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Akimi4 [234]
3 years ago
9

You are standing on a skateboard, initially at rest. A friend throws a very heavy ball towards you. You can either catch the obj

ect or deflect the object back towards your friend (such that it moves away from you with the same speed as it was originally thrown). What should you do in order to MINIMIZE your speed on the skateboard?
Choose one of the following three answers.
A) Your final speed on the skateboard will be the same regardless whether you catch the ball or deflect the ball.
B) You should catch the hall.
B) You should deflect the ball
Physics
1 answer:
kicyunya [14]3 years ago
5 0

Answer:

B) You should catch the ball.

Explanation:

In order to minimize the speed on the skateboard, you should catch the ball.

Let 'M' be the mass of the ball, 'm' the mass of the skater, 'V0' the speed the ball was thrown and 'Vs' the speed of the skater.

By conservation of momentum, if the skater catches the ball:

M*V_{0} + m*0 = (m+M)*V_{s}\\V_{s} = \frac{M*(V_{0} - V_{s})}{(m)}\\

If the skater deflects the ball, it will assume a new velocity with direction opposite to that of the skater, let it be 'Vb':

M*V_{0} + m*0 = m*V_{s}-M*V_{b}\\V_{s} = \frac{M*V_{0}+ M*V_{b}}{(m)}\\V_{s} = \frac{M(V_{0}+ V_{b})}{m}

Since:

V_{0} - V_{s} < V_{0}+ V_{b}

A lower velocity would be achieved by catching the ball.

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Explanation:

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3 years ago
A solid ball is released from rest and slides down a hillside that slopes downward at an angle 51.0 ∘ from the horizontal. what
Romashka [77]
In order for the object not to slip, the component of the weight parallel to the surface must be equal to the frictional force (which acts in the opposite direction):
F_{//}= F_a

The parallel component of the weight is:
F_{//} = mg \sin \alpha
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The frictional force is
F_a = \mu m g \cos \alpha
where \mu is the coefficient of static friction.

Equalizing the two forces, we have
mg \sin \alpha = \mu m g \cos \alpha
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3 years ago
Every winter i fly to Chicago.vit takes 3 hours. what is my average speed​
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3 0
4 years ago
A cat with a mass of 5 kg sits on a branch that is 26 m off the ground. What is
Ostrovityanka [42]

Answer:

1274 J.

Explanation:

P. E= mgh

where m = 5 kg

g = 9.8m/s²

h = 26m

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4 0
4 years ago
A stationary 500 kg tank fires a 20 kg miegile at 100 m/s. What is the velocity of the tank after the missile is fired? Assume t
dedylja [7]

Answer:

v₁ = 4 [m/s].

Explanation:

This problem can be solved by using the principle of conservation of linear momentum. Where momentum is preserved before and after the missile is fired.

P=m*v

where:

P = linear momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

(m_{1}*v_{1})=(m_{2}*v_{2})

where:

m₁ = mass of the tank = 500 [kg]

v₁ = velocity of the tank after firing the missile [m/s]

m₂ = mass of the missile = 20 [kg]

v₂ = velocity of the missile after firing = 100 [m/s]

(500*v_{1})=(20*100)\\v_{1}=2000/500\\v_{1}=4[m/s]

8 0
3 years ago
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