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Maslowich
3 years ago
13

A 70.0-kg ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. Suppos

e the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came. What would their final velocities be in this case?
Physics
1 answer:
brilliants [131]3 years ago
4 0

Answer:

v₂=- 34 .85 m/s

v₁=0.14 m/s

Explanation:

Given that

m₁=70 kg ,u₁=0 m/s

m₂=0.15 kg ,u₂=35 m/s

Given that collision is elastic .We know that for elastic  collision

Lets take their final speed is v₁ and v₂

From momentum conservation

m₁u₁+m₂u₂=m₁v₁+m₂v₂

70 x 0+ 0.15 x 35 = 70 x v₁ + 0.15 x v₂

70 x v₁ + 0.15 x v₂=5.25                   --------1

v₂-v₁=u₁-u₂            ( e= 1)

v₂-v₁ = -35        --------2

By solving above equations

v₂=- 34 .85 m/s

v₁=0.14 m/s

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A cable that weighs 8 lb/ft is used to lift 900 lb of coal up a mine shaft 650 ft deep. Find the work done. Show how to approxim
Harrizon [31]

Answer:

2275000 lb.ft

Explanation:

Let work done on the cable be denoted by: W_ca

Let work done on the coal be denoted by: W_co

Now, dividing the cable into segments, let x represent the length from top of the mine shaft to the segment.

Meanwhile let δx be the length of the segment.

We are told the cable weighs 8 lb/ft. Thus;

Work done on one segment = 8 × δx × x = 8x•δx

Therefore, work done on cable is;

W_ca = ∫8x•δx between the boundaries of 0 and 650

Thus;

W_ca = 4x² between the boundaries of 0 and 650

W_ca = 4(650²) - 4(0²)

W_ca = 1,690,000 lb.ft

Workdone on the 900 lb of coal will be calculated as;

W_co = 900 × 650

W_co = 585000 lb.ft

Thus,

Total work done = W_ca + W_co

Total workdone = 1690000 + 585000

Total workdone = 1690000 + 585000

Total workdone = 2275000 lb.ft

6 0
3 years ago
angie is moving a couch with a mass of 48 kg. she moves it in 120 seconds using 240 j of work. what is her power output
natta225 [31]
Power =(work done) / (time to do the work) = 240 J / 120 sec = 2 watts.
8 0
3 years ago
Find the ratio of the diameter of aluminium to copper wire, if they have the same
kicyunya [14]

Answer:

1.24

Explanation:

The resistivity of copper\rho_1=2.65\times 10^{-8}\ \Omega-m

The resistivity of Aluminum,\rho_2=1.72\times 10^{-8}\ \Omega-m

The wires have same resistance per unit length.

The resistance of a wire is given by :

R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi (\dfrac{d}{2})^2}\\\\\dfrac{R}{l}=\rho \dfrac{1}{\pi (\dfrac{d}{2})^2}

According to given condition,

\rho_1 \dfrac{1}{\pi (\dfrac{d_1}{2})^2}=\rho_2 \dfrac{1}{\pi (\dfrac{d_2}{2})^2}\\\\\rho_1 \dfrac{1}{{d_1}^2}=\rho_2 \dfrac{1}{{d_2}^2}\\\\(\dfrac{d_2}{d_1})^2=\dfrac{\rho_1}{\rho_2}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{\rho_1}{\rho_2}}\\\\\dfrac{d_2}{d_1}=\sqrt{\dfrac{2.65\times 10^{-8}}{1.72\times 10^{-8}}}\\\\=1.24

So, the required ratio of the diameter of Aluminum to Copper wire is 1.24.

3 0
3 years ago
How can we use magnetic fields to transport objects from one location to another?
sergejj [24]
Wow this is a tough one
7 0
3 years ago
A pair of slits in a double slit experiment are illuminated with monochromatic light of wavelength 480 nm. The slits are separat
Romashka-Z-Leto [24]

Using the appropriate approximations:

dx/L = mλ

d = slit separation

x = fringe spacing

L = distance between slits and screen

m = some integer, used to determine the distance from the central bright fringe to another bright fringe


We don't really need a value for m because we're calculating the distance between any pair of consecutive fringes. Let's just set m = 1


Given values:

d = 1.0mm

L = 2.0m

λ = 480nm


Substitute the terms in the equation with our given values and solve for x:

1.0*10⁻³*x/2 = 480*10⁻9

<h3>x = 0.96mm</h3>
8 0
3 years ago
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