Physics College

As a torque activity, your Physics TA sets up the arrangement decribed below. A uniform rod of mass mr = 158 g and length L = 100.0 cm is attached to the wall with a pin as shown. Cords are attached to the rod at the r1 = 10.0 cm and r2 = 90.0 cm mark, passed over pulleys, and masses of m1 = 281 g and m2 = 177 g are attached. Your TA asks you to determine the following: (a) The position r3 on the rod where you would suspend a mass m3 = 200 g in order to balance the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Use standard angle notation to determine the direction of the force the pin exerts on the rod. Express the direction of the force the pin exerts on the rod as the angle F, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). r3 =  Fp = F =  (b) Let's now remove the mass m3 and determine the new mass m4 you would need to suspend from the rod at the position r4 = 20.0 cm in order to balance the rod and keep it horizontal if released from a horizontal position. In addition, for this case, what force (magnitude and direction) does the pin exert on the rod? Express the direction of the force the pin exerts on the rod as the angle F, measured with respect to the positive x-axis (counterclockwise is positive and clockwise is negative). m4 =  Fp =  F =  (c) Let's now remove the mass m4 and determine the mass m5 you would suspend from the rod in order to have a situation such that the pin does not exert a force on the rod and the location r5 from which you would suspend this mass in order to balance the rod and keep it horizontal if released from a horizontal position. m5 =  r5 = 

As A Torque Activity, Your Physics TA Sets Up The Arrangement Decribed Below. A Uniform Rod Of Mass Mr

Answers

Answer 1

a) Recall, the net torque on the rod must be zero. Thus,

Σt = 0

where

t represents torque

Thus,

t1 + t2 - tr - t3 = 0

t = rF

where

F = force

r = distance

r1F1 + r2F2 - rrFr - r3F3 = 0

r3F3 = r1F1 + r2F2 - rrFr

r3 = (r1F1 + r2F2 - rrFr)/F3

Note,

F1 = T1 = m1g

F2 = T2 = m2g

F3 = T3 = m3g

Thus,

r3 = (r1m1g + r2m2g - rrmrg)/m3g

g cancels out

r3 = (r1m1 + r2m2 - rrmr)/m3

From the information given,

r1 = 10 cm = 10/100 = 0.1 m

r2 = 90 cm = 90/100 = 0.9 m

rr = 100/2 = 50 cm = 50/100 = 0.5 m

m1 = 281 g = 281/1000 = 0.281 kg

m2 = 177g = 0.177 kg

mr = 158g = 0.158 kg

m3 = 200g = 0.2kg

By substituting these values into the equation,

r3 = (0.1 x 0.281 + 0.9 x 0.177 - 0.5 x 0.158)/0.2

r3 = 0.542 m

The force exerted by the pin, Fp = mg

g = 9.8

Fp = (m3 - mr - m1 - m2)g

Fp = (0.2 + 0.158 - 0.281 - 0.177)9.8

Fp = - 0.981

Taking the absolute value,

IFpI = 0.981 N

F = - 90 degrees

b) r1F1 + r2F2 - rrFr - r4F4 = 0

r4F4 = r1F1 + r2F2 - rrFr = 0

F4 = (r1F1 + r2F2 - rrFr)/r4

Note,

F1 = T1 = m1g

F2 = T2 = m2g

F3 = T3 = m3g

F4 = T4 = m4g

Thus,

m4g = (r1m1g + r2m2g - rrmrg)/r4

m4g = (r1m1 + r2m2 - rrmr)/r4

r4 = 0.2

By substituting these values into the equation,

m4 = (0.1 x 0.281 + 0.9 x 0.177 - 0.5 x 0.158)/0.2

m4 = 0.542 kg

The force exerted by pin is

Fp = (m4 + mr - m1 - m2(g

Fp = (0.542 + 0.158 - 0.281 - 0.177)9.8

Fp = 2.37 N

Fp = 2.37 N

F = 90 degrees

c) When the pin does not exert a force,

Fp = 0

F1 + F2 - Fr = F5

m1 + m2 - mr = m5

m5 = 0.281 + 0.177 - 0.158

m5 = 0.3 kg

Since the net torque on the rod is zero,

t1 + t2 - tr - t5

t5 = t1 + t2 - tr - t5

t5 = t1 + t2 - tr - t5

r5 = r1F1 + r2F2 - ffFr)/F5

r5 = (r1m1 + r2m2 - rrmr)/m5

r5 = (0.1 x 0.281 + 0.9 x 0.177 - 0.5 x 0.158)/0.3

r5 = 0.36

Related Questions


Two blocks of mass M₁ and M₂ are connected by a massless
string that passes over a massless pulley as shown in the
figure. M₁ has a mass of 3.75 kg and rests on an incline of
0₁ = 63.5°. M2 rests on an incline of 0₂ = 15.5°. Find the
mass of block M₂ so that the system is in equilibrium (i.e.,
not accelerating). All surfaces are frictionless

Answers

The correct answer is 58.58 Kg. (Mass of M_2)

What is mass string and friction system?

A spring-mass system in simple calculation can be described as a spring system where a block is hung or attached at the free end of the spring. If the surface is frictionless so µ = 0 (we can assume)

To just begin to slide up the friction will be kinetic friction

Applying free body diagram on blocks (as diagram is not given in question so assumption is the basis on given data only)

Given, M_1 = 3.75 Kg., M_2 =?

O_1 = 63.5◦ and O_2 = 15.5◦, g = 9.8 m/s2

So, if we require to keep the system in equilibrium position

Then we can write an equation as follows:

M_1x g x Sin63.5◦ = M_2 x g x Cos 15.5◦ (To be in Equilibrium)

 63.5 x 9.8 x 0.89101 = M_2 x 9.8 x 0.9659

 M_2 = 63.5 x 0.89101 / 0.9659

 M_2 = 58.58 Kg. (Mass of M_2)

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What is the initial velocity of an automobile acquiring a final velocity of 32 m/s with an acceleration of 4.0m/s ²

Answers

Answer:

Explanation:

Given:

V = 32 m/s

a = 4.0 m/s²

__________

V₀ - ?

V = V₀ + a*t

V₀ = V - a*t = 32 - 4*t

Time is not set according to the condition of the problem!

There's  not enough given information t o answer the question. It depends on how long the car has been accelerating.

it could be 28 m/s 1 second ago.

it could be 16 m/s 4 seconds ago.

it could be 10 m/s 5.5 seconds ago.

etc.

i'll take a wild guess and speculate that the question actually tells how long the car has been accelerating, but you didn't copy that part.

the spaceship is flying through space far from planets and stars with the engines firing.
The astronaut shuts the engines off.

The spaceship will….

a. stop moving immediately

b. slow down gradually and stop

c. continue with whatever speed it had when the engines were cut off

d. speed up for just a little while, then slow down

Answers

c. continue with whatever speed it had when the engines were cut off

PLEASE HELP

Which is not an accurate statement about Earth's gravitational pull?

A) Earth's gravitational pull helps keep it in orbit.

B) Earth's gravitational pull is the same as Jupiter's gravitational pull.

C) Earth's gravitational pull is 9.8 m/s2.

D) Earth's gravity helps keep people from floating outside of the planet.

Answers

B would be your right answer. Because Jupiter’s gravity pull is much stronger than earths

The image shows street lights powered by solar panels. Which sequence shows the energy transformations taking place in these lights?

Picture of three solar panels street light on a sunny day with blue background

A.
gravitational potential energy → vibrational energy → chemical potential energy
B.
radiant energy → chemical potential energy → motion energy
C.
radiant energy → electric energy → radiant energy
D.
sound energy → chemical potential energy → radiant energy
E.
gravitational potential energy → motion energy → radiant energy
Reset Next

Answers

The sequence that shows the energy transformations taking place in these lights are radiant energy → electric energy → radiant energy.

What is law of conservation of conservation of energy?

The principle or law of conservation of energy states that energy can neither be created nor destroyed but can be converted from one form to another.

Based on this law, the energy of a substance can be converted from one form to another

For example, energy can be converted as follows;

potential energy to kinetic energychemical energy to electric energyelectrical energy to sound energyetc,

The sequence of energy that takes place on street lights powered by solar panels is given as follows;

Radiant energy (light energy from sun) to electrical energy (converted by photo voltaic cell of the panels) to radiant energy (light given by the street lights).

Thus, the sequence that shows the energy transformations taking place in these lights are radiant energy → electric energy → radiant energy.

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Answer:

C

Explanation:

I did the test

How to do Projectile Motion?

Answers

A thrown ball undergoes projectile motion so throwing a ball in the air is an example of projectile motion.

What is Projectile Motion?

Projectile motion is the motion of an object pitched (projected) into the air. After the starting force that launches the object, the only occurrence of the force of gravity in the object is called a projectile motion, and its path is called its trajectory. Projectile motion is a form of motion in which an object or particle ( called a projectile, is thrown near the earth's surface and moves along a curved path under the action of gravity only. Throwing a ball or a cannonball. The motion of a billiard ball on the billiard table.  t. The motion of the earth around the un-projectile motion is a special case of two-dimensional motion. A particle in motion at a vertical level with an initial velocity and experiencing a free-fall (downward) acceleration, displays projectile motion.

So we can conclude that Projectile motion is applicable in both throwing and hitting. A thrown ball undergoes projectile motion.

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An object is projected or flung into the air, and only gravity's acceleration affects the object's velocity. A projectile is what it is, and its trajectory is what it took to get there.

What is Projectile motion?

An item or particle that is projected toward the surface of the Earth and moves along a curved path only under the influence of gravity is said to be experiencing projectile motion. Galileo demonstrated that this curving path was a parabola, however it can also be a straight line in the unique situation where it is hurled straight up.

Ballistics is the study of such motions, and this trajectory is a ballistic trajectory. Gravity, which works downward and gives the item a downward acceleration toward the Earth's center of mass, is the sole force of mathematical significance that is actively acting on it.

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For an object spinning around a central point, what will happen if its distance from the center is decreased

Answers

Answer:

Its a acceleration will increase

Explanation:

The force required to keep an object in a circular motion is given by

[tex]F=\frac{mv^2}{R}[/tex]

where v is the radial velocity and R is the radius of the object with mass m.

Now our question is what happens to the above equation as we decrease R?

We can see that as R decreases the quantity mv^2 /R increases (since R is getting smaller ).

Hence, we conclude that F increases. But what if F? it is the centripetal force.

Since centripetal force has increased, so has the quantity v^2 /R (called the acceleration ).

Meaning an increase in centripetal force implies an increase n acceleration.

Since in the answer choices we are not given the option to increase our centripetal force, the next best choice is to choose 'acceleration will increase. '

An archery bow is drawn a distance d = 0.39 m and loaded with an arrow of mass m = 0.088 kg. The bow acts as a spring with a spring constant of k = 195 N/m, and the arrow flies with negligible air resistance. To simplify your work, let the gravitational potential energy be zero at the initial height of the arrow. If the arrow is shot at an angle of θ = 45° above the horizontal, how high, in meters above the initial height, will the arrow be when it reaches its peak?

Answers

The maximum height reached by the arrows is determined as 8.6 m.

What is the initial speed of the arrow?

The initial velocity of the arrow is calculated by applying the principle of conservation of energy as shown below;

K.E = U

where;

K.E is the kinetic energy of the arrowU is the elastic potential energy of the bow

¹/₂mv² = ¹/₂kx²

mv² = kx²

v² = kx²/m

v = √(kx²/m)

where;

k is spring constant of the bowm is the mass of the arrowx is the extension of the bow

v = √(195 x 0.39²/0.088)

v = 18.36 m/s

The maximum height reached by the arrow is calculated as follows;

H = (v² sin²θ) / (2g)

where;

θ is angle of projection of the arrowg is acceleration due to gravity

H = (18.36² (sin45)²) / (2 x 9.8)

H = 8.6 m

Thus, the height of the arrow above the ground when it reaches its peak is 8.6 m.

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Three people are driving their cars in different directions, in an open field. At one point, while they are all driving, they each measure the other drivers’ velocities. When they compare measurements afterward, they notice that they all got different measurements from each other. Why do their measurements not match?

Answers

Answer:

because of their change in momentum

Please help me with this!

Answers

The balloon goes 25 m east in first 10 s.

Then the wind blows the balloon 35 m west in 8 s.

a

The balloon travels 25 m in 10s.

b

the balloon travels a distance 35 m in next 8 s.

c

thhe total distance travelled by the balloon is,

[tex]\begin{gathered} d=25+35 \\ =60\text{ m} \end{gathered}[/tex]

d

Average speed in 10 s is,

[tex]\begin{gathered} s=\frac{25}{10} \\ =2.5\text{ m/s} \end{gathered}[/tex]

e

The average speed in next 8 s is,

[tex]\begin{gathered} s_8=\frac{35}{8} \\ =4.37\text{ m/s} \end{gathered}[/tex]

f

the average speed for the entire trip is,

[tex]\begin{gathered} s_{av}=\frac{25+35}{10+8} \\ =3.33\text{ m/s} \end{gathered}[/tex]

g

displacement during the first 10 s is.

[tex]d_1=(25m)\hat{i}[/tex]

h.

displacement during next 8 s is,

[tex]d_2=(-35m)\hat{i}[/tex]

i

The total displacement is,

[tex]\begin{gathered} d_1+d_2=(25-35)\hat{i}_{} \\ =-(10m)\hat{i} \end{gathered}[/tex]

j

the average velocity in 10 s is,

[tex]\begin{gathered} v_1=\frac{25m}{10}\hat{i} \\ =(2.5m)\hat{i} \end{gathered}[/tex]

k

The average velocity in 8 s is,

[tex]\begin{gathered} v_2=\frac{-35\text{ m}}{8\text{ s}}\hat{i} \\ =-(4.37m)\hat{i} \end{gathered}[/tex]

l

The average velocity entire the whole trip is,

[tex]\begin{gathered} v_{av}=\frac{d_1+d_2}{18} \\ =\frac{-10m\hat{i}}{18\text{ s}} \\ =-(0.55\hat{m/s)i} \end{gathered}[/tex]

A steel cable on a bridge has a linear mass density of 15 kg/m. If the cable has been pulled taunt with a tension of 5536 N, what is the speed of a wave on it?

Answers

[tex]\begin{gathered} T=5,536N \\ \mu=15\text{ kg/m} \\ v=? \\ v=\sqrt{\frac{T}{\mu}} \\ v=\sqrt{\frac{5,536N}{15\text{ kg/m}}} \\ v=19.21\text{ m/s} \\ The\text{ speed of a wave on the cable is 19.21m/s} \end{gathered}[/tex]

A 2.0 microF capacitor is connected across a 60 Hz voltage source, and a current of 2.0 mA is measured on a VOM. What is the capacitive reactance of the circuit?

Answers

Let's write down and name the variables we know.

C: capacitance; C = 2 μF = 2*10^-6 F

f: frequency of voltage source; f = 60 Hz

Xc: capacitive reactance of circuit (we are solving for this)

We also know that ω = 2πf = 120π.

From this information, we can use the following equation:

Xc = 1/(ωC)

And we can solve for Xc.

Xc = 1/(120π*2*10^-6)

Xc = 1326.291 Ω

what energy is gotten from wind​

Answers

Kinetic energy is gotten from wind which is converted into rotational energy.

How energy is produced from the wind​?

The wind is used to produce electricity using the kinetic energy created by air in motion wind turbines convert the kinetic energy in the wind into mechanical power. This mechanical ability can be used for particular tasks (such as grinding grain or forcing water) or can be converted into electricity by a generator. into electricity. In present wind turbines, wind rotates the rotor blades, which change kinetic energy into rotational energy. Wind turbines labor on an easy principle: in lieu of using electricity to make wind like a fan wind turbines use the wind to make electricity. The wind turns the rotter-like blades of a turbine around a rotor, which spins a generator, which produced electricity.

So we can conclude that Wind rotates the rotor blades that convert kinetic energy into rotational energy.

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