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ruslelena [56]
2 years ago
8

An air-track cart with mass m1=0.28kg and initial speed v0=0.75m/s collides with and sticks to a second cart that is at rest ini

tially. if the mass of the second cart is m2=0.43kg, how much kinetic energy is lost as a result of the collision?
Physics
1 answer:
arsen [322]2 years ago
7 0
Kinetic energy is calculated through the equation,

   KE = 0.5mv²

At initial conditions,

  m₁:  KE = 0.5(0.28 kg)(0.75 m/s)² = 0.07875 J

  m₂ : KE = 0.5(0.45 kg)(0 m/s)² = 0 J

Due to the momentum balance,

   m₁v₁ + m₂v₂ = (m₁ + m₂)(V)

Substituting the known values,

   (0.29 kg)(0.75 m/s) + (0.43 kg)(0 m/s) = (0.28 kg + 0.43 kg)(V)

   V = 0.2977 m/s

The kinetic energy is,
   KE = (0.5)(0.28 kg + 0.43 kg)(0.2977 m/s)²
   KE = 0.03146 J

The difference between the kinetic energies is 0.0473 J. 
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What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56×104v/m and a magnetic fiel
sashaice [31]

1) 3.38\cdot 10^6 m/s

When both the electric field and the magnetic field are acting on the electron normal to the beam and normal to each other, the electric force and the magnetic force on the electron have opposite directions: in order to produce no deflection on the electron beam, the two forces must be equal in magnitude

F_E = F_B\\qE = qvB

where

q is the electron charge

E is the magnitude of the electric field

v is the electron speed

B is the magnitude of the magnetic field

Solving the formula for v, we find

v=\frac{E}{B}=\frac{1.56\cdot 10^4 V/m}{4.62\cdot 10^{-3} T}=3.38\cdot 10^6 m/s

2) 4.1 mm

When the electric field is removed, only the magnetic force acts on the electron, providing the centripetal force that keeps the electron in a circular path:

qvB=m\frac{v^2}{r}

where m is the mass of the electron and r is the radius of the trajectory. Solving the formula for r, we find

r=\frac{mv}{qB}=\frac{(9.1 \cdot 10^{-31} kg)(3.38\cdot 10^6 m/s)}{(1.6\cdot 10^{-19} C)(4.62\cdot 10^{-3}T)}=4.2\cdot 10^{-3} m=4.1 mm

3) 7.6\cdot 10^{-9}s

The speed of the electron in the circular trajectory is equal to the ratio between the circumference of the orbit, 2 \pi r, and the period, T:

v=\frac{2\pi r}{T}

Solving the equation for T and using the results found in 1) and 2), we find the period of the orbit:

T=\frac{2\pi r}{v}=\frac{2\pi (4.1\cdot 10^{-3} m)}{3.38\cdot 10^6 m/s}=7.6\cdot 10^{-9}s

7 0
2 years ago
What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?
alukav5142 [94]

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

5 0
2 years ago
A crane uses a block and tackle to lift a 2200N flagstone to a height of 25m
Cloud [144]

Remember the headline:  ENERGY IS NEVER CREATED OR DESTROYED

The amount of energy before and after are always equal.  All we ever do with energy is move it around from one place to another.

a). A crane can't create energy.  Lifting the same rock in 20 different ways always takes the <u><em>same amount of work</em></u>.  It doesn't matter whether one person picks the rock straight up, or 50 people get around it and lift it, or roll it up a ramp, or lift it with 16 pulleys and a mile of rope, or use a giant steam crane.

You want to lift a 2200N weight up 25m, you're going to have to supply

(2200N) x (25m) = <em>55,000 Joules</em> of work.

c). YOU put out 55,000 Joules of energy.  It had to GO someplace. Where is it now ? ===>  It's the potential energy the rock has now, from being 25m higher than it was before.  That <em>55,000 Joules</em> is NOW the potential energy  of the rock.

No energy was created or destroyed.  It just got moved around.  

55,000 Joules of energy began as nuclear energy in the core of the sun. Solar radiation carried it to the Earth. Plants absorbed it, and stored it as chemical energy.  You ... or a cow that you ate later ... ate the plants and took the chemical energy.  One way or the other, the chemical energy got stored in your blood and fat.  When you needed to put it out somewhere, you moved it into your muscles, and they converted it into mechanical energy.  Then you used the mechanical energy to exert forces.  Today, you used the original 55,000 joules to lift the flagstone, and NOW that energy is in the flagstone, 25 meters up off the ground !

6 0
2 years ago
If a current of 2.4 a is flowing in a cylindrical wire of diameter 2.0 mm, what is the average current density in this wire?
Gnom [1K]

The average current density in the wire is given by:

J=\frac{I}{A}

where I is the current intensity and A is the cross-sectional area of the wire.


The cross-sectional area of the wire is given by:

A=\pi r^2

where r is the radius of the wire. In this problem, r=\frac{d}{2}=\frac{2.0 mm}{2}=1.0 mm=0.001 m, so the cross-sectional area is

A=\pi (0.001 m)^2=3.14 \cdot 10^{-6} m^2


and the average current density is

J=\frac{I}{A}=\frac{2.4 A}{3.14 \cdot 10^{-6} m^2}=7.64 \cdot 10^5 A/m^2

8 0
2 years ago
Read 2 more answers
Tyler stands at rest on a skateboard. He has a mass of 120 kg. His friend (m = 60 kg) jumps into his arms at a speed of 2 m/s. I
Andrews [41]
Momentum question. This is an inelastic collision, so 

m1v1+m2v2=Vf(m1+m2)
Vf=(m1v1+m2v2)/(m1+m2)=[(120kg)(0m/s)+(60kg)(2m/s)] / (120kg+60kg)
Vf=120kg m/s  /   180kg
Vf=0.67m/s

0.67m/s
5 0
2 years ago
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