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dlinn [17]
2 years ago
8

A girl is standing still in the middle of a baseball field. The girl catches the baseball and begins to move in the same directi

on as the thrown ball. What type of collision is being described? inelastic, since the girl moves in the same direction as the thrown ball inelastic, since the girl moves in the opposite direction of the thrown ball elastic, since the girl moves in the same direction as the thrown ball elastic, since the girl moves in the opposite direction of the thrown ball
Physics
2 answers:
dalvyx [7]2 years ago
6 0

Answer:

inelastic, since the girl moves in the same direction as the thrown ball

Explanation:

When two bodies collide then after collision if two bodies separate from each other and move with two different velocities then this is an example of elastic collision.

While if two bodies collide and after collision they stick with each other and moves in same direction with same velocity then this is an example of perfectly inelastic collision.

So here in the given case ball is caught by the girl and that the ball and girl move together with same speed in the initial direction of motion of ball. This shows the situation of perfectly inelastic collision

So here correct answer will be

inelastic, since the girl moves in the same direction as the thrown ball

Diano4ka-milaya [45]2 years ago
4 0
The answer to this question is the first option which states that, inelastic since the girl moves in the same direction as the thrown ball. Inelastic collision involves two or more bodies sticking together and moving as one mass after the collision. In this collision the kinetic energy is not conserved but the momentum is conserved. 
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You and your friend Peter are putting new shingles on a roof pitched at 20degrees . You're sitting on the very top of the roof w
Anit [1.1K]

Answer:

v₀ =3.8 m/s

Explanation:

Newton's second law of the box:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Known data

m=2.1 kg  mass of the box

d= 5.4m  length of the roof

θ = 20° angle θ of the roof with respect to the horizontal direction

μk= 0.51 : coefficient of kinetic friction between the box and the roof  

g = 9.8 m/s² : acceleration due to gravity

Forces acting on the box

We define the x-axis in the direction parallel to the movement of the box on the roof  and the y-axis in the direction perpendicular to it.

W: Weight of the box  : In vertical direction

N : Normal force : perpendicular to the direction the  roof

fk : Friction force: parallel to the direction to the roof

Calculated of the weight  of the box

W= m*g  =  (2.1 kg)*(9.8 m/s²)= 20.58 N

x-y weight components

Wx= Wsin θ= (20.58)*sin(20)° =7.039 N

Wy= Wcos θ =(20.58)*cos(20)°= 19.34 N

Calculated of the Normal force

∑Fy = m*ay    ay = 0

N-Wy= 0

N=Wy =19.34 N

Calculated of the Friction force:

fk=μk*N= 0.51* 19.34 N = 9.86 N

We apply the formula (1) to calculated acceleration of the block:

∑Fx = m*ax  ,  ax= a  : acceleration of the block

Wx-f = ( 2.1)*a

7.039 - 9.86  = ( 2.1)*a

-2.821 = ( 2.1)*a

a=(-2.821) /( 2.1)

a= -1.34  m/s²

Kinematics of the box

Because the box moves with uniformly accelerated movement we apply the following formula to calculate the final speed of the block :

vf²=v₀²+2*a*d Formula (2)

Where:  

d:displacement  = 5.4 m

v₀: initial speed  

vf: final speed  = 0

a : acceleration of the box = -1.34  m/s²

We replace data in the formula (2)

0²=v₀²+2*(-1.34)*(5.4)

2*(1.34)*(5.4)= v₀²

v_{o} =\sqrt{14.472}

v₀ = 3.8 m/s

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2 years ago
Two runners ran side by side each holding one end of a horizontal pole. What would most likely happen if one of the runners bega
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Answer:

See explanation.

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a 4357 kg roller coaster car starts from rest at the top of a 36.5 m high track. determine the speed of the car at the top of a
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A 5.00-g bullet is shot through a 1.00-kg wood block suspended on a string 2.00 m long. The center of mass of the block rises a
o-na [289]

Answer:395.6 m/s

Explanation:

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mass of wood block M=1 kg

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let v_1 and v_2 be the velocity of bullet and block after collision

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Now after the collision block rises to an height of 0.38 cm

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5\times 450=5\times v_1+1000\times 0.272

v_1=395.6 m/s

4 0
2 years ago
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