Ask question

Ask question

All categories

2 years ago

12

Puck B has twice the mass of puck A. Starting from rest, both pucks are pulled the same distance across frictionless ice by stri

ngs with the same tension.a. Compare the final kinetic energies of pucks A and B. b. Compare the final speeds of pucks A and B.

1 answer:

gogolik [260]2 years ago

8 0

Answer:

(a) 1 : 2

(b) same

Explanation:

Let the mass of puck A is m and the mass of puck B is 2 m.

initial speed for both the pucks is same as u and the distance is same for both is s.

let the tension is T for same.

The kinetic energy is given by

$K = 0.5 mv^2$

(a) As the speed is same, so the kinetic energy depends on the mass.

So, kinetic energy of A : Kinetic energy of B = m : 2m = 1 : 2

(b) A the distance s same so the final velocities are also same.

You might be interested in

Help me please I only have a few minutes to get this doneplease and thank q

DIA [1.3K]

A: particles are more spread out in gas

6 0

2 years ago

A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 110,000 kg of scrap metal in

erica [24]

Explanation:

Given that,

Mass of a freight car, $m_1=30,000-kg$

Speed of a freight car, $u_1=0.85\ m/s$

Mass of a scrap metal, $m_2=110,000\ kg$

(a) Let us assume that the final velocity of the loaded freight car is V. The momentum of the system will remain conserved as follows :

$30000\times 0.85=(30000+110000)V\\\\V=\dfrac{30000\times 0.85}{30000+110000}\\\\=0.182\ m/s$

So, the final velocity of the loaded freight car is 0.182 m/s.

(b) Lost on kinetic energy = final kinetic energy - initial kinetic energy

$\Delta K=\dfrac{1}{2}[(m_1+m_2)V^2-m_1u_1^2)]\\\\=\dfrac{1}{2}\times [(30,000+110,000 )0.182^2-30000(0.85)^2]\\\\=-8518.82\ J$

Lost in kinetic energy is 8518.82. Negative sign shows loss.

6 0

2 years ago

An inverse-time circuit breaker (CB) is used for branch-circuit short-circuit and ground-fault protection for a 30-horsepower, 2

Maru [420]

Answer:

Explanation:

Motor rating is given in horsepower (hp), it will be converted in watt (W).

Standard to install circuit breaker for an electric circuit is usually 20% ~ 25% more than the Rated Current of the circuit

while

Standard to install overload relay for an electric circuit is usually 20% ~ 25% more than the Running Current of the circuit.

So, to find the maximum capacity of the circuit breaker, rated current of the motor will be multiplied by 1.2 ~ 1.25

Step by Step Explanation:

30hp = 22371W (as 1hp = 745.7)

Assuming unity power factor (cosФ=1) and 208V phase to phase voltage:

Rated Power (watt) = √3 . V.I. cosФ

<em>{if 208V is phase to neutral voltage, then use following formula:</em>

<em> Rated Power (watt) = 3 . V.I. cosФ}</em>

$\frac{22371}{\sqrt{3} * 208 * 1} = I$

Rated Current = <u>62.169A</u>

So, required maximum rating for circuit breaker is:

20% to 25% of the rated current = 62.17*1.2 ~ 62.17*1.25

=74.6A ~ 77.7A

Hence, any breaker between the above mentioned rating will be appropriate.

3 0

3 years ago

I think I know this one but I don’t at the same time

galben [10]

Answer: B.) 6

Explanation:

To answer this problem you get the number of students who attended Wednesday (18) and the number of students who attended Tuesday (12) and subtract 18 - 12 = 6

Answer = B.) 6

8 0

3 years ago

You push your couch a distance of 3.9 m across the living room floor with a horizontal force of 220.0 n. the force of friction i

Mrac [35]

The general formula to calculate the work is:

$W=Fd \cos \theta$

where F is the force, d is the displacement of the couch, and $\theta$ is the angle between the direction of the force and the displacement. Let's apply this formula to the different parts of the problem.

(a) Work done by you: in this case, the force applied is parallel to the displacement of the couch, so $\theta=0^{\circ}$ and $\cos \theta=1$ , therefore the work is just equal to the product between the horizontal force you apply to push the couch and the distance the couch has been moved:

$W=Fd=(220.0 N)(3.9 m)=858 J$

(b) work done by the frictional force: the frictional force has opposite direction to the displacement, therefore $\theta=180^{\circ}$ and $\cos \theta=-1$ . Therefore, we must include a negative sign when we calculate the work done by the frictional force:

$W=-Fd=-(144.0 N)(3.9 m)=-561.6 J$

(c) The work done by gravity is zero. In fact, gravity (which points downwards) is perpendicular to the displacement of the couch (which is horizontal), therefore $\theta=90^{\circ}$ and $\cos \theta=0$ : this means

$W=0$ .

(d) Work done by the net force:

The net force is the difference between the horizontal force applied by you and the frictional force:

$F=220 N-144 N=76 N$

And the net force is in the same direction of the displacement, so $\theta=0^{\circ}$ and $\cos \theta=1$ and the work done is

$W=Fd=(76 N)(3.9 m)=296.4 J$

4 0

3 years ago