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nirvana33 [79]
3 years ago
7

Two person A and B of weight 60kg and 40kg respectively stand facing each other, and pull on a light rope streched between them.

Find the ratio of distance covered by them when they meet if their intial separation was 5 metre
Physics
1 answer:
goblinko [34]3 years ago
6 0

To be able to successfully pull the rope with balance, the work done by person A should be just equal to the work done by person B.

<span>workA = workB  ----> 1</span>

Where: work = force * distance

In this case, force = mass * gravity

So work = mass * gravity * distance

Rewriting equation 1:

<span>massA * gravity * distanceA = massB * gravity * distanceB  ---->2</span>

We know that distanceA + distanceB = 5 m, therefore distanceA = 5 – distanceB. Substituting this to equation 2 and cancelling constant g:

massA * distanceA = massB * distanceB

60 kg * (5 – distanceB) = 40 kg * distanceB

distanceB = 3 m

distanceA = 5 – distanceB = 2 m

 

<span>ratio is: distanceA / distanceB = 2/3 = 0.67</span>

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Answer:

The angle through which the wheel turned is 947.7 rad.

Explanation:

initial angular velocity, \omega _i = 33.3 rad/s

angular acceleration, α = 2.15 rad/s²

final angular velocity, \omega_f = 72 rad/s

angle the wheel turned, θ = ?

The angle through which the wheel turned can be calculated by applying the following kinematic equation;

\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\\theta = \frac{\omega_f^2\   -\  \omega_i^2}{2\alpha } \\\\\theta = \frac{(72)^2\   -\  (33.3)^2}{2(2.15)}\\\\\theta = 947.7 \ rad

Therefore, the angle through which the wheel turned is 947.7 rad.

5 0
2 years ago
A certain quantity of steam has a temperature of 100.0 oC. To convert this steam into ice at 0.0 oC, energy in the form of heat
KonstantinChe [14]

Answer:

2452.79432 m/s

Explanation:

m = Mass of ice

L_s = Latent heat of steam

s_w = Specific heat of water

L_i = Latent heat of ice

v = Velocity of ice

\Delta T = Change in temperature

Amount of heat required for steam

Q_1=mL_s\\\Rightarrow Q_1=m(2.256\times 10^6)

Heat released from water at 100 °C

Q_2=ms_w\Delta T\\\Rightarrow Q_2=m4186\times (100-0)\\\Rightarrow Q_2=m0.4186\times 10^6

Heat released from water at 0 °C

Q_3=mL_i\\\Rightarrow Q_3=m(333.5\times 10^3)\\\Rightarrow Q_3=m(0.3335\times 10^6)

Total heat released is

Q=Q_1+Q_2+Q_3\\\Rightarrow Q=m(2.256\times 10^6)+m0.4186\times 10^6+m(0.3335\times 10^6)\\\Rightarrow Q=3008100m

The kinetic energy of the bullet will balance the heat

K=Q\\\Rightarrow \frac{1}{2}mv^2=3008100m\\\Rightarrow v=\sqrt{2\times 3008100}\\\Rightarrow v=2452.79432\ m/s

The velocity of the ice would be 2452.79432 m/s

6 0
3 years ago
How would decreasing the volume of the reaction vessel affect each of the following equilibria?2NOBr(g)⇌2NO(g)+Br2(g)
mafiozo [28]

Answer:

The equilibrium position will shift towards the left hand side or reactants side

Explanation:

Decreasing the volume (increasing the pressure) of the system will shift the equilibrium position towards the lefthand side or reactants side. This is because, decreasing the volume (increasing the pressure) implies shifting the equilibrium position towards the side having the least number of moles.

There are two moles of reactants and a total of three moles of products(total). Hence decreasing the volume and increasing the pressure of the gas phase reaction will shift the equilibrium position towards the lefthand side.

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It was  <em>(1/2) (Net force on the cart) m/s²) </em>.

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ale4655 [162]

Answer:

TRUE

Explanation:

The answer is true.

Balance forces acting on a body will not change the motion of the body because the body experiences no net resultant force in one direction. When any body experiences equal forces with opposite directions, the net force or the resultant force experience by the body is zero.

In case of an unbalanced forces, there is a net force acting in one direction and so it causes the body to change in its state of motion in the direction of the net force.

4 0
3 years ago
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