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SSSSS [86.1K]
3 years ago
8

A 1.72-kg block of soft wood is suspended by two strings from the ceiling. The wood is free to rotate in pendulum-like fashion w

hen a force is exerted upon it. A 8.50-g bullet is fired into the wood. The bullet enters the wood at 431 m/s and exits the opposite side shortly thereafter. If the wood rises to a height of 13.8 cm, then what is the exit speed of the bullet?
Physics
1 answer:
romanna [79]3 years ago
4 0

Answer:

Vf_{2} = 98.21 m/s                    

Explanation:

The block has a final potential energy

U = mgh

U = (1.72)(9.8)(0.138)

    = 2.3261 J

How after of the collision the block's mechanical energy is conserved, then we can calculate the block's initial velocity

     2.3261 = \frac{mv^{2} }{2}

     2.3261 = \frac{1.72v^{2} }{2}

            v = 1.6446 m/s

In the collision is conserved the lineal momentum of the system then:

m_{1}Vo_{1} + m_{2}Vo_{2} = m_{1}Vf_{1} + m_{2}Vf_{2}

(1.72)(0) + (0.0085)(431) = (1.72)(1.6446) + (0.085)Vf_{2}

3.6635 = 2.8287 + 0.0085Vf_{2}

Vf_{2} = 98.21 m/s                    

 

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Riley is cleaning his room(a once a year activity) and pulls the Hoover with a force of 8 N a total distance of 20 metres. How m
Radda [10]

Answer:

160J

Explanation:

Given force = 8N and total distance = 20 meters

Workdone = force x distance

= 8 x 20

= 160J

Therefore, workdone by Riley in pulling the hoover is 160J

7 0
3 years ago
1. You place an object 63 cm in front of a converging lens, with a 40 cm focal length.
NikAS [45]

Answer:

1.

109.6 cm ,  - 1.74 , real

2.

1.5

Explanation:

1.

d₀ = object distance = 63 cm

f = focal length of the lens = 40 cm

d = image distance = ?

using the lens equation

\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}

\frac{1}{40} = \frac{1}{63} + \frac{1}{d}

d = 109.6 cm

magnification is given as

m = \frac{-d}{d_{o}}

m = \frac{-109.6}{63}

m = - 1.74

The image is real

2

d₀ = object distance = a

d = image distance = - (a + 5)

f = focal length of lens = 30 cm

using the lens equation

\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}

\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}

a = 10

magnification is given as

m = \frac{-d}{d_{o}}

m = \frac{- (- (a +5))}{a}

m = \frac{(5 + 10)}{10}

m = 1.5

6 0
3 years ago
What is the density of a block of wood with a mass of 120 g and a volume of 200 cm
creativ13 [48]
D=m/V therefore the answer is 120/200 or 0.6
7 0
3 years ago
A wave with a greater amplitude will transfer . . . . \
BabaBlast [244]

less mass is more mass but less energy in more mass. less mass has more energy

6 0
3 years ago
A baseball is batted from a height of 1.09 m with a speed of
kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s

<h3>Resultant speed of the ball and crosswind</h3>

V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0
2 years ago
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