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jasenka [17]
4 years ago
10

A 5.00 g object moving to the right at 20.0 cm/s makes an elastic head-on collision with a 10.0 g object that is initially at re

st.
(a) Find the velocity of each object after the collision. cm/s (5.00 g object) cm/s (10.0 g object)
(b) Find the fraction of the initial kinetic energy transferred to the 10.0 g object. %
Physics
1 answer:
marusya05 [52]4 years ago
3 0

Answer: a) 6.67cm/s b) 1/2

Explanation:

According to law of conservation of momentum, the momentum of the bodies before collision is equal to the momentum of the bodies after collision. Since the second body was initially at rest this means the initial velocity of the body is "zero".

Let m1 and m2 be the masses of the bodies

u1 and u2 be their velocities respectively

m1 = 5.0g m2 = 10.0g u1 = 20.0cm/s u2 = 0cm/s

Since momentum = mass × velocity

The conservation of momentum of the body will be

m1u1 + m2u2 = (m1+m2)v

Note that the body will move with a common velocity (v) after collision which will serve as the velocity of each object after collision.

5(20) + 10(0) = (5+10)v

100 + 0 = 15v

v = 100/15

v = 6.67cm/s

Therefore the velocity of each object after the collision is 6.67cm/s

b) kinectic energy of the 10.0g object will be 1/2MV²

= 1/2×10×6.67²

= 222.44Joules

kinectic energy of the 5.0g object will be 1/2MV²

= 1/2×5×6.67²

= 222.44Joules

= 111.22Joules

Fraction of the initial kinetic transferred to the 10g object will be

111.22/222.44

= 1/2

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3 years ago
Suppose a straight 1.00-mm-diameter copper wire could just "float" horizontally in air because of the force due to the Earth’s m
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To solve this problem it is necessary to apply the concepts related to the Force since Newton's second law, as well as the concept of Electromagnetic Force. The relationship of the two equations will allow us to find the magnetic field through the geometric relations of density and volume.

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B = Magnetic Field

I = Current

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From Newton's second law we know that the relation of Strength and weight is determined as

F_g = mg

Where,

m = Mass

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For there to be balance the two forces must be equal therefore

F_{mag} = F_g

BIL = mg

Our values are given as,

Diameter (d) = 1.0mm = 1*10^{-3}m

Radius (r) = \frac{d}{2} = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m

Magnetic Field (B) = 5.0*10^{-5} T

From the relationship of density another way of expressing mass would be

\rho = \frac{m}{V} \rightarrow m = \rho V

At the same time the volume ratio for a cylinder (the shape of the wire) would be

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Replacing this two expression at our first equation we have that:

BIL = mg

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BIL = ( \rho \pi r^2 L)g

Re-arrange to find I

I = \frac{( \rho \pi r^2 L)g}{BL}

I = \frac{( \rho \pi r^2 )g}{B}

We have for definition that the Density of copper is 8.9*10^3 Kg/m^3, gravity acceleration is 9.8m/s^2 and the values of magnetic field (B) and the radius were previously given, then:

I = \frac{( (8.9*10^3 ) \pi (0.5*10^{-3})^2 )(9.8)}{5.0*10^{-5}}

I = 1370.05A

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3 years ago
An aquarium open at the top has 30-cm-deep water in it. You shine a laser pointer into the top opening so it is incident on the
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Answer:

You must add 8cm of water to the tank

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In order to find how much the height is we will use the Snell Refraction law

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This law relates the index of refraction of the water (n2), the index of refraction of the air (n1), the incidence angle relative to the vertical (theta1) and the refraction angle relative to the vertical (theta2) by using the next equation:

(n1)*(sin(theta1))=(n2)*(sin(theta2))

Then we will find the refraction angle relative to the vertical this way:

(n1/n2)*(sin(theta1))=sin(theta2)

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Now that we have this information we can imagine a triangle with a 30cm height and a 32.12° angle. This way we can find how much X is, this X will be the distance between the vertical line and the spot the beam hits the bottom, so we can use some trigonometry to find it, this way:

tan(32.12)=(X/30cm)

X=(tan(32.12))*(30cm)

Then, X=18.8cm, we can approximate it to 19cm

Once we have X we will add 5cm to it which is how much the beam needs to be moved, then the new X will be 24cm

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tan(32.12)=(24cm/Y)

Y=(24cm/tan(32.12))

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3 years ago
Can the resistors in an "unbalanced" Wheatstone bridge circuit be treated as a combination of series and/or parallel resistors?
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Answer:

Explanation:

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3 years ago
A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.2m/s and the hatchback ca
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Answer:

The pickup truck and hatchback will meet again at 440.896 m

Explanation:

Let us assume that both vehicles are at origin at the start means initial position is zero i.e. s_{o} = 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.

Truck:

v_{i} = 33.2 m/s, a = 0 (since the velocity is constant), s_{o} = 0

Using s =s_{o}+v_{i}t+1/2at^{2}

s = 33.2t .......... eq (1)

Hatchback:

a=5m/s^{2}, v_{i} = 0 m/s (since initial velocity is zero), s_{o} = 0

Using s =s_{o}+v_{i}t+1/2at^{2}

putting in the data we will get

s=(1/2)(5)t^{2}

now putting 's' value from eq (1)

2.5t^{2}-33.2t=0

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t = 13.28 s

so both vehicles will meet up gain after 13.28 sec.

putting t = 13.28 in eq (1) will give

s = 440.896 m

So, both vehicles will meet up again at 440.896 m.

7 0
3 years ago
Read 2 more answers
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