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2 years ago

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A hockey puck of mass m traveling along the x axis at 4.5 m/s hits another identical hockey puck at rest. If after the collision

the second puck travels at a speed of 3.5 m/s at an angle of 30° above the x axis, is this an elastic collision?
A) Yes, since momentum is conserved.
B) No, since momentum is not conserved.
C) Yes, since kinetic energy is conserved.
D) No, since kinetic energy is not conserved.
E) Not enough information.

1 answer:

TEA [102]2 years ago

5 0

Answer:

D) No, since kinetic energy is not conserved.

Explanation:

Since momentum is always conserved in all collision

so in Y direction we can say

$0 = m(3.5 sin30) - mv_y$

$v_y = 1.75 m/s$

Now similarly in X direction we will have

$m(4.5) = m(3.5 cos30 ) + mv_x$

$v_x = 1.47 m/s$

now final kinetic energy of both puck after collision is given as

$KE_f = \frac{1}{2}m(3.5^2) + \frac{1}{2}m(1.75^2 + 1.47^2)$

$KE_f = 8.73 m$

initial kinetic energy of both pucks is given as

$KE_i = \frac{1}{2}m(4.5^2) + 0$

$KE_i = 10.125 m$

since KE is decreased here so it must be inelastic collision

D) No, since kinetic energy is not conserved.

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A group of science and engineering students embarks on a quest to make an electrostatic projectile launcher. For their first tri

vekshin1

Electric charge on the plastic cube: $1.3\cdot 10^{-7}C$

Explanation:

The electric potential around a charged sphere (such as the Van der Graaf) generator is given by

$V(r)=\frac{kQ}{r}$

where

k is the Coulomb's constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

Here we have:

V = 200,000 V on the surface of the sphere, so at r = 12.0 cm

We need to find the voltage V' at 2.0 cm from the edge of the sphere, so at

r' = 12.0 + 2.0 = 14.0 cm

Since the voltage is inversely proportional to r, we can use:

$Vr=V'r'\\V'=\frac{Vr}{r'}=\frac{(200,000)(12.0)}{14.0}=171,429 V$

This is the potential at the location of the plastic cube.

Now we can use the law of conservation of energy, which states that the initial electric potential energy of the cube is totally converted into kinetic energy when the plastic cube is at infinite distance from the generator. So we can write:

$qV' = \frac{1}{2}mv^2$

where:

q is the charge on the plastic cube

V' is the potential at the location of the cube

m = 5.0 g = 0.005 kg is the mass of the cube

v = 3.0 m/s is the final speed of the cube

Solving for q, we find the charge on the cube:

$q=\frac{mv^2}{2V'}=\frac{(0.005)(3.0)^2}{2(171,429)}=1.3\cdot 10^{-7}C$

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2 years ago

1.5 kg of air within a piston-cylinder assembly executes a Carnot power cycle with maximum and minimum temperatures of 800 K and

liraira [26]

Answer:

Explanation:

Carton cycle consists of four thermodynamic processes . The first is isothermal expansion at higher temperature , then adiabatic expansion which lowers the temperature of gas . The third process is isothermal compression at lower temperature and the last process is adiabatic compression which increases the temperature of the gas to its original temperature .

So the given process of isothermal compression must have been done at the temperature of 300K , keeping the temperature constant .

Work done on gas at isothermal compression is equal to heat transfer .

work done on gas = 80 x 10³ J

work done on gas = n RT ln v₁ / v₂

n is number of moles v₁ and v₂ are initial and final volume

molecular weight of gas = 28.97 g

1.5 kg = 1500 / 28.97 moles

= 51.77 moles

work done on gas = n RT ln v₁ / v₂

Putting the values in the equation above

80 x 10³ = 51.78 x 8.31 x 300 x ln v₁ / .2

ln v₁ / .2 = .62

v₁ / .2 = 1.8589

v₁ = 0.37 m³

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2 years ago

Two movers use a rope system to lift a box to a third-story apartment. They do 1,200 J of work on the rope system, and the rope

xeze [42]

Efficiency is defined as the measure of the amount of work or energy is conserved in a certain process. At all times, in every process, work or energy is always lost or wasted due to certain interference. Not all work given is converted to useful work or energy. Thus , efficiency is calculated by dividing the energy or work output to the energy or work input then the value is multiplied by 100 to express efficiency as percentage.

Efficiency = work output / work input
Efficiency = (1020 J / 1200 J) = 85%

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2 years ago

Read 2 more answers

A 0.40-kg cart with charge 4.0 x 10-5 C starts at rest on a horizontal frictionless surface 0.50 m from a fixed object with char

maw [93]

Answer:

26.82m/s

Explanation:

Given

Mass = m= 0.4kg

Initial Velocity = u = 0

Charge = 4.0E-5C

Distance= d = 0.5m

Object Charge = 2E-4C

First, we'll calculate the initial energy (E)

E = Potential Energy

PE = kQq / d

Where k = coulomb constant = 8.99E9Nm²/C²

Energy is then calculated by;

PE = 8.99E9 * 4E-5 * 2E-4 / 0.5

PE = 143.84J

Energy = Potential Energy = Kinetic Energy

K.E = ½mv² = 143.84J

½mv² = ½ * 0.40 * v² = 143.85

0.2v² = 143.85

v² = 143.85/0.2

v² = 719.25

v = √719.25

v = 26.81883666380777

v = 26.82m/s

Hence, the object is 26.82m/s fast when the cart moving is very far (infinity) from the fixed charge

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2 years ago

A 100-N force causes an object to<br> accelerate at 2 m/s/s. What is the<br> mass of the object?

Kitty [74]

Answer:

$50\; \rm kg$ .

Explanation:

By Newton's Second Law, the acceleration $a$ of an object is proportional to the net force $\sum F$ on it. In particular, if the mass of the object is $m$ , then

$\sum F = m \cdot a$ .

Rewrite this equation to obtain:

$\displaystyle m = \frac{\sum F}{a}$ .

In this case, the assumption is that the $100\; \rm N$ force is the only force that is acting on the object. Hence, the net force $\sum F$ on the object would also be

Make sure that all values are in their standard units. Forces should be in Newtons (same as $\rm kg \cdot m \cdot s^{-2}$ , and the acceleration of the object should be in meters-per-second-squared ( $\rm m \cdot s^{-2}$ ). Apply the equation $\displaystyle m = \frac{\sum F}{a}$ to find the mass of the object.

$\displaystyle m = \frac{100\; \rm N}{2\; \rm m \cdot s^{-2}} = 50\; \rm kg$ .

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2 years ago