Ask question

Ask question

All categories

2 years ago

14

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.

You might be interested in

rodikova [14]

Answer:

See the explanation below.

Explanation:

12) When an object is falling, how does the objects velocity change? what formula do you use?

The speed of a falling object is increased by a value of 9.81 meters per second per second. That is if we throw any body regardless of mass from a considerable height, its speed in the first second will be 9.81[ m/ s] , in the next second will be equal to 19.62 [m/s] in the next will be equal to 29.43 [m/ s].

The formula is:

$v=v_{0}+g*t$

where:

vo = initial velocity = 0

g = gravity = 9.81[m/s^2]

t = time [s]

13)

what is a falling speed at 6s, 9s, 112s?

v = 0 + (9.81*6) = 58.86[m/s]

v = 0 + (9.81*9) = 88.29 [m/s]

v = 0 + (9*112) = 1098.72 [m/s]

14)

If an object is falling at 65 [m/s]. How long has it been falling ? what is the formula that you use?

The formula is the same:

$v=v_{o}+g*t$

65 = 0 + 9.81*t

t = 65/9.81

t = 6.62[s]

15)

What formula is used to determine the distance an object is falling ?

$y = y_{o}+v_{o}*t + 0.5*9.81*t^{2}$

where:

y = distance [m]

yo = initial distance, in most of the cases and depending of the reference point it will be eqaul to zero

vo = initial velocity, if it is free fall, then = 0

t = time [s]

g = gravity = 9.81[m/s^2]

This equation will be reduce to:

$y = 0.5*g*t^{2}$

16)

using the times given in problem 13. Determine the distance fallen for each.

y = 0.5*9,81*(6)^2 = 176.58 [m]

y = 0.5*9,81*(9)^2 = 397.3 [m]

y = 0.5*9,81*(112)^2 = 61528.3 [m]

17)

If an object has fallen a distance of 87.3 [m]. How long was it falling?

87.3 = 0.5*9.81*t^2

$t=\sqrt{\frac{87.3}{0.5*9.81} }\\ t=4.21[s]$

4 0

2 years ago

While practicing S-turns, a consistently smaller half-circle is made on one side of the road than on the other, and this turn is

erica [24]

Answer:

The answer is "4-5-6 because the bank is growing too quickly in the beginning of its turn".

Explanation:

The S-turns is also known as the reference technique, in which the ground track of the aircraft on both sides of a defined ground-based straight-line distance represents 2 different but equivalent circles.

Throughout the S-turns, on either side of the road, a progressively smaller half-circle is formed, and this turn does not stop until the road or reference line is crossed during the early part of the turn, the twists increase too quickly.

5 0

1 year ago

5.16 An insulated container, filled with 10 kg of liquid water at 20 C, is fitted with a stirrer. The stirrer is made to turn by

Anna007 [38]

Answer:

a) W=2.425kJ

b) $\Delta E=2.425kJ$

c) $T_f=20.06^{o}C$

d) Q=-2.425kJ

Explanation:

a)

First of all, we need to do a drawing of what the system looks like, this will help us visualize the problem better and take the best possible approach. (see attached picture)

The problem states that this will be an ideal system. This is, there will be no friction loss and all the work done by the object is transferred to the water. Therefore, we need to calculate the work done by the object when falling those 10m. Work done is calculated by using the following formula:

$W=Fd$

Where:

W=work done [J]

F= force applied [N]

d= distance [m]

In this case since it will be a vertical movement, the force is calculated like this:

F=mg

and the distance will be the height

d=h

so the formula gets the following shape:

$W=mgh$

so now e can substitute:

$W=(25kg)(9.7 m/s^{2})(10m)$

which yields:

W=2.425kJ

b) Since all the work is tansferred to the water, then the increase in internal energy will be the same as the work done by the object, so:

$\Delta E=2.425kJ$

c) In order to find the final temperature of the water after all the energy has been transferred we can make use of the following formula:

$\Delta Q=mC_{p}(T_{f}-T_{0})$

Where:

Q= heat transferred

m=mass

$C_{p}$ =specific heat

$T_{f}$ = Final temperature.

$T_{0}$ = initial temperature.

So we can solve the forula for the final temperature so we get:

$T_{f}=\frac{\Delta Q}{mC_{p}}+T_{0}$

So now we can substitute the data we know:

$T_{f}=\frac{2 425J}{(10000g)(4.1813\frac{J}{g-C})}+20^{o}C$

Which yields:

$T_{f}=20.06^{o}C$

d)

For part d, we know that the amount of heat to be removed for the water to reach its original temperature is the same amount of energy you inputed with the difference that since the energy is being removed this means that it will be negative.

$\Delta Q=-2.425kJ$

3 0

2 years ago

Calculate the volume of a liquid with a density of 5.45 g/ml and a mass of 65g

katrin [286]

Density=mass/volume
5.45g/ml=65g/V
V=65g/5.42g/ml
V=11.92ml

5 0

2 years ago

8.4-1 Consider a magnetic field probe consisting of a flat circular loop of wire with radius 10 cm. The probe’s terminals corres

Vlad1618 [11]

Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

$\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}$

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

$\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)$

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

$\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T$

hence, B_o = 1.013μT

6 0

2 years ago