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N76 [4]
2 years ago
10

An astronaut hits a golf ball of mass m on the Moon, where there is no atmosphere and the acceleration due to gravity is g 6 , w

here g is the acceleration due to gravity on Earth. Assume that the golf club is in contact with the ball for a time t. Just after losing contact with the club, the ball has an initial speed v directed at an angle T above the horizontal. 6. What is the magnitude of the average force exerted by the club on the ball during contact
Physics
1 answer:
sasho [114]2 years ago
5 0

Answer:

The magnitude of the average force exerted by the club on the ball during contact = mv/t

Explanation:

Impulse exerted on the ball = Momentum of the ball = mass * velocity = m*v

As we know,  

m*v = Integration of F.dt with limits 0 to T

Ft = mv

F = mv/t

The magnitude of the average force exerted by the club on the ball during contact = mv/t

You might be interested in
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined
MrRa [10]

Incomplete question as the car's  speed is missing.I have assumed car's  speed as 6.0m/s.The complete question is here

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s

Answer:

F_{B}=-5755N

Explanation:

Set up force equation

∑F=ma

∑F=W+FB

\frac{mv^{2} }{R}=W+F_{B}\\  F_{B}=\frac{mv^{2} }{R}-W\\F_{B}=\frac{(W/g)v^{2} }{R}-W\\F_{B}=\frac{(6000N/9.8m/s^{2} )(6m/s)^{2} }{(15m)}-6000N\\F_{B}=-5755N

The minus sign for downward direction

6 0
2 years ago
An object of mass m has these three forces acting on it (there is no normal force, "no surface"). F1 = 1 N, F2 = 9 N, and F3 = 5
hammer [34]

Answer:

solved

Explanation:

a) F_net = (F2 - F3)i - F1 j

b) |Fnet| = sqrt( (F2 - F3)^2 + F1^2)

= sqrt( (9- 5)^2 + 1^2)

= 4.123 N

c) θ = tan^-1( (Fnet_y/Fnet_x)

= tan^-1( -1/(9-5) )

= -14.036°

7 0
2 years ago
A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the
Oksana_A [137]

Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

Distance= S= 12 cm = 0.12 m

using

2aS= Vf2 - Vi2

==> 2a ×0.12 = o- 30 × 30

==> a = 900 ÷ 0.24 = 3750 m/sec2

Now we know;

F = ma

F= -Kx

==> ma= -kx

==> 20 × 3750 = -K × 0.12

==> k = 625000 N/ m

5 0
2 years ago
Two objects, X and Y, move toward one another and eventually collide. Object X has a mass of 2M and is moving at a speed of 2v0
Sergeu [11.5K]

Answer:

c. The force exerted by X on Y is F to the right, and the force exerted by Yon X is F to the left.

Explanation:

If we take both objects as one single system, during the collision, assuming no external forces acting, the only forces present are the one that object X exerts on object Y, and the force that object Y does on object X.

These two forces, according to Newton's 3rd Law, form an action-reaction pair, and consequently, are equal in magnitude, acting in opposite directions.

As the object X is moving to the right, the force that produces, (F), is in the same direction (on Y), while for object Y, moving to the left, the force that produces (F also in magnitude) is in the same direction (on X), so the right answer is c.

The effect of the forces is different, due to masses are different, according Newton's 2nd Law.

6 0
2 years ago
The weight of a 72.0 kg astronaut on the Moon, where g = 1.63 m/s2 is (5 points) Select one: a. 112 N b. 117 N c. 135 N d. 156 N
kipiarov [429]

Answer: The weight of a 72.0 kg astronaut on the Moon is 117.36 N.

Explanation:

Mass of the astronaut on the moon , m= 72 kg

Acceleration due to gravity on moon,g  = 1.63 m/s^2

According to Newton second law of motion: F = ma

This will changes to = Weight = mass × g

Weight=72 kg\times 1.63m/s^2=117.36 N

The weight of a 72.0 kg astronaut on the Moon is 117.36 N.

7 0
2 years ago
Read 2 more answers
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