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lina2011 [118]
3 years ago
9

A 10.0 g bullet moving at 300m/s is fired into a 1.00 kg block at rest. The bullet emerges (the bullet does not get embedded in

the block) with half of its orginal speed. What is the velocity of the block right after the collision?
Physics
1 answer:
chubhunter [2.5K]3 years ago
5 0

Answer:

v' = 1.5 m/s

Explanation:

given,

mass of the bullet, m = 10 g

initial speed of the bullet, v = 300 m/s

final speed of the bullet after collision, v' = 300/2 = 150 m/s

Mass of the block, M = 1 Kg

initial speed of the block, u = 0 m/s

velocity of the block after collision, u' = ?

using conservation of momentum

 m v + Mu = m v' + M u'

 0.01 x 300 + 0 = 0.01 x 150 + 1 x v'

v' = 0.01 x 150

v' = 1.5 m/s

Speed of the block after collision is equal to v' = 1.5 m/s

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shtirl [24]

Consider the motion of the car before brakes are applied:

v₀ = maximum initial velocity of the car before the brakes are applied

t = reaction time = 0.50 s

x₀ = distance traveled by the car before brakes are applied

since car moves at constant speed before brakes are applied

Using the equation

x₀ = v₀ t

x₀ = v₀  (0.50)


Consider the motion after brakes are applied :

v₀ = initial velocity of the car before the brakes are applied

a = acceleration = - 10 m/s²

v = final velocity of the car after it comes to stop = 0 m/s

x = stopping distance = initial distance - distance traveled before applying the brakes = 38 - x₀ = 38 - v₀  (0.50)

Using the equation

v² = v²₀ + 2 a x

inserting the values

0² = v²₀ + 2 (- 10) (38 - v₀  (0.50))

v²₀ = 20 (38 - v₀  (0.50))

v₀ = 23 m/s



3 0
3 years ago
2.
gizmo_the_mogwai [7]

Answer:

a) P1=100kpa

V1=6m³

V2=?

P2=50kpa

rearranging mathematically the expression for Boyle's law

V2=(P1V1)/P2=(100×6)/50=12m³

b) same apartment as in (a) but only the value of P2 changes

=> V2=(100×6)/40=15m³

Explanation:

since temperature is not changing we use Boyle's law. mathematically expressed as P1V1=P2V2

4 0
3 years ago
Use your knowledge of waves and the graph shown to determine the frequency of wave C. What does it tell you about speed of the w
stealth61 [152]

Answer

10

Explanation:

goes up by 10 each time 10 to 20 to 30

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2 years ago
a container of water is knocked off a 10.0 meter high ledge with a horizontal velocity of 1.00 meters/second. calculate the time
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Answer:

1.43 s

Explanation:

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The vertical distance covered by an object in free fall is given by

S=ut + \frac{1}{2}at^2

where

u = 0 is the initial vertical speed

t is the time

a= g = 9.8 m/s^2 is the acceleration

since u=0, it can be rewritten as

S=\frac{1}{2}gt^2

And substituting S=10.0 m, we can solve for t, to find the duration of the fall:

t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(10.0 m)}{9.8 m/s^2}}=1.43 s

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