ago
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d1i1m1o1n [39]

Answer:

The mass and velocity for kinetic energy. Potential Energy: How high an object is and the mass in kilograms or it is the weight in and how high an object is. There are two formulas to calculate potential energy, but the one with grams is used more often.

Explanation:

Hope this helps!

5 0
3 years ago
Melvin is traveling south on I-95 at 29 m/s (65 mph) when a deer jumps into his path, 50 m ahead. a. If his reaction time is 0.1
aleksandr82 [10.1K]

Answer:

a. 5.22 meters

b. 2.9 seconds

c. No, Melvin does not hit the deer

Explanation:

The parameters with which Melvin is travelling are as follows;

The speed of Melvin's motion, u = 29 m/s

The distance from Melvin at which the deer jumps into the path = 50 m

a. Distance, d = Velocity, u × Time, t

The time it takes Melvin to react = 0.18 seconds

The distance, "d₁" Melvin travels before his foot hits the break = The velocity with which Melvin was traveling, "u" × The time duration it takes Melvin to hit the brakes, "t₁"

∴ d₁ = 29 m/s × 0.18 s = 5.22 m

The distance, Melvin travels before his foot hits the break = d₁ = 5.22 m

b. Melvin's acceleration after his foot hits the brakes, a = -10 m/s²

Therefore, we have;

The time it takes "t₂" it takes for him to come to a complete stop given as follows;

y = u + a × t₂

Where;

v = The final velocity after Melvin comes to a complete stop = 0 m/s

By substituting the known values, we have;

0 = 29 m/s + (-10 m/s²) × t₂ = 29 m/s - 10 m/s² × t₂

∴ 29 m/s = 10 m/s² × t₂

t₂ = (29 m/s)/(10 m/s²) = 2.9 s

The time it takes it takes for him to come to a complete stop = t₂ = 2.9 s

c. The distance, "d₂", Melvin reaches while accelerating (decelerating) at -10 m/s² to come to a complete stop is given as follows;

v² = u² + 2·a·d₂

Therefore, we have;

0² = (29 m/s)² + 2 × (-10 m/s) × d₂ = (29 m/s)² - 2 × 10 m/s × d₂

∴  (29 m/s)² = 2 × 10 m/s × d₂

d₂ = ((29 m/s)²)/(2 × 10 m/s²) = (841 m²/s²)/(20 m/s²) = 42.05 m

The distance, Melvin reaches while accelerating (decelerating) at -10 m/s² to come to a complete stop = d₂ = 42.05 m

Given that d₂ = 42.05 m < 50 m (The distance separating Melvin's initial location and the deer, Melvin does not hit the deer.

3 0
3 years ago
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