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bixtya [17]
3 years ago
10

two teams are playing tug of war. team a pulls to the right with a force of 450n .team b pulls to the left with a force of 415 n

. what is the net force on the rope and what is its direction?
Physics
1 answer:
Alex_Xolod [135]3 years ago
4 0

Explanation:

It is given that, two teams are playing tug of war.

Force applied by Team A, F_A=450\ N

Force applied by Team B, F_B=415\ N

We need to find the net force acting on the rope. It is equal to :

F_{net}=F_A-F_B

F_{net}=450-415

F_{net}=35\ N

So, the net force acting on the rope is 35 N and it is acting toward right. Hence, this is the required solution.

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A large balloon of mass 210 kg is filled with helium gas until its volume is 329 m3. Assume the density of air is 1.29 kg/m3 and
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(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

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- The weight of the balloon, labelled with W, whose magnitude is

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where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

B=\rho_a V g

where \rho_a is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

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B=\rho_a V g

where \rho_a is the air density, V is the volume of the balloon and g the acceleration due to gravity.

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\rho_a = 1.29 kg/m^3 is the air density

V=329 m^3 is the volume of the balloon

g = 9.8 m/s^2 is the acceleration due to gravity

So the buoyant force is

B=(1.29 kg/m^3)(329 m^3)(9.8 m/s^2)=4159 N

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m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg

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And so, the net force on the balloon is

F=B-W=4159 N-2635 N=1524 N

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where m' is the additional mass. Re-arranging the equation for m', we find

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B=\rho_a V g

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