Consider a simple tension member that carries an axial load of P=22.44N. Find the total elongation in the member due to the load. Assume that the member is made of steel, which has a modulus of elasticity of E=204.00 N/mm2. Also assume that the member is 3048 mm long and has a cross-sectional area of 1290 mm2

The unknown capacitance C connected parallelly with a 30-μF capacitor initially charged at 80 V is calculated to be 90 μF. After being connected, the total potential difference across both capacitors reduces to 20 V.

The question deals with capacitors and how they function in a circuit. A capacitor is an electronic component that stores electrical energy and releases it in the circuit when necessary. When a charged 30-μF capacitor is connected across an initially uncharged capacitor, the voltage or potential difference across the connected capacitors will equalize. In this case, the final potential difference is noted as 20 V.

We know that for capacitors connected in parallel, the total charge stored is the sum of the charges stored in each capacitor, i.e., Q = Q₁ + Q₂, where Q is the total charge, and Q₁ and Q₂ are the charges stored in capacitor 1 (the 30-μF capacitor) and 2 (unknown capacitance C), respectively.

By using the formula Q = CV (charge = capacitance × voltage), where C is the capacitance and V is the voltage, we understand that Q₁ (charge on the 30 uF capacitor) is initially 30 μF × 80 V = 2400 μC (micro coulombs). After connecting the uncharged capacitor, the voltage drops to 20V, thus the final charge on the 30 uF capacitor becomes 30 μF × 20 V = 600 μC. The remaining charge must then be stored in the previously uncharged capacitor.

So, the charge on the unknown capacitor C would be the difference i.e., 2400 μC - 600 μC = 1800 μC. Now using the Q = CV formula, we can calculate the unknown capacitance C. V here is 20V as capacitors connected in parallel have the same voltage. So, C = Q / V = 1800 μC / 20 V = 90 μF. Thus, the unknown capacitance is 90 μF.

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Answer:

Option B is the correct answer.

Explanation:

Period of a spring is given by the expression

            [tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]

Here, spring constant, k = 110 N/m

         Mass = 0.41 kg

Substituting,

        [tex]T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{0.41}{110}}=0.38s[/tex]

Option B is the correct answer.

Answer: 1.00

Explanation:

The index of refraction [tex]n[/tex] is a number that describes how fast light propagates through a medium or material.

Being its equation as follows:

[tex]n=\frac{c}{v}[/tex]  (1)

Where [tex]c=3(10)^{8}m/s[/tex] is the speed of light in vacuum and [tex]v[/tex] its speed in the other medium and [tex]n=1.55[/tex].

So, from (1) we can find the velocity at which the signal travels and then the time it requires to travel:

[tex]v=\frac{c}{n}[/tex]  (2)

[tex]v=\frac{3(10)^{8}m/s}{1.55}[/tex]  (3)

[tex]v=193548387.1m/s[/tex]  (4)

Now, knowing the velocity [tex]v[/tex] is the distance [tex]d=0.35m[/tex] traveled in a time [tex]t[/tex]:

[tex]v=\frac{d}{t}[/tex]  (5)

We can isolate [tex]t[/tex] from (5) and find the value of the required time:

[tex]t=\frac{d}{v}[/tex]  (6)

[tex]t=\frac{0.35m}{193548387.1m/s}[/tex]  (7)

[tex]t=0.000000001s=1(10)^{-9}s=1ns[/tex]  (8) This is the time it takes the signal to travel through the optical fiber: 1 nanosecond.

Final answer:

The time for a signal to travel 0.35 m through an optical fiber with an index of refraction of 1.55 is approximately 1.81 nanoseconds.

Explanation:

To calculate the time required for a signal to travel through an optical fiber with an index of refraction of 1.55, we use the equation that relates the speed of light in a vacuum (c) to the speed of light in a material (v) using the index of refraction (n): v = c/n. Given that the speed of light in a vacuum is approximately 3.00×108 m/s, we can find the speed of light in the optical fiber.

The speed of light in the fiber (v) is:
v = (3.00 × 108 m/s) / 1.55 = 1.935×108 m/s.

Now, we calculate the time (t) it takes for light to travel 0.35 m in the fiber, using the formula
t = distance/speed:

t = 0.35 m / (1.935 × 108 m/s).

The calculated time t is found to be approximately 1.81 nanoseconds, after converting from seconds by multiplying by 109.

Answer:

Coefficient of static friction between the road and the car's tire is 0.81

Explanation:

It is given that,

Velocity of the car, v = 40 m/s

Radius of the curve, r = 200 m

We need to find the minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding. Let it is equal to μ.

The centripetal force of the car is balanced by the force of friction as :

[tex]\dfrac{mv^2}{r}=\mu mg[/tex]

[tex]\mu=\dfrac{v^2}{rg}[/tex]

[tex]\mu=\dfrac{(40\ m/s)^2}{200\ m\times 9.8\ m/s^2}[/tex]

[tex]\mu=0.81[/tex]

So, the coefficient of static friction between the road and the car's tire is 0.81 Hence, this is the required solution.

The minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding is 0.816.

Given the data in the question;

For a not car not to slide of the road, the frictional force and centripetal force should balance each other.

That is; Frictional force = Centripetal force

[tex]Frictional\ Force = uF = umg \\\\Centripetal\ Force = \frac{mv^2}{r}[/tex]

So,

[tex]u_{min}mg = \frac{mv^2}{r} \\\\u_{min}g =\frac{v^2}{r}\\\\u_{min} =\frac{v^2}{gr}\\[/tex]

Where [tex]u_{min}[/tex] is the minimum coefficient friction, v is the velocity, r is the radius and g is acceleration due to gravity( [tex]g = 9.8m/s^2\\[/tex])

We substitute our values into the equation

[tex]u_{min} = \frac{(40.0m/s)^2}{9.8m/s^2\ *\ 200m}\\\\ u_{min} = \frac{1600m^2/s^2}{1960m^2/s^2} \\\\u_{min} = 0.81[/tex]

Therefore, the minimum coefficient of static friction between the road and the car's tires that will allow the car to travel at this speed without sliding is 0.816

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The magnitude of the instantaneous acceleration of the object is 12.3 m/s².

The magnitude of the instantaneous acceleration of the object can be calculated using the formula:

a = v² / r

Where v is the speed and r is the radius of the circle. In this case, the speed is the distance traveled divided by the time taken, which is equal to the circumference of the circle divided by the time taken:

v = (2πr) / t

Substituting the values given in the question, we can calculate the acceleration:

a = [(2πr) / t]² / r = (4π²r²) / t² = (4π² * (2.5)²) / (4)² = 12.3 m/s²

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The instantaneous acceleration of a particle moving in a circular path of diameter 5.0 m at a constant speed, completing the path in 4.0 s, is approximately 6.17 m/s².

The subject of this question is the instantaneous acceleration of a particle moving in a circle, which is a topic within physics, specifically, kinematics. This is a case of uniform circular motion, which implies a constant speed, but changing direction, thereby resulting in centripetal (towards the center) acceleration.

The radius of the circle is half of its diameter: r = 5.0 m/2 = 2.5 m. As the particle completes a full circle in 4.0 s, the speed can be calculated using circumference (2πr) and time: v = 2πr/t = 2π * 2.5 m/4.0 s ≈ 3.92 m/s. The centripetal acceleration (the instantaneous acceleration for a particle in a uniform circular motion) can be calculated as: a = v²/r = (3.92 m/s)²/2.5 m ≈ 6.17 m/s².

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Answer:4.08 N

Explanation:

Given data

superball dropped from a height of 10  cm

Mass of ball[tex]\left ( m\right )[/tex]=0.05kg

time of contact[tex]\left ( t\right )[/tex]=34.3[tex]\times 10^{-3}[/tex] s

Now we know impulse =[tex]Force\times time\ of\ contact[/tex]=Change in momentum

[tex]F_{average}\times t[/tex]=[tex]m\left ( v-(-v)\right )[/tex]

and velocity at the bottom is given by

v=[tex]\sqrt{2gh}[/tex]

[tex]F_{average}\times 34.3\times 10^{-3}[/tex]=[tex]0.05\left ( 1.4-(-1.4)\right )[/tex]

[tex]F_{average}[/tex]=4.081N

To find the average force exerted on the superball by the table, we calculate its change in momentum during the bounce and divide by the contact time. The result is an approximate average force of 4.08 N upward.

The question involves a superball and a clay ball, both with the same mass of 0.05 kg, dropped from a height of 10 cm. The superball bounces back to its original height, while the clay ball sticks to the table. We need to calculate the average force exerted on the superball by the table.

Step-by-Step Solution

Calculate the velocity of the superball just before hitting the table: Using the equation for free fall, v = √(2gh), where g is 9.8 m/s² (acceleration due to gravity) and h is 0.10 m.

v = √(2 * 9.8 * 0.10)

v = √(1.96)

v ≈ 1.40 m/s

Determine the change in momentum: Before impact, the momentum is [tex]P_{before[/tex] = m * v. Since the superball bounces back with the same speed, the momentum after impact is [tex]P_{after[/tex] = -m * v because the direction changes.

[tex]P_{before[/tex] = 0.05 kg * 1.40 m/s = 0.07 kg·m/s

[tex]P_{after[/tex] = 0.05 kg * (-1.40 m/s) = -0.07 kg·m/s

Calculate the impulse: Impulse is the change in momentum, so Impulse = [tex]P_{after[/tex] - [tex]P_{before[/tex].

Impulse = -0.07 kg·m/s - 0.07 kg·m/s

Impulse = -0.14 kg·m/s

Calculate the average force: The impulse-momentum theorem states that Impulse = [tex]F_{avg[/tex] * Δt, where [tex]F_{avg[/tex] is the average force and Δt is the time of contact (34.3 ms = 0.0343 s). Solving for [tex]F_{avg[/tex] ,

[tex]F_{avg[/tex] = Impulse / Δt

[tex]F_{avg[/tex] = -0.14 kg·m/s / 0.0343 s

[tex]F_{avg[/tex] ≈ -4.08 N

The magnitude of the average force exerted on the ball by the table is approximately 4.08 N in the upward direction.

Explanation:

The object is moving along the parabola y = x² and is at the point (√2, 2).  Because the object is changing directions, it has a centripetal acceleration towards the center of the circle of curvature.

First, we need to find the radius of curvature.  This is given by the equation:

R = [1 + (y')²]^(³/₂) / |y"|

y' = 2x and y" = 2:

R = [1 + (2x)²]^(³/₂) / |2|

R = (1 + 4x²)^(³/₂) / 2

At x = √2:

R = (1 + 4(√2)²)^(³/₂) / 2

R = (9)^(³/₂) / 2

R = 27 / 2

R = 13.5

So the centripetal force is:

F = m v² / r

F = m (5)² / 13.5

F = 1.85 m

The force on an object moving along a parabola at a point can be determined by calculating the acceleration at that point (derived from the velocity and its change), and then applying Newton's second law of motion (F=ma). The computations involve complex physics and calculus concepts.

An object of mass m moves along a parabolic path y=x2 with constant speed, but the direction of its velocity is continuously changing, which should be considered as an acceleration and hence results in a force according to Newton's second law (F=ma).

The force on the object at a particular point (21/2 , 2) due to its acceleration can be determined by first calculating the acceleration at that point and then using Newton's second law. The details of these calculations involve some complex high school level physics and calculus concepts, but essentially involve calculating the derivative of the object's velocity with respect to time at the given point, then multiplying that by the object's mass.

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Answer:

Force of gravity, F = 0.74 N

Mass of an object, m = 1.2 kg

Distance above earth's surface, [tex]d=1.9\times 10^7\ m[/tex]

Mass of Earth, [tex]M=5.97\times 10^{24}\ kg[/tex]

Radius of Earth, [tex]r=6.37\times 10^6\ m[/tex]

We need to find the force of gravity above the surface of Earth. It is given by :

[tex]F=G\dfrac{mM}{R^2}[/tex]

R = r + d

R = 25370000 m

[tex]F=6.67\times 10^{-11}\times \dfrac{5.97\times 10^{24}\ kg\times 1.2\ kg}{(25370000\ m)^2}[/tex]

F = 0.74 N

So, the force of gravity on the object is 0.74 N. Hence, this is the required solution.

The force of gravity exerted on the 1.2-kg mass located four Earth radii from Earth's center is 0.059 N. This result showcases the inverse-square law as the gravitational force is significantly less than the same mass at Earth’s surface (11.76 N).

The force of gravity on an object depends on its mass and its distance from the center of the Earth. The general equation to calculate this force is F = GMm/r², where F is the gravitational force, G is the universal gravitational constant (6.67x10^-11 N(m/kg)²), M is the mass of the Earth (5.98x10^24 kg), m is the mass of the object (in this case, 1.2 kg), and r is the distance from the center of the Earth.

Put strictly, the stated distance must be added to Earth's radius (6.37x10^6 m) to find the total distance from Earth's center. So, r = Earth’s radius + the object’s height from the surface, which gives us r = 6.37x10^6 m + 1.9x10^7 m = 2.537x10^7 m.

Substituting these values into the equation gives F = (6.67x10^-11 N(m/kg)² * 5.98x10^24 kg * 1.2 kg) / (2.537x10^7 m)² = 0.059 N.To put this into perspective, the weight of the same mass (1.2 kg) on the surface of the Earth would be 1.2 kg * 9.8 m/s² (acceleration due to gravity at Earth’s surface) = 11.76 N. This demonstrates the inverse-square law of gravity, where the gravitational force decreases with the square of the distance from the source.

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Answer:

a)

840 N

b)

10920 J

c)

- 10192 J

d)

4.3 m/s

Explanation:

a)

T = tension force in the cable in upward direction = ?

a = acceleration of the person in upward direction = 0.70 m/s²

m = mass of the person being lifted = 80 kg

Force equation for the motion of person in upward direction is given as

T - mg = ma

T = m (g + a)

T = (80) (9.8 + 0.70)

T = 840 N

b)

d = distance traveled in upward direction = 13 m

[tex]W_{t}[/tex] = Work done by tension force

Work done by tension force is given as

[tex]W_{t}[/tex] = T d

[tex]W_{t}[/tex] = (840) (13)

[tex]W_{t}[/tex] = 10920 J

c)

d = distance traveled in upward direction = 13 m

[tex]W_{g}[/tex] = Work done by person's weight

Work done by person's weight is given as

[tex]W_{g}[/tex] = - mg d

[tex]W_{g}[/tex] = - (80 x 9.8) (13)

[tex]W_{g}[/tex] = - 10192 J

d)

[tex]F_{net}[/tex] = Net force on the person = ma = 80 x 0.70 = 56 N

v₀ = initial speed of the person = 0 m/s

v = final speed

Using work-energy theorem

[tex]F_{net}[/tex] d = (0.5) m (v² - v₀²)

(56) (13) = (0.5) (80) (v² - 0²)

v = 4.3 m/s

Answer:

[tex]V_{A}= 11.88 m/s[/tex]

Explanation:

given data:

water level at point A = 8 m

diameter of orifice = 0.5 cm

velocity at point A  =  0

h1 =0.8 m

h2 = 8 - h1 = 8 - 0.8 = 7.2 m

Applying Bernoulli  theorem between point A and B

[tex]P_{o}+\rho _{water}gh_{2}+\frac{1}{2}\rho v_{B}^{2}+\rho _{water}gh_{1}=P_{o} +\frac{1}{2}\rho v_{A}^{2}+\rho _{water}gh_{1}[/tex]

[tex]V_{A}=\sqrt{2gh_{2}}[/tex]

[tex]V_{A}=\sqrt{2*9.81*7.2}[/tex]

[tex]V_{A}= 11.88 m/s[/tex]

Answer:

[tex]\omega = 2.56 rad/s[/tex]

Explanation:

As the cylinder rotates the centripetal force on all the passengers is due to normal force due to the wall

So here we can say

[tex]N = m\omega^2 R[/tex]

now when floor is removed all the passengers are safe because here friction force on the passenger is counter balanced by the weight of the passengers

so we can say

[tex]F_f = mg[/tex]

[tex]\mu_s F_n = mg[/tex]

[tex]\mu_s (m\omega^2 R) = mg[/tex]

[tex]\mu_s \omega^2 R = g[/tex]

[tex]\omega = \sqrt{\frac{g}{\mu_s R}}[/tex]

for minimum rotational speed we have

[tex]\omega = \sqrt{\frac{9.8}{0.60(2.5)}[/tex]

[tex]\omega = 2.56 rad/s[/tex]

The minimum rotational frequency for the ride to be safe is approximately 8.28 rpm.

To determine the minimum rotational frequency for which the ride is safe, we need to consider the static coefficient of friction between clothing and steel. Since the passengers are standing inside a hollow cylinder, they will experience a centrifugal force pushing them against the wall of the cylinder. To prevent sliding, the static friction force needs to be greater than or equal to the gravitational force pulling them downward. The formula to calculate the static friction force is Fs = μs * N, where μs is the coefficient of static friction and N is the normal force.

For clothing against steel, the coefficient of static friction ranges from 0.60 to 1.0. Assuming the worst-case scenario with μs = 0.60, we can calculate the minimum rotational frequency:

Centrifugal force = m * g = m * ω^2 * R, where m is the mass of the passengers, g is the acceleration due to gravity, ω is the angular velocity in radians per second, and R is the radius of the cylinder.
Static friction force = μs * m * g
Equating these two forces, we get μs * m * g = m * ω^2 * R
Simplifying the equation, we find ω = sqrt(μs * g / R)

Converting the angular velocity to revolutions per minute (rpm), we have rpm = 60 * ω / (2 * π)
Substituting the values, the minimum rotational frequency for the ride to be safe is approximately 8.28 rpm.

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Answer:

x= 2.7734 m

Explanation:

Given the angle of incidence θ= 33.0°

distance from laser point and mirror = 1.10 m

laser point is 1.80 m above the floor

let the distance between mirror base and laser beam striking the floor after reflection = x  

figure has been attached and from the from the figure we can write

[tex]tan\theta= \frac{1.80}{x}[/tex]

[tex]tan33\degree= \frac{1.80}{x}[/tex]

[tex]tan33\degree= 0.649[/tex]

putting value and solving we get

x= 2.7734 m

Final answer:

An additional charge of 2100 μC flows from the battery into the capacitor when it is immersed in transformer oil with a dielectric constant of 4.5.

Explanation:

To determine how much additional charge flows from the battery when a 6.0-μF air capacitor is immersed in transformer oil with a dielectric constant of 4.5, we need to examine the effect of the dielectric on the capacitor's capacitance.

Initially, the capacitance of the air capacitor Cair is 6.0 μF. The charge Qinitial on the capacitor when connected to a 100-V battery is given by:

Qinitial = Cair × Vbattery

Qinitial = 6.0 μF × 100 V

Qinitial = 600 μC

When the capacitor is immersed in oil, the capacitance increases due to the dielectric constant (κ) of the oil:

Coil = κ × Cair

Coil = 4.5 × 6.0 μF = 27.0 μF

Since the battery remains connected, the voltage across the capacitor stays at 100 V, so the new charge Qfinal becomes:

Qfinal = Coil × Vbattery

Qfinal = 27.0 μF × 100 V

Qfinal = 2700 μC

The additional charge ΔQ that flows from the battery is the difference between Qfinal and Qinitial:

ΔQ = Qfinal - Qinitial

ΔQ = 2700 μC - 600 μC

ΔQ = 2100 μC

Therefore, an additional charge of 2100 μC flows from the battery into the capacitor.

Answer:

28.1 mph

Explanation:

The force of friction acting on the car provides the centripetal force that keeps the car in circular motion around the curve, so we can write:

[tex]F=\mu mg = m \frac{v^2}{r}[/tex] (1)

where

[tex]\mu[/tex] is the coefficient of friction

m is the mass of the car

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

v is the maximum speed of the car

r is the radius of the trajectory

On the snowy day,

[tex]\mu=0.50\\v = 20 mph = 8.9 m/s[/tex]

So the radius of the curve is

[tex]r=\frac{v^2}{\mu g}=\frac{(8.9)^2}{(0.50)(9.8)}=16.1 m[/tex]

Now we can use this value and re-arrange again the eq. (1) to find the maximum speed of the car on a sunny day, when [tex]\mu=1.0[/tex]. We find:

[tex]v=\sqrt{\mu g r}=\sqrt{(1.0)(9.8)(8.9)}=12.6 m/s=28.1 mph[/tex]

The maximum speed at which the car can take the same curve on a sunny day is about 28 mph

[tex]\texttt{ }[/tex]

Centripetal Acceleration can be formulated as follows:

[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

[tex]\texttt{ }[/tex]

Centripetal Force can be formulated as follows:

[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

coefficient of friction on a snowy day = μs₁ = 0.50

maximum speed of the car on a snowy day = v₁ = 20 mph

coefficient of friction on a sunny day = μs₂ = 1.0

Asked:

maximum speed of the car on a snowy day = v₂ = ?

Solution:

Firstly , we will derive the formula to calculate the maximum speed of the car:

[tex]\Sigma F = ma[/tex]

[tex]f = m \frac{v^2}{R}[/tex]

[tex]\mu N = m \frac{v^2}{R}[/tex]

[tex]\mu m g = m \frac{v^2}{R}[/tex]

[tex]\mu g = \frac{v^2}{R}[/tex]

[tex]v^2 = \mu g R[/tex]

[tex]\boxed {v = \sqrt { \mu g R } }[/tex]

[tex]\texttt{ }[/tex]

Next , we will compare the maximum speed of the car on a snowy day and on the sunny day:

[tex]v_1 : v_2 = \sqrt { \mu_1 g R } : \sqrt { \mu_2 g R }[/tex]

[tex]v_1 : v_2 = \sqrt { \mu_1 } : \sqrt { \mu_2 }[/tex]

[tex]20 : v_2 = \sqrt { 0.50 } : \sqrt { 1.0 }[/tex]

[tex]20 : v_2 = \frac{1}{2} \sqrt{2}[/tex]

[tex]v_2 = 20 \div \frac{1}{2} \sqrt{2}[/tex]

[tex]v_2 = 20 \sqrt{2} \texttt{ mph}[/tex]

[tex]\boxed{v_2 \approx 28 \texttt{ mph}}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Circular Motion

Answer:

0.182 m/s

Explanation:

m1 = 30,000 kg, m2 = 110,000 kg, u1 = 0.85 m/s

let the velocity of loaded freight car is v

Use the conservation of momentum

m1 x u1 + m2 x 0 = (m1 + m2) x v

30,000 x 0.85 = (30,000 + 110,000) x v

v = 0.182 m/s

Answer:

9.89 m/s

Explanation:

d = diameter of the space station = 20.0 m

r = radius of the space station

radius of the space station is given as

r = (0.5) d

r = (0.5) (20.0)

r = 10 m

a = acceleration produced at outer rim = 9.80 m/s²

v = speed at which it rotates

acceleration is given as

[tex]a = \frac{v^{2}}{r}[/tex]

[tex]9.80 = \frac{v^{2}}{10}[/tex]

v = 9.89 m/s

Answer:

Option C is the correct answer.

Explanation:

Absolute  pressure is sum of gauge pressure and atmospheric pressure.

That is

               [tex]P_{abs}=P_{gauge}+P_{atm}[/tex]

We have

          [tex]P_{abs}=550.280 kPa\\\\P_{atm}=1atm=101325Pa=101.325kPa[/tex]

Substituting

         [tex]P_{abs}=P_{gauge}+P_{atm}\\\\550.280=P_{gauge}+101.325\\\\P_{gauge}=448.955kPa[/tex]

Option C is the correct answer.

Explanation:

It is given that,

Weight of the person, W = F = 846 N

When the person steps onto a spring scale in the bathroom, the spring compresses by 0.574 cm, x = 0.00574 m

(a) The force acting on the spring is is given by Hooke's law as :

[tex]F=-kx[/tex]

[tex]k=\dfrac{F}{x}[/tex]

[tex]k=\dfrac{846\ N}{0.00574\ m}[/tex]

k = 147386.7 N/m

(b) If the spring is compressed by 0.314 cm or 0.00314 m, weight of the person is given by again Hooke's law as :

[tex]F=kx[/tex]

[tex]F=147386.7\ N/m\times 0.00314\ m[/tex]

F = 462.7 N

Hence, this is the required solution.

Explanation:

33×10^-6 ×74 ×(0.86 - 0.41)

For a direct current resistor-capacitor circuit where the capacitor is initially uncharged, the charge stored on one of the capacitor's plates is given by:

Q(t) = Cℰ(1-e^{-t/(RC)})

Q(t) is the charge, t is time, ℰ is the battery's terminal voltage, R is the resistor's resistance, and C is the capacitor's capacitance.

The time constant of the circuit τ is the product of the resistance and capacitance:

τ = RC

Q(t) can be rewritten as:

Q(t) = Cℰ(1-e^{-t/τ})

We want to know how much charge is stored when one time constant has elapsed, i.e. what Q(t) is when t = τ. Let us plug in this time value:

Q(τ) = Cℰ(1-e^{-τ/τ})

Q(τ) = Cℰ(1-1/e)

Q(τ) = Cℰ(0.63)

Given values:

C = 5.0×10⁻⁶F

ℰ = 12V

Plug in these values and solve for Q(τ):

Q(τ) = (5.0×10⁻⁶)(12)(0.63)

Q(τ) = 3.8×10⁻⁵C

The charge on either plate of given capacitor after one time constant has elapsed is 3.8×10⁻⁵C.

The charge stored on one of the capacitor's plates can be calculated by,  

[tex]\bold{Q(t) = C\epsilon(1-e^{-t/(RC)})}[/tex]

Where,

Q(t) - the charge,

t - time,

ℰ- the battery's terminal voltage,

R - the resistor's resistance,

C -  the capacitor's capacitance.

The time constant of the circuit τ is equal to the product of the resistance R and capacitance C :  

τ = RC  

The amount of charge Q(t) when t = τ. put the values,  

Q(τ) = Cℰ(1-e^{-τ/τ})  

Q(τ) = Cℰ(1-1/e)  

Q(τ) = Cℰ(0.63)

Given values:  

C = 5.0×10⁻⁶F  

ℰ = 12V

Put the values in the formula and solve for Q(τ):  

Q(τ) = (5.0×10⁻⁶)(12)(0.63)  

Q(τ) = 3.8×10⁻⁵C

Therefore, the charge on either plate of given capacitor after one time constant has elapsed is 3.8×10⁻⁵C.

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Answer:

209.3 seconds

Explanation:

P = 300 W, m =  0.250 kg, T1 = 10 degree C, T2 = 70 degree C

c = 4186 J / kg C

Heat given to water = mass x specific heat of water x rise in temperature

H = 0.250 x 4186 x (70 - 10)

H = 62790 J

Power = Heat / Time

Time, t = heat / Power

t = 62790 / 300 = 209.3 seconds

It took approximately [tex]\( 209.3 \)[/tex] seconds to heat the water from [tex]\( 10.0^\circ \text{C} \)[/tex] to [tex]\( 70.0^\circ \text{C} \)[/tex].

To determine how many seconds it took to heat the water using the immersion heater, we need to calculate the amount of energy required and then use the power of the heater to find the time.

First, calculate the change in temperature of the water:

[tex]\[ \Delta T = 70.0^\circ \text{C} - 10.0^\circ \text{C} = 60.0^\circ \text{C} \][/tex]

Next, calculate the energy [tex]\( Q \)[/tex] required to heat the water using the specific heat capacity of water [tex]\( C = 4186 \text{ J/kg}^\circ \text{C} \)[/tex]:

[tex]\[ Q = mc\Delta T \][/tex]

where [tex]\( m \)[/tex] is the mass of water and [tex]\( c \)[/tex] is the specific heat capacity of water.

Given:

[tex]\[ m = 0.250 \text{ kg} \][/tex]

[tex]\[ c = 4186 \text{ J/kg}^\circ \text{C} \][/tex]

[tex]\[ \Delta T = 60.0^\circ \text{C} \][/tex]

[tex]\[ Q = 0.250 \times 4186 \times 60.0 \][/tex]

[tex]\[ Q = 62790 \text{ Joules} \][/tex]

Now, calculate the time t required using the power P  of the heater:

[tex]\[ P = 300.0 \text{ W} \][/tex]

The time [tex]\( t \)[/tex] is given by:

[tex]\[ t = \frac{Q}{P} \][/tex]

[tex]\[ t = \frac{62790}{300.0} \][/tex]

[tex]\[ t = 209.3 \text{ seconds} \][/tex]

Answer:

The magnification is -6.05.

Explanation:

Given that,

Focal length = 34 cm

Distance of the image =2.4 m = 240 cm

We need to calculate the distance of the object

[tex]\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}[/tex]

Where, u = distance of the object

v = distance of the image

f = focal length

Put the value into the formula

[tex]\dfrac{1}{u}=\dfrac{1}{34}-\dfrac{1}{240}[/tex]

[tex]\dfrac{1}{u}=\dfrac{103}{4080}[/tex]

[tex]u =\dfrac{4080}{103}[/tex]

The magnification is

[tex]m = \dfrac{-v}{u}[/tex]

[tex]m=\dfrac{-240\times103}{4080}[/tex]

[tex]m = -6.05[/tex]

Hence, The magnification is -6.05.

Answer:

The minimum uncertainty in the energy state of an atom is [tex]1.0557\times10^{-35}\ J[/tex]

Explanation:

Given that,

Time t = 10 s

We need to calculate the minimum uncertainty in the energy state

Using Hisen burg of uncertainty principle equation

[tex]\Delta E\Delta t=\dfrac{h}{2\pi}[/tex]

[tex]\Delta E=\dfrac{h}{2\pi}\times\dfrac{1}{\Delta t}[/tex]

Where, E = energy

t = time

Put the value into the formula

[tex]\Delta E=\dfrac{6.63\times10^{-34}}{2\times3.14}\times\dfrac{1}{10}[/tex]

[tex]\Delta E=1.0557\times10^{-35}\ J[/tex]

Hence, The minimum uncertainty in the energy state of an atom is [tex]1.0557\times10^{-35}\ J[/tex]

Final answer:

The minimum uncertainty in the energy state of an atom can be found using the Heisenberg Uncertainty Principle. If an electron remains in a state for 10 seconds, the minimum uncertainty in its energy is approximately 5.3x10^-25 J. The uncertainty principle states that there is a fundamental limit to how precisely we can simultaneously know certain pairs of physical properties, such as energy and time.

Explanation:

The minimum uncertainty in the energy state of an atom can be found using the Heisenberg Uncertainty Principle. The uncertainty in energy (AE) is given by the equation AEAt ≥ h/4π, where At is the uncertainty in time and h is Planck's constant. In this case, if the electron remains in the state for 10 s (At = 10 s), the minimum uncertainty in the energy can be calculated. Substituting the known values into the equation, we find the minimum uncertainty in energy to be approximately 5.3x10-25 J.

The uncertainty principle states that there is a fundamental limit to how precisely we can simultaneously know certain pairs of physical properties, such as energy and time. The uncertainty in energy is inversely proportional to the uncertainty in time. Therefore, the longer the electron remains in the energy state, the smaller the uncertainty in energy will be. This means that the more stable the state, the more accurately we can determine its energy value.

It is important to note that the uncertainty in energy is relatively small compared to typical excitation energies in atoms, which are on the order of 1 eV. Therefore, the uncertainty principle has a minimal effect on the accuracy with which we can measure the energy of such states.

Given:

L = 1 mH = [tex]1\times 10^{-3}[/tex] H

total Resistance, R = 11 [tex]\Omega[/tex]

current at t = 0 s,

[tex]I_{o}[/tex] = 2.8 A

Formula used:

[tex]I = I_{o}\times e^-{\frac{R}{L}t}[/tex]

Solution:

Using the given formula:

current after t = 0.5 ms = [tex]0.5\times 10^{-3} s[/tex]

for the inductive circuit:

[tex]I = 2.8\times e^-{\frac{11}{1\times 10^{-3}}\times 0.5\times 10^{-3}}[/tex]

[tex]I =   2.8\times e^-5.5[/tex]

I =0.011 A

Answer:

10 seconds

Explanation:

x = x₀ + v₀ t + ½ at²

250 = 0 + (0) t + ½ (5) t²

250 = 2.5 t²

t² = 100

t = 10

It takes 10 seconds to land from a height of 250 ft.

Answer:

10 seconds

Explanation:

It takes 10 seconds for an object to fall from an elevation of 250 ft.

250 = 0 + (0) t + ½ (5) t²

Answer:

170.36 amu

Explanation:

Ion makes 5 revolutions in 1.11 ms

Frequency = 5 / (1.11 x 10^-3) = 4504.5 rps

B = 50 mT = 0.05 T

q = 1.6 x 10^-19 C

Let m be the mass in kg

Time period is given by

T = (2 π m) / (B q)

Frequency is the reciprocal of time period.

f = 1 / T = B q / 2πm

So,

m = B q / 2 π f

m = (0.05 x 1.6 x 10^-19) / ( 2 x 3.14 x 4504.5) = 2.828 x 10^-25 kg

as we know that

1 amu = 1.66 x 10^-27 kg

So, m = (2.828 x 10^-25 ) / (1.66 x 10^-27) = 170.36 amu

Final answer:

The peak emf generated by a 1000-turn, 42.0 cm diameter coil initially perpendicular to the Earth's 5.00 × 10^-5 T magnetic field and rotated to be parallel in 12.0 ms is approximately 1.1 V.

Explanation:

Calculating the Peak EMF in a Rotating Coil

To calculate the peak emf generated by a rotating coil in a magnetic field, we can use Faraday's Law of electromagnetic induction. The formula derived from Faraday's Law for a coil with multiple turns is:

emf = -N * (change in magnetic flux)/change in time

The magnetic flux (Φ) is given by the product of the magnetic field (B), the area (A) of the coil, and the cosine of the angle (θ) between the magnetic field and the normal to the surface of the coil:

Φ = B * A * cos(θ)

The question states that the coil with 1000 turns and a 42.0 cm diameter is initially perpendicular to the Earth's magnetic field of 5.00 × 10-5 T, and then rotated to become parallel. This change goes from cos(90°), which is 0, to cos(0°), which is 1, over a time interval of 12.0 ms.

The area A of the coil is π * (radius)2, where radius is half the diameter. The radius is 42.0 cm / 2 = 21.0 cm = 0.21 m. Thus:

A = π * (0.21 m)2

Now, we plug the values into the equation to find the peak emf:

Peak emf = -(1000) * (5.00 × 10-5 T * π * (0.21 m)2 - 0) / (12.0 × 10-3 s)

After calculation:

Peak emf ≈ 1.1 V

This is the maximum emf induced in the coil during its rotation.

The peak emf generated by rotating the given coil in the Earth's magnetic field is approximately 3.63 V.

To find the peak emf generated by rotating a coil in a magnetic field, we use Faraday's Law of Induction. The peak emf (ε) can be calculated using the formula:

ε = NABω

where N is the number of turns, A is the area of the coil, B is the magnetic field strength, and ω is the angular velocity in radians per second. First, we need to find the area (A) of the coil:

A = πr²

Given the diameter of the coil is 42.0 cm, the radius (r) is 21.0 cm or 0.21 m. So,

A = π(0.21)² = 0.1385 m²

Next, we determine the angular velocity (ω) given one full rotation in 12.0 ms:

ω = 2π / T

where T is the period (12.0 ms or 0.012 s),

ω = 2π / 0.012 = 523.6 rad/s

Now, substituting the values into the emf formula:

ε = 1000 × 0.1385 m² × 5.00 × 10⁻⁵ T × 523.6 rad/s = 3.629 V

Thus, the peak emf generated is approximately 3.63 V.

Answer:

Maximum speed = 19.81 m/s

Explanation:

Maximum speed can the car have without flying off the road at the top of the hill.

For this condition to occur we have

         Centripetal force ≥ Weight of car.

          [tex]\frac{mv^2}{r}\geq mg[/tex]

For maximum speed without flying we have

        [tex]\frac{mv^2}{r}=mg\\\\\frac{v^2}{r}=g\\\\v=\sqrt{rg}=\sqrt{40\times 9.81}=19.81m/s[/tex]

Maximum speed = 19.81 m/s

The maximum speed of the car on top of hill is 19.8 m/s.

The given parameters;

radius of the hill, r = 40 m

The maximum speed of the car on top of hill is calculated as follows;

the centripetal force must be equal or greater than weight of the car.

[tex]F_c = mg\\\\\frac{mv^2}{r} = mg\\\\\frac{v^2}{r} = g\\\\v^2 = rg\\\\v = \sqrt{rg} \\\\v = \sqrt{40 \times 9.8} \\\\v = 19.80 \ m/s[/tex]

Thus, the maximum speed of the car on top of hill is 19.8 m/s.

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Answer:

[tex]T = 3.5^0 C[/tex]

Explanation:

Heat given by water + cup  = Heat absorbed by the ice

here we can say that let the final temperature of the system is "T"

so we will have heat absorbed by the ice given as

[tex]Q = mL + ms\Delta T[/tex]

[tex]Q = (32.1)(335) + (32.1)(4.186)(T - 0)[/tex]

[tex]Q_{in} = 10753.5 + 134.4 T[/tex]

now we will have heat given by cup + water as

[tex]Q = m_w s_w(23.5 - T) + m_c s_c(23.5 - T)[/tex]

[tex]Q = 120(4.186)(23.5 - T) + 64.2(0.900)(23.5 - T)[/tex]

[tex]Q_{out} = 560.1(23.5 - T)[/tex]

now we have

[tex]Q_{in} = Q_{out}[/tex]

[tex]10753.5 + 134.4T = 560.1(23.5 - T)[/tex]

[tex]10753.5 + 134.4T = 13162.4 - 560.1 T[/tex]

[tex]694.5T = 2409[/tex]

[tex]T = 3.5^0 C[/tex]

Answer:

6 x (10)^26 electrons.

Explanation:

1 mole = 18 gr

1 gm =1/18 mole

1 kg = 1000/18 mole

Now , 1 mole of any compound = 6.022 x (10)^23 atoms.

Therefore, 1 kg of H20= (1000/18)*(6.022 x (10)^23) atoms

Roughly , 3.34 x (10)^25 molecules

And each molecule has 18 electrons

Therefore, 6 x (10)^26 electrons.

Thank you.