PHYSICS COLLEGE

Interactive Solution 9.1 presents a model for solving this problem. The wheel of a car has a radius of 0.380 m. The engine of the car applies a torque of 456 N·m to this wheel, which does not slip against the road surface. Since the wheel does not slip, the road must be applying a force of static friction to the wheel that produces a countertorque. Moreover, the car has a constant velocity, so this countertorque balances the applied torque. What is the magnitude of the static frictional force?

Answers

Answer 1

Answer:

Answer:

The magnitude of the static frictional force is 1200 N

Explanation:

given information :

radius, r = 0.380 m

applied-torque, τ1 = 456 N

The car has a constant velocity, thus the acceleration is zero

α = 0

Στ = I α

τ1 - τ2 = I α

τ2 = counter-torque

τ1 - τ2 = 0

τ1 = τ2

r x $F_(s)$ = τ1

$F_(s)$ = the static frictional force (N)

$F_(s)$ = τ1 /r

= 456 N/0.380 m

= 1200 N

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An electron is a subatomic particle (m = 9.11 x 10-31 kg) that is subject to electric forces. An electron moving in the +x direction accelerates from an initial velocity of +6.18 x 105 m/s to a final velocity of 2.59 x 106 m/s while traveling a distance of 0.0708 m. The electron's acceleration is due to two electric forces parallel to the x axis: = 8.87 x 10-17 N, and , which points in the -x direction. Find the magnitudes of (a) the net force acting on the electron and (b) the electric force .

An F-35 stealth jet takes off from the aircraft carrier Ronald Reagan. Starting from rest, the jet accelerated with a constant acceleration of 55.3 m/s2 along a straight line on the deck. What is the displacement of the jet when it reaches a speed of 181 m/s?

Answers

Answer:

When the jet reaches a speed of 181 m/s, its displacement is 296 m.

Explanation:

Hi there!

The equation of position and velocity of an object traveling with constant acceleration along a straight line are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the object at time t.

If we place the origin of the frame of reference at the point where the jet starts moving, then, x0 = 0. Since the jet starts from rest, v0 is also zero. Then the equations get reduced to the following:

x = 1/2 · a · t²

v = a · t

We know the acceleration and the final velocity of the jet. So, using the equation of velocity, we can find the time it takes the jet to reach that velocity. Then, we can calculate the position of the jet at that time. Since the initial position is zero, the final position of the jet will be equal to the displacement (because displacement = final position - initial position).

v = a · t

v/a = t

181 m/s / 55.3 m/s² = t

t = 3.27 s

The final position of the jet will be:

x = 1/2 · a · t²

x = 1/2 · 55.3 m/s² · (3.27 s)²

x = 296 m

When the jet reaches a speed of 181 m/s, its displacement is 296 m.

The displacement of the F-35 jet when it reaches a speed of 181 m/s is 16515 m.

To find displacement using constant acceleration,

we can use the following equation:

displacement = (final velocity)^2 - (initial velocity)^2 / 2 * acceleration.

In this case, the initial velocity is 0 m/s and the final velocity is 181 m/s.

The acceleration is given as 55.3 m/s^2.

Plugging in these values, we get:

displacement = (181)^2 - (0)^2 / 2 * 55.3 = 16515 m.

The displacement of the F-35 jet when it reaches a speed of 181 m/s is 16515 m.

Learn more about displacement here:

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Occasionally, people can survive falling large distances if the surface they land on is soft enough. During a traverse of Eiger's infamous Nordvand, mountaineer Carlos Ragone's rock anchor gave way and he plummeted 516 feet to land in snow. Amazingly, he suffered only a few bruises and a wrenched shoulder. Assuming that his impact left a hole in the snow 3.6 ft deep, estimate the magnitude of his average acceleration as he slowed to a stop (that is while he was impacting the snow).

Answers

Answer:

4611.58 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.174 ft/s²

Equation of motion

$v^2-u^2=2as\n\Rightarrow v=√(2as+u^2)\n\Rightarrow v=√(2* 32.174* 516+0^2)\n\Rightarrow v=182.218\ ft/s$

$v^2-u^2=2as\n\Rightarrow a=(v^2-u^2)/(2s)\n\Rightarrow a=(0^2-182.218^2)/(2* 3.6)\n\Rightarrow a=-4611.58\ ft/s^2$

Magnitude of acceleration while stopping is 4611.58 ft/s²

1. If a net force of 412 N is required to accelerate an object at 5.82 m/s2, what must theobject's mass be?

Answers

Answer:

The mass of the object is approximately 70.79 kilograms

Explanation:

We use Newton's second law to solve this problem. This law states that the net force on an object equals the product of its mass times the acceleration:

$F_(net)=m\,a$

Therefore, for this case, since the net force on the object and its acceleration are given, we can use the equation above to solve for the unknown mass:

$F_(net)=m\,a\n412\,N=m\,(5.82\,(m)/(s^2) )\nm=(412\,N)/(5.82\,(m)/(s^2) ) \nm=70.79\.kg$

If 3.00 ✕ 10−3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.59 h, what is the current in the cell during that period? Assume the gold ions carry one elementary unit of positive charge.

Answers

Answer:

0.158 A.

Explanation:

Mass of gold deposited = 3 x 10^-3 kg

= 3 g

Molar mass = 196 g/mol

Number of moles = 3/196

= 0.0153 mol.

Faraday's constant,

1 coloumb = 96500 C/mol

Quantity of charge, Q = 96500 * 0.0153

= 1477.04 C.

Remember,

Q = I * t

t = 2.59 hr

= 2.59 * 3600 s

= 9324 s

Current, I = 1477.04/9324

= 0.158 A.

Answer:

0.158A

Explanation:

Using Faraday's first law of electrolysis which states that the mass(m) of a substance deposited or liberated at any electrode is directly proportional to the quantity of charge or electricity (Q) passed. i.e

m ∝ Q

m = Z Q

Where;

Z is the proportionality constant called electrochemical equivalent.

Faraday also observed that when 1 Faraday of electricity is equivalent to 96500C of charge.

Also,

Quantity of charge (Q), which is the product of current (I) passing through and the time taken (t) for the electrolysis, is given by;

Q = I x t; ----------------------(i)

With all of these in place, now let's go answer the question.

Since the gold ions carry one elementary unit of positive charge, now let's write the cathode-half reaction for gold (Au) as follows;

Au⁺ + e⁻ = Au ---------------------(ii)

From equation (ii) it can be deduced that when;

1 Faraday (96500C) of electricity is passed, 1 mole of Au forms ( = 197 grams of Au) [molar mass of Au = 197g]

Then, 3.00 x 10⁻³ kg (= 3 g of Au) will be formed by 3g x 96500C / 197g = 1469.5C

Therefore, the quantity of charge (Q) deposited is 1469.5C

Substitute this value (Q = 1469.5C) and time t = 2.59h (= 2.59 x 3600 s) into equation (i);

Q = I x t

1469.5 = I x 2.59 x 3600

1469.5 = I x 9324

Solve for I;

I = 1469.5 / 9324

I = 0.158A

Therefore, the current in the cell during that period is 0.158A

Note:

1 mole of gold atoms = 176g

i.e the molar mass of gold (Au) is 176g

Suppose that a particle accelerator is used to move two beams of particles in opposite directions. In a particular region, electrons move to the right at 6020 m/s and protons move to the left at 1681 m/s. The particles are evenly spaced with 0.0476 m between electrons and 0.0662 m between protons. Assuming that there are no collisions and that the interactions between the particles are negligible, what is the magnitude of the average current in this region

Answers

Answer:

$2.429783984* 10^(-14)\ A$

Explanation:

Velocity of electron = 6020 m/s

Velocity of proton = 1681 m/s

Electron space = 0.0476 m

Proton space = 0.0662 m

e = Charge of particle = $1.6* 10^(-19)\ C$

Number of electrons passing per second

$n_e=(6020)/(0.0476)\n\Rightarrow n_e=126470.588$

Number of protons passing per second

$n_p=(1681)/(0.0662)\n\Rightarrow n_p=25392.749$

Current due to electrons

$I_e=n_ee\n\Rightarrow I_e=126470.588* 1.6* 10^(-19)\n\Rightarrow I_e=2.0235* 10^(-14)\ A$

Current due to protons

$I_p=n_pe\n\Rightarrow I_p=25392.749* 1.6* 10^(-19)\n\Rightarrow I_p=4.06283984* 10^(-15)\ A$

Total current

$I=2.0235* 10^(-14)+4.06283984* 10^(-15)\n\Rightarrow I=2.429783984* 10^(-14)\ A$

The average current is $2.429783984* 10^(-14)\ A$

A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to move it 4.00 m to the side.(a) What is the magnitude of F when the crate is in this final position? During the crate's displacement, what are
(b) the total work done on it,
(c) the work done by the gravitational force on the crate, and
(d) the work done by the pull on the crate from the rope?
(e) Knowing that the crate is motionless before and after its displacement, use the answers to (b), (c), and (d) to find the work your force F does on the crate.
(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)?