Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle. The horizontal surface on which block A (mass 2.10 kg) moves is frictionless. The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s. a)What is the tension force that the rope exerts on block B? b)What is the tension force that the rope exerts on block A? c)What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?

Answers

Answer 1

Answer:

The tension force that the rope exerts on block B is 62.3 N, the tension force that the rope exerts on block A is 1.89 N, and the moment of inertia of the pulley for rotation about the axle on which it is mounted is $\rm 0.430 \; kg\;m^2$ .

Given :

Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope.
The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle.
The horizontal surface on which block A (mass 2.10 kg) moves is frictionless.
The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s.

a) First, determine the acceleration of the B block.

$\rm s = ut + (1)/(2)at^2$

$\rm 1.8 = (1)/(2)* a* (2)^2$

$\rm a = 0.9\; m/sec^2$

Now, apply Newton's second law of motion in order to determine the tension force that the rope exerts on block B.

$\rm \sum F=ma$

$\rm mg-T_b=ma$

$\rm T_b = m(g-a)$

$\rm T_b = 7* (9.8-0.9)$

$\rm T_b = 62.3\;N$

b) Now, again apply Newton's second law of motion in order to determine the tension force that the rope exerts on block A.

$\rm \sum F=ma$

$\rm T_a=ma$

$\rm T_a = 2.1* 0.9$

$\rm T_a = 1.89\;N$

c) The sum of the torque in order to determine the moment of inertia of the pulley for rotation about the axle on which it is mounted.

$\rm \sum \tau = I\alpha$

$\rm T_br-T_ar = I\alpha$

$\rm I = ((T_b-T_a)r^2)/(a)$

Now, substitute the values of the known terms in the above expression.

$\rm I = ((62.3-1.89)(0.080)^2)/(0.90)$

$\rm I = 0.430 \; kg\;m^2$

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Answer 2

Answer:

Answer:

(a) 62.3 N

(b) 1.89 N

Explanation:

(a) Find the acceleration of block B.

Δy = v₀ t + ½ at²

1.80 m = (0 m/s) (2.00 s) + ½ a (2.00 s)²

a = 0.90 m/s²

Draw a free body diagram of block B. There are two forces:

Weight force mg pulling down,

and tension force Tb pulling up.

Sum of forces in the -y direction:

∑F = ma

mg − Tb = ma

Tb = m (g − a)

Tb = (7.00 kg) (9.8 m/s² − 0.90 m/s²)

Tb = 62.3 N

(b) Draw a free body diagram of block A. There are three forces:

Weight force mg pulling down,

Normal force N pushing up,

and tension force Ta pulling right.

Sum of forces in the +x direction:

∑F = ma

Ta = ma

Ta = (2.10 kg) (0.90 m/s²)

Ta = 1.89 N

Tension force Tb pulling down,

and tension force Ta pulling left.

Sum of torques in the clockwise direction:

∑τ = Iα

Tb r − Ta r = Iα

(Tb − Ta) r = I (a/r)

I = (Tb − Ta) r² / a

I = (62.3 N − 1.89 N) (0.080 m)² / (0.90 m/s²)

I = 0.430 kg m²

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Answers

Answer:

Explanation:

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You have a battery marked " 6.00 V ." When you draw a current of 0.361 A from it, the potential difference between its terminals is 5.07 V . What is the potential difference when you draw 0.591 A

Answers

Answer:

Explanation:

Battery voltage is 6V

A current of 0.361A is draw the voltage reduces to, 5.07V

This shows that the appliances resistance that draws the currents is

Using KVL

The battery has an internal resistance r

V=Vr+Va

Vr is internal resistance voltage

Va is appliance voltage

6=5.07+Va

Va=6-5.07

Va=0.93

Using ohms law to the resistance of the appliance

Va=iR

R=Va/i

R=0.93/0.361

R=2.58ohms

Then if the circuit draws a current of 0.591A

Then the voltage across the load is

V=iR

Va=0.591×2.58

Va=1.52V

Then the voltage drop at the internal resistance is

V=Vr+Va

Vr=V-Va

Vr=6-1.52

Vr=4.48V

Answer:

V = 4.48 V

Explanation:

• As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.

• We can calculate this loss, applying Ohm's law to the internal resistance, as follows:

$V_(rint) = I* r_(int)$

• The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

$V = V_(b) - V_(rint) = 5.07 V = 6.00 V - 0.361 A * r_(int)$

• We can solve for rint, as follows:

$r_(int) =(V_(b)- V_(rint) )/(I) = (6.00 V - 5.07 V)/(0.361A) = 2.58 \Omega$

• When the circuit draws from battery a current I of 0.591A, we can find the potential difference between the terminals of the battery, as follows:

$V = V_(b) - V_(rint) = 6.00 V - 0.591 A * 2.58 \Omega = 4.48 V$

• As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.

How does the force of gravity change as the mass of one object doubles?

Answers

The force of gravity changes as the mass of one object doubles. As the mass of one object is doubled then the force between the objects also gets doubled.

What is Force?

Force is an influence which can change the motion of an object through the application of an external force. A force can cause an object with the mass to change its velocity, that is the object undergo acceleration.

Force is directly proportional to the mass of the object and the acceleration of the object. If we double the mass of one of the objects, then we double the strength of the force. If we double the masses of both the objects, then we quadruple the strength of force.

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If the mass of one of the objects is doubled, then the force of gravity between them is doubled. ... Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces.

A playground merry-go-round of radius R = 2.20 m has a moment of inertia I = 275 kg · m2 and is rotating at 9.0 rev/min about a frictionless vertical axle. Facing the axle, a 23.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round? rev/min

Answers

Answer:

6.4 rpm

Explanation:

$I_(m)$ = moment of inertia of merry-go-round = 275 kgm²

m = mass of the child = 23 kg

R = radius of the merry-go-round = 2.20 m

$I_(c)$ = moment of inertia of child after jumping on merry-go-round = mR² = (23) (2.20)² = 111.32 kgm²

Total moment of inertia after child jumps is given as

$I_(f)$ = $I_(m)$ + $I_(c)$ = 275 + 111.32 = 386.32 kgm²

Total moment of inertia before child jumps is given as

$I_(i)$ = $I_(m)$ = 275 kgm²

$w_(i)$ = initial angular speed = 9 rpm

$w_(f)$ = final angular speed

using conservation of angular momentum

$I_(i)$ $w_(i)$ = $I_(f)$ $w_(f)$

(275) (9) = (386.32) $w_(f)$

$w_(f)$ = 6.4 rpm

A horizontal cylindrical tank 8.00 ft in diameter is half full of oil (60.0 Ib/ft3). Find the force on one end

Answers

Answer:

Assuming h as the height of the cylindrical tank

$F=480\pi h \,g\,\, (lb)/(ft)$

Explanation:

Assuming that the height is $h$ we can find the volume of the cylindrical tank, then:

$V=\pi*r^2*h$

The diameter is 8.00 ft then $r=4.00 ft$ the total volume of the tank is:

$V=\pi (4.00 ft)^2 h=16\pi h\,\, ft^2$

But the tank is half full of oil, then we need half of the volume. For that reason the volume of oil is:

$V_(oil)=(16\pi h)/(2)ft^2=8\pi h \,\,ft^2$

We know the density of the oil $\rho=60.0\,lb/ft^3$ , with this we can fing the mass of oil that we have because:

$\rho=(m)/(V)$ then $m=\rho V$

Then the mass of oil that we have is:

$m=(60.0(lb)/(ft^3))(8\pi h\,\,ft^2)$

$m=480\pi h (lb)/(ft)$

Note that with the value of h we have the mass in correct units.

Finally to find the force we now that $F=mg$ then we just need to multiply the mass by the gravity.

$F=480\pi h \,g\,\, (lb)/(ft)$

Hearing the siren of an approaching fire truck, you pull over to side of the road and stop. As the truck approaches, you hear a tone of 460 Hz; as the truck recedes, you hear a tone of 410 Hz. How much time will it take to jet from your position to the fire 5.00 km away, assuming it maintains a constant speed?