An astronaut leaves Earth in a spaceship at a speed of 0.96 c relative to an observer on Earth. The astronaut's destination is a star system 14.4 light-years away (one light-year is the distance light travels in one year.) According to the astronaut, how long does the trip take?

Answer:

0.3 N

Explanation:

mass of cork = 0.01 kg, density of cork = 250 kg/m^3

density of water = 1000 kg/m^3, g = 10 m/s^2

Tension in the rope = Buoyant force acting on the cork - Weight of the cork

Buoyant force = volume of cork x density of water x g

                        = mass x density of water x g / density of cork

                       = 0.01 x 1000 x 10 / 250 = 0.4 N

Weight of cork = mass of cork x g = 0.01 x 10 = 0.1 N

Thus, the tension in the rope = 0.4 - 0.1 = 0.3 N

Answer:

Tension = 0.3 N

Explanation:

As we know that the cork is inside water

so the buoyancy force on the cork is counter balanced by tension force in string and weight of the block

So the force equation is given as

[tex]F_b = T + mg[/tex]

now we will have

[tex]Volume = \frac{mass}{density}[/tex]

[tex]V = \frac{0.01}{250} = 4 \times 10^{-5} m^3[/tex]

now buoyancy force on the block is given by

[tex]F_b = \rho V g[/tex]

[tex]F_b = 1000(4 \times 10^{-5})(10)[/tex]

[tex]F_b = 0.4 N[/tex]

now by force balance equation

[tex]0.4 = T + 0.01(10)[/tex]

[tex]T = 0.4 - 0.1 = 0.3 N[/tex]

Answer:

c) 359%

Explanation:

k = spring constant of the spring

x₁ = initial stretch of the spring = 7 cm = 0.07 m

x₂ = final stretch of the spring = 15 cm = 0.15 m

Percentage increase in energy is given as

[tex]\frac{\Delta U \times 100}{U_{1}}=\frac{((0.5)k{x_{2}}^{2} - {(0.5)k x_{1}}^{2})(100)}{(0.5)k{x_{1}}^{2}}[/tex]

[tex]\frac{\Delta U \times 100}{U_{1}}=\frac{({x_{2}}^{2} - {x_{1}}^{2})(100)}{{x_{1}}^{2}}[/tex]

[tex]\frac{\Delta U \times 100}{U_{1}}=\frac{({0.15}^{2} - {0.07}^{2})(100)}{{0.07}^{2}}[/tex]

[tex]\frac{\Delta U \times 100}{U_{1}}[/tex] = 359%

Answer:

Motorcycle travel 39.75 m while coming to a stop

Explanation:

We have equation of motion

        v = u + at

   Final velocity, v = 0 m/s, u = 15 m/s, t = 5.3 s

        0 = 15 + a x 5.3

         a = -2.83 m/s²

 Now we have the other equation of motion

        v² = u² + 2as

        0² = 15² - 2 x 2.83 x s

         s = 39.75 m

Motorcycle travel 39.75 m while coming to a stop

Answer:

Gravitational force = 89.18 x 10^-31 N

Electric force = 2.4 x 10^-17 N

Magnetic force = 7.68 x 10^-17 N

Explanation:

B = 80 micro tesla = 80 x 10^-6 T  north

E = 150 N/C downward

v = 6 x 10^6 m/s east

Gravitational force = m g = 9.1 x 10^-31 x 9.8 = 89.18 x 10^-31 N

Electric force = q E = 1.6 x 10^-19 x 150 = 2.4 x 10^-17 N

Magnetic force = q v B Sin 90 = 1.6 x 10^-19 x 6 x 10^6 x 80 x 10^-6

                          = 7.68 x 10^-17 N

Answer:

15.7 N

Explanation:

Draw a free body diagram.  The block has four forces acting on it.  Gravity pulling down, normal force pushing perpendicular to the plane, friction pointing up the plane, and applied force F pushing up the plane.

Sum of the forces normal to the plane:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces parallel to the plane:

∑F = ma

Nμ + F − mg sin θ = 0

F = mg sin θ − Nμ

Substituting:

F = mg sin θ − mgμ cos θ

F = mg (sin θ − μ cos θ)

Given mg = 87.0 N, θ = 24.1°, and μ = 0.25 (because the block is not moving):

F = 87.0 N (sin 24.1° − 0.25 cos 24.1°)

F = 15.7 N

The minimum magnitude of the force F required to prevent the block from slipping is approximately 14.114 N.

The minimum magnitude of the force F required to prevent the block from slipping is given by the equation:

[tex]\[ F = w \sin(\theta) - \mu_s w \cos(\theta) \][/tex]

where w is the weight of the block, [tex]\( \theta \)[/tex] is the angle of inclination, and [tex]\mu_s \)[/tex] is the coefficient of static friction.

Given:

-[tex]\( w = 87.0 \, \text{N} \)[/tex]

- [tex]\( \theta = 24.1^\circ \)[/tex]

- [tex]\( \mu_s = 0.25 \)[/tex]

First, convert the angle from degrees to radians because the sine and cosine functions in most calculators require the input to be in radians:

[tex]\[ \theta_{\text{rad}} = \theta \times \frac{\pi}{180^\circ} \][/tex]

[tex]\[ \theta_{\text{rad}} = 24.1^\circ \times \frac{\pi}{180^\circ} \approx 0.420 \text{ radians} \][/tex]

Now, calculate the minimum force F using the given formula:

[tex]\[ F = w \sin(\theta_{\text{rad}}) - \mu_s w \cos(\theta_{\text{rad}}) \][/tex]

[tex]\[ F = 87.0 \, \text{N} \times \sin(0.420) - 0.25 \times 87.0 \, \text{N} \times \cos(0.420) \][/tex]

[tex]\[ F \ =87.0 \, \text{N} \times 0.412 - 0.25 \times 87.0 \, \text{N} \times 0.910 \][/tex]

[tex]\[ F \ = 35.864 \, \text{N} - 21.75 \, \text{N} \][/tex]

[tex]\[ F \ =14.114 \, \text{N} \][/tex]

Answer:

6.03 x 10^-3 C/Kg

Explanation:

E = 3.8 x 10^4 N/C, u = 2.32 m/s, s = 5.98 cm = 0.0598 m, t = 0.2 s, g = 9.8 m/s^2

Acceleration on object is a .

Use second equation of  motion.

S = u t + 1/2 a t^2

0.0598 = 2.32 x 0.2 + 0.5 x a x 0.2 x 0.2

0.0598 = 4.64 + 0.02 x a

a = - 229 m/s^2

Now, F = ma = qE

q / m = a / E = 229 / (3.8 x 10000)

q / m = 6.03 x 10^-3 C/Kg

Calculate the charge-to-mass ratio of the object given initial speed, distance, time, and electric field strength.

Given: Electric field magnitude = 3.80 x 10^4 N/C, initial speed = 2.32 m/s, distance traveled = 5.98 cm, time = 0.200 s, acceleration due to gravity = 9.80 m/s^2.

To find: The object's charge-to-mass ratio (q/m).

Calculations: Determine the acceleration using the provided data, then relate the forces acting on the object (gravity and electric field) to find q/m.

Answer:

a)  Tangential speed = 3.57 m/s

b) Centripetal acceleration = 17.79 m/s²

c) Maximum tangential speed the ball can have = 5.41 m/s

Explanation:

a) Tangential speed, v = rω

   Radius, r = 0.86 m

   Angular speed, ω = 0.660 rev/s = 0.66 x 2 x π = 4.15 rad/s

  Tangential speed, v = rω = 0.86 x 4.15 = 3.57 m/s

b) Centripetal acceleration,

          [tex]a=\frac{v^2}{r}=\frac{3.57^2}{0.86}=14.79m/s^2[/tex]

c) Maximum tension in horizontal circular motion

           [tex]T=\frac{mv^2}{r}\\\\136=\frac{4\times v^2}{0.86}\\\\v=5.41m/s[/tex]

  Maximum tangential speed the ball can have = 5.41 m/s

The questions involve concepts of circular motion: tangential speed, centripetal acceleration, and the tension in the rope maintaining the motion. Tangential speed is derived from the radius of the circle and the angular velocity, centripetal acceleration is derived from the square of tangential speed divided by the radius, and the maximum tangibility speed before the rope breaks can be calculated from its tension.

This is a problem of circular motion, and it's crucial to understand the concepts of tangential speed, centripetal acceleration, and the tension in the rope.

The tangential speed of a body moving along a circular path is its normal linear speed for that moment. It can be calculated by using the formula v = rω, where v is the tangential speed, r is the radius of the circle, and ω is the angular velocity. Make sure to convert the angular speed from rev/s to rad/s (ω = 0.660 rev/s * 2π rad/rev).

Centripetal acceleration is the rate of change of tangential velocity, and it always points toward the center of the circle. It can be calculated using the formula ac = v²/r.

The tension in the rope is the force exerting on the ball to keep it moving circularly. If it exceeds its maximum limit (136N in this case), the rope will break. The maximum tangential speed can be derived from the equation T = m * v² / r.

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Answer:

amplitude, a = 3.076 mm

frequency, f = 10.35 Hz

Explanation:

Vmax = 200 mm /s = 0.2 m/s

Amax = 13 m/s

Let f be the frequency and a be the amplitude.

Use the formula of maximum velocity.

Vmax = ω a

0.2 = ω a    ..... (1)

Use the formula of maximum acceleration.

Amax = ω^2 x a

13 = ω^2 a    ..... (2)

Divide equation (2) by (1)

ω = 13 / 0.2 = 65 rad/s

Put in equation (1)

0.2 = 65 x a

a = 3.076 x 10^-3 m

a = 3.076 mm

Let f be the frequency

ω = 2 π f

f = 65 / (2 x 3.14) = 10.35 Hz  

Answer:

Option (D)

Explanation:

In the projectile motion

Horizontal distance = horizontal velocity × time

3.3 = u × 0.66

u = 5 m/s

Muzzle velocity is 5 m/s

Ham como assim pq ta em ingles

Answer:

5.062 x 10^-9 mg

Explanation:

Charge on glass rod = 89 uC

Charge on silk cloth = - 89 uC

mass of scarf, M = 100 g

Number of excess electrons on silk cloth = charge / charge of one electron

n = (89 x 10^-6) / ( 1.6 x 10^-19)

n = 5.5625 x 10^14

mass of one electron = 9.1 x 10^-31 kg

Mass of  5.5625 x 10^14 electrons =  5.5625 x 10^14 x 9.1 x 10^-31 kg

ΔM = 5.062 x 10^-16 kg

Δ M / M = (5.062 x 10^-16) / 0.1 = 5.062 x 10^-15 kg = 5.062 x 10^-9 mg

It is not noticeable.

Answer:

YES THERE ARE VERY MANY BOOKS THAT YOU CAN BUY OR BORROW FROM YOUR LOCAL LIBRARY, HOPE THIS HELPS! HAVE A GREAT DAY!

Explanation:

Answer:

Vertically upwards

Explanation:

Fleming's left hand rule: it states taht when we spread our fore finger, middle finger and thumb in such a way that they are mutually perpendicular to each other, theN thumb indicates the direction of force, fore finger indicates the direction of magnetic field and the middle finger indicates the direction of current.

So according to this rule the direction of force is vertically upwards.

Final answer:

To find the index of refraction of the glass, we can use Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities of the light in the two media.

Explanation:

To find the index of refraction of the glass, we can use Snell's law, which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities of the light in the two media:

n1 * sin(θ1) = n2 * sin(θ2)

Given that the incident angle is 50.0° and the refracted angle is 27.7°, we can plug in the values to find the index of refraction:

n1 * sin(50.0°) = n2 * sin(27.7°)

The index of refraction of the glass is the ratio of the index of refraction of the glass to the index of refraction of air, which is approximately 1:

n2 / n1 = 1 / sin(27.7°)

R = 0.407Ω.

The resistance  R of a particular conductor is related to the resistivity ρ of the material by  the equation R = ρL/A, where ρ is the material resistivity, L is the length of the material and A is the cross-sectional area of ​​the material.

To calculate the resistance R of a wire made of a material with resistivity of 3.2x10⁻⁸Ω.m, the length of the wire is 2.5m and its diameter is 0.50mm.

We have to use the equation R = ρL/A but first we have to calculate the cross-sectional area of the wire which is a circle. So, the area of a circle is given by A = πr², with r = d/2. The cross-sectional area of the wire is A = πd²/4.  Then:

R =[(3.2x10⁻⁸Ω.m)(2.5m)]/[π(0.5x10⁻³m)²/4]

R = 8x10⁻⁸Ω.m²/1.96x10⁻⁷m²

R = 0.407Ω

To determine the resistance of this wire is equal to 0.4082 Ohms.

Given the following data:

To determine the resistance of this wire:

First of all, we would determine the area of the wire by using this formula:

[tex]A =\frac{\pi d^2}{4} \\\\A =\frac{3.142 \times (5\times 10^{-4})^2}{4}\\\\A = 1.96 \times 10^{-7}\;m^2[/tex]

Now, we can determine the resistance of this wire:

[tex]R=\frac{\rho L}{A} \\\\R=\frac{3.2 \times 10^{-8} \times 2.5}{1.96 \times 10^{-7}}[/tex]

Resistance = 0.4082 Ohms.

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Final answer:

The student must use Coulomb's law to calculate the net electrostatic force between two charged red blood cells, and then use energy conservation to determine the initial speed needed for the cells to overcome this force and just touch. So, the initial speed of each cell is approximately [tex]\( 662.8 \, \text{m/s} \)[/tex].

Explanation:

Coulomb's Law and Initial Speed Calculation

The student's problem involving two red blood cells carrying negative charges requires the use of Coulomb's law to determine the electrostatic force of repulsion between them. Since the cells are modeled as spheres, the charges are assumed to be uniformly distributed. To find the initial speed needed for the cells to just touch, one must consider energy conservation principles, where the initial kinetic energy of the cells should be equivalent to the electrostatic potential energy at the point of closest approach (when they just touch).

The formula for Coulomb's law is
F = k * |q1 * q2| / r^2, where F is the force between the charges, q1 and q2 are the charges, r is the separation distance, and k is Coulomb's constant (8.9875 x 10^9 Nm^2/C^2). The potential energy (U) due to electrostatic force is given by U = k * |q1 * q2| / r. By setting the kinetic energy (1/2 * m * v^2) for both cells equal to U, and solving for v, we can find the initial speed of each cell.

To find the initial speed of each cell using the potential energy and kinetic energy, you can follow these steps:

Given:

[tex]- Charges: \( q_1 = -2.50 \, \text{pC} \), \( q_2 = -3.30 \, \text{pC} \)\\- Separation distance: \( r = 3.75 \times 10^{-6} \, \text{m} \)\\- Coulomb's constant: \( k = 8.9875 \times 10^9 \, \text{Nm}^2/\text{C}^2 \)\\- Mass: \( m = 9.05 \times 10^{-14} \, \text{kg} \)\\[/tex]

1. Calculate the potential energy \( U \) using Coulomb's law:

  [tex]\[ U = \frac{k \cdot |q_1 \cdot q_2|}{r} \] \[ U = \frac{8.9875 \times 10^9 \, \text{Nm}^2/\text{C}^2 \cdot |(-2.50 \times 10^{-12} \, \text{C}) \cdot (-3.30 \times 10^{-12} \, \text{C})|}{3.75 \times 10^{-6} \, \text{m}} \][/tex]

2. Calculate the potential energy \( U \):

  [tex]\[ U = \frac{8.9875 \times 10^9 \, \text{Nm}^2/\text{C}^2 \cdot 8.25 \times 10^{-24} \, \text{C}^2}{3.75 \times 10^{-6} \, \text{m}} \] \[ U = \frac{7.465625 \times 10^{-14} \, \text{Nm}}{3.75 \times 10^{-6} \, \text{m}} \] \[ U = 1.989 \times 10^{-8} \, \text{J} \][/tex]

3. Set the kinetic energy equal to the potential energy:

 [tex]\[ \frac{1}{2} m v^2 = U \][/tex]

4. Solve for \( v \):

  [tex]\[ v = \sqrt{\frac{2U}{m}} \] \[ v = \sqrt{\frac{2 \times 1.989 \times 10^{-8} \, \text{J}}{9.05 \times 10^{-14} \, \text{kg}}} \][/tex]

5. Calculate \( v \):

 [tex]\[ v \approx \sqrt{\frac{2 \times 1.989 \times 10^{-8} \, \text{J}}{9.05 \times 10^{-14} \, \text{kg}}} \] \[ v \approx \sqrt{4.391 \times 10^5 \, \text{m}^2/\text{s}^2} \] \[ v \approx 662.8 \, \text{m/s} \][/tex]

So, the initial speed of each cell is approximately [tex]\( 662.8 \, \text{m/s} \)[/tex].

Since there is no viscous drag, the conservation of energy provides the necessary framework to equate the initial kinetic energy to the electrostatic potential energy. Accounting for the fact that each blood cell has a different charge, the calculation would involve determining the net repulsive force and then the corresponding speed for each cell to overcome this force and just barely touch.

The initial speed required for each red blood cell to just barely touch is approximately [tex]\( 1.479 \times 10^{3} \)[/tex]m/s.

Given:

- Mass of each red blood cell, [tex]\( m = 9.05 \times 10^{-14} \)[/tex] kg

- Charge on the first cell, [tex]\( q_1 = -2.50 \) pC \( = -2.50 \times 10^{-12} \) C[/tex]

- Charge on the second cell, [tex]\( q_2 = -3.30 \) pC \( = -3.30 \times 10^{-12} \) C[/tex]

- Radius of each cell, [tex]\( r = 3.75 \times 10^{-6} \) m[/tex]

 When the cells just barely touch, the distance between their centers will be 2r , since each cell has a radius r.

The electrostatic force F between the two charges can be calculated using Coulomb's law:

[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]

where \( k \) is Coulomb's constant, [tex]\( k = 8.9875 \times 10^9 \)[/tex] N m²/C².

Substituting the values, we get:

[tex]\[ F = (8.9875 \times 10^9) \frac{|(-2.50 \times 10^{-12})(-3.30 \times 10^{-12})|}{(2 \times 3.75 \times 10^{-6})^2} \][/tex]

Now, we need to find the initial kinetic energy ( KE ) of each cell, which must be equal to the work done by the electrostatic force when the cells move from infinity to a distance ( 2r ) apart. The work done by the electrostatic force is equal to the potential energy U at a distance 2r:

[tex]\[ U = k \frac{|q_1 q_2|}{2r} \][/tex]

The initial kinetic energy of each cell is given by:

[tex]\[ KE = \frac{1}{2} m v^2 \][/tex]

where v is the initial speed we want to find.

Since there are no other forces doing work (no viscous drag), the total mechanical energy is conserved, and the initial kinetic energy must equal the potential energy at the point of touching:

[tex]\[ \frac{1}{2} m v^2 = k \frac{|q_1 q_2|}{2r} \][/tex]

Solving for v, we get:

[tex]\[ v = \sqrt{\frac{2k |q_1 q_2|}{m (2r)}} \][/tex]

Substituting the values, we have:

[tex]\[ v = \sqrt{\frac{2 \times (8.9875 \times 10^9) \times |(-2.50 \times 10^{-12})(-3.30 \times 10^{-12})|}{9.05 \times 10^{-14} \times (2 \times 3.75 \times 10^{-6})}} \][/tex]

Now, let's calculate the value of \( v \):

[tex]\[ v = \sqrt{\frac{2 \times (8.9875 \times 10^9) \times (2.50 \times 10^{-12} \times 3.30 \times 10^{-12})}{9.05 \times 10^{-14} \times (2 \times 3.75 \times 10^{-6})}} \] \[ v = \sqrt{\frac{2 \times (8.9875 \times 10^9) \times (8.25 \times 10^{-24})}{9.05 \times 10^{-14} \times (7.5 \times 10^{-6})}} \] \[ v = \sqrt{\frac{2 \times (7.4234375 \times 10^{-14})}{6.79875 \times 10^{-20}}} \][/tex]

[tex]\[ v = \sqrt{\frac{14.846875 \times 10^{-14}}{6.79875 \times 10^{-20}}} \] \[ v = \sqrt{2.18375 \times 10^{6}} \] \[ v \approx 1.479 \times 10^{3} \text{ m/s} \][/tex]

Final answer:

The question revolves around the final temperature of a mixture of ice and water in a thermally insulated system. The specific heats and enthalpy of fusion are required to solve, but the proper method involves using the conservation of energy.

Explanation:

The student is asking about the final temperature of a system consisting of a 0.0811 kg ice cube at 0°C and 0.397 kg of water at 14.8°C. To solve this, we can use the principle of conservation of energy, stating that the heat lost by the water will equal the heat gained by the ice cube as it warms up, melts, and then possibly continues to heat up as water. However, without knowing the specific heat capacities and the enthalpy of fusion, we cannot provide a numerical answer. Given the specific heats and the heat of fusion, one would set up an equation where the heat lost by the water equals the heat gained by the ice, and solve for the final temperature. The specific heat of water, the specific heat of ice, and the enthalpy of fusion of ice would be necessary information to perform actual calculations.

Answer:

[tex]3.0\cdot 10^{-5} N/m^2[/tex]

Explanation:

The intensity of the radiation at the location of the satellite is given by:

[tex]I=\frac{P}{4\pi r^2}[/tex]

where

[tex]P=3.8\cdot 10^{26}W[/tex] is the power of the emitted radiation

[tex]4\pi r^2[/tex] is the area over which the radiation is emitted (the surface of a sphere), with

[tex]r=5.8\cdot 10^{10}m[/tex] being the radius of the orbit

Substituting,

[tex]I=\frac{3.8\cdot 10^{26}}{4\pi (5.8\cdot 10^{10})^2}=8989 W/m^2[/tex]

Now we can find the pressure of radiation, that for a totally absorbing surface (such as the satellite) is:

[tex]p=\frac{I}{c}[/tex]

where

[tex]c=3.0\cdot 10^8 m/s[/tex]

is the speed of light. Substituting,

[tex]p=\frac{8989}{3.0\cdot 10^8}=3.0\cdot 10^{-5} N/m^2[/tex]

Answer:

980 kJ

Explanation:

Work = change in energy

W = mgh

W = (1000 kg/m³ × 5.0 m³) (9.8 m/s²) (20 m)

W = 980,000 J

W = 980 kJ

The pump does 980 kJ of work.

The work done by the pump against gravity in transfering 5 m³ of water to a tank 20 m above the lake is 980000 J

We'll begin by calculating the mass of the water. This can be obtained as follow:

Volume of water = 5 m³

Density of water = 1000 Kg/m³

Mass = Density × Volume

Mass of water = 1000 × 5

Mass of water (m) = 5000 Kg

Height (h) = 20 m

Acceleration due to gravity (g) = 9.8 m/s²

Work done = mgh

Work done = 5000 × 9.8 × 20

Thus, the work done by the pump is 980000 J

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Answer:

1.2826 x 10^-13 m

Explanation:

[tex]\lambda  = \frac{h}{\sqrt{2 m K}}[/tex]

Here, k be the kinetic energy and m be the mass

K = 50 KeV = 50 x 1.6 x 10^-16 J = 80 x 10^-16 J

m = 1.67 x 10^-27 kg

[tex]\lambda  = \frac{6.63 \times  10^{-34}}{\sqrt{2 \times 1.67\times 10^{-27}\times 80\times 10^{-16}}}[/tex]

λ = 1.2826 x 10^-13 m

Answer:

Intensity of a wave is Power per unit area.

Explanation:

The intensity of wave is defined as the power per unit area in the direction of motion of wave i.e.

[tex]I=\dfrac{P}{A}[/tex]

Also, the intensity of a wave is directly proportional to the square of its amplitude. Let a is the amplitude of a wave. So,

[tex]I\propto a^2[/tex]

The SI unit of power is watt and the SI unit of area is m². So, the SI unit of intensity is W/m². Hence, the correct option is (a) "Intensity of a wave is Power per unit area".

Answer:

Time period of oscillations is 0.62 s

Explanation:

Due to suspension of weight the change in the length of the spring is given as

[tex]\Delta L = L_f - L_i[/tex]

[tex]\Delta L = 21.5 - 11.9 = 9.6 cm[/tex]

now we know that spring is stretched due to its weight so at equilibrium the force due to weight is counter balanced by the spring force

[tex]mg = kx[/tex]

[tex]0.037 (9.81) = k(0.096)[/tex]

[tex]k = 3.78 N/m[/tex]

Now the period of oscillation of spring is given as

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

Now plug in all values in it

[tex]T = 2\pi \sqrt{\frac{0.037}{3.78}}[/tex]

[tex]T = 0.62 s[/tex]

The period of the oscillation for the spring with a spring constant of 3.78 N/m will be 0.7316 sec.

We know that the formula of the spring force is written as,

[tex]F = k \times \delta[/tex]

Given to us,

weight on the spring, F = 37 g = (9.81 x 0.037) = 0.36297 N

Deflection in spring, δ = (21.5 - 11.9) = 9.6 cm = 0.096 m

Substitute the value,

[tex]0.36297 = k \times 0.096\\k= 3.78\rm\ N/m[/tex]

We know that the weight is been pulled down therefore, an external force is been applied to the spring to extend it further to 25.2 cm.

We know that the formula of the spring force is written as,

[tex]F = k \times \delta[/tex]

Given to us

Deflection in the spring, δ = (25.2 - 11.9) = 13.3 cm = 0.133 m

Spring constant, k = 3.78 N/m

Substitute the values,

[tex]F = 3.78 \times 0.133\\\\F = 0.5\\\\m \times g = 0.5\\\\m = 0.05125\rm\ kg[/tex]

We know the formula for the period of oscillation is given as,

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

Substitute the values,

m = 0.05125 kg

k = 3.78 N/m

[tex]T = 2\pi \sqrt{\dfrac{0.05125}{3.78}}\\T = 0.7316\rm\ sec[/tex]

Hence, the period of the oscillation for the spring with a spring constant of 3.78 N/m will be 0.7316 sec.

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Answer:

option c) 0.88c is the correct answer

Explanation:

using the Lorrentz equation we have

[tex]t=\frac{d}{v}\sqrt{1-(\frac{v}{c})^2}[/tex]

where,

t = time taken to cover the distance

d = Distance

v = velocity

c = speed of light

given

d = 15 light years

Now,

[tex]8years=\frac{15years\times c}{v}\sqrt{1-(\frac{v}{c})^2}[/tex]

or

[tex](\frac{v}{c})^2\times (\frac{8}{15})^2=1-(\frac{v}{c})^2[/tex]

or

[tex](\frac{v}{c})^2\times 0.284=1-(\frac{v}{c})^2[/tex]

or

[tex](\frac{v}{c})^2\times 0.284 + (\frac{v}{c})^2 =1[/tex]

or

[tex](1.284\times \frac{v}{c})^2 =1[/tex]

or

[tex] (\frac{v}{c}) =\sqrt{\frac{1}{0.284}}[/tex]

or

[tex]v=0.88c[/tex]

Answer:

d = 506.25 ft

Explanation:

As we know by kinematics that

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

here we know that initially the stone is dropped from rest from the edge of the roof

so here initial speed will be zero

now we have

[tex]v_i = 0[/tex]

also the acceleration of the stone is due to gravity which is given as

[tex]g = 32 ft/s^2[/tex]

now we have

[tex]v_f = 180 ft/s[/tex]

so from above equation

[tex]180^2 - 0 = 2(32)d[/tex]

[tex]d = 506.25 ft[/tex]

Using the equation of motion, we find that the height of the roof from which the stone was dropped is approximately 506.25 feet.

The subject question involves a physics concept relating to the motion of objects under the influence of gravity, in particular, the calculation of displacement during free fall. We can solve this problem using one of the fundamental equations of motion: final velocity2 = initial velocity2 + 2*a*d. In this case, initial velocity (u) = 0 (as the stone was dropped), final velocity (v) = -180 feet per second (negative as the stone is moving downwards), and acceleration (a) due to gravity is -32 feet/second2 (negative as gravity acts downward).

Substituting these values, we get:
(-180)2 = (0)2 + 2*(-32)*d.

This simplifies to:
32400 = -64d.

Solving for d (which represents the height of the roof), we get:
d = -32400 / -64 = 506.25 feet. Hence, the roof is approximately 506.25 feet high.

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Explanation:

a) The equivalent resistance for resistors in series is the sum:

R = R₁ + R₂ + R₃

R = 10 Ω + 20 Ω + 30 Ω

R = 60 Ω

b) The resistors are in series, so they have the same current.

I = V/R

I = 12 V / 60 Ω

I = 0.2 A

c) The voltage drop can be found with Ohm's law:

V₁ = I R₁ = (0.2 A) (10 Ω) = 2 V

V₂ = I R₂ = (0.2 A) (20 Ω) = 4 V

V₃ = I R₃ = (0.2 A) (30 Ω) = 6 V

d) The power lost in each resistor is current times voltage drop:

P₁ = I V₁ = (0.2 A) (2 V) = 0.4 W

P₂ = I V₂ = (0.2 A) (4 V) = 0.8 W

P₃ = I V₃ = (0.2 A) (6 V) = 1.2 W

(a) The equivalent resistance of the circuit is 60 Ω.

(b) The current in each resistor is 0.2 A.

(c) The voltage across each resistor is 2 V, 4 V, and 6 V respectively.

(d) The power lost in each resistor is 0.4 W, 0.8 W and 1.2 W.

The equivalent resistance of the circuit is determined as follows;

Rt = R₁ + R₂ + R₃

Rt = 10 + 20 + 30

Rt = 60 Ω

The current in the resistors is calculated as follows;

V = IR

I = V/R

I = 12/60

I = 0.2 A

V1 =IR1 = 0.2 x 10 = 2 V

V2 = IR2  = 0.2 x 20 = 4 V

V3 = 0.2 x 30 = 6 V

P1 = I²R1 = (0.2)² x 10 = 0.4 W

P2 = I²R2 = (0.2)² x 20 = 0.8 W

P3 = I²R3 = (0.2)² x 30 = 1.2 W

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Answer:

Explanation: r₂ = 15 x 10⁻³ m ; r = 12 x 10⁻³m ; r₁ = 10 x 10⁻³ m

     E = Q /4π∈r²x ( r³ - r₁ ³) / ( r₂³ - r₁³ )

         = 15 x 10⁻⁶ x 8.99 x 10⁹x( 12³ - 10³ ) / ( 15³- 10³ ) X 12³

          = 23.92 N/C

The electric field is the force acting on a charge located in a region in space. The electric field at a distance r from the center of the shell is 2.8x10⁶N/C

The electric field (E) can be defined as a region in the space that interacts with electric charges or charged bodies through electric forces.

It is a vectorial field in which an electric charge (q) suffers the effect of the electric force (F). In other words, it represents the force that is acting on a charge located in a certain region.

E = F/q

It is expressed in newton/coulomb (N/C)

In the exposed example, we have the following data,

We need to get the value of the electric field at the given distance r.

E = Q/4πεr² = kQ/r²

Since the r value is located between r1 and r2, we need to get the electric field at that intermedia distance, r. This is a new ratio.

The total charge Q is distributed in the whole area of the shell thickness or density, delimited by r1 and r2.

But to get the electric field at the intermedia distance, r, we only need to get the charge q lower than Q and distributed between the r and r1, which represents a volume v lower than the total volume V of the shell.

So to get the q value, we need to consider the charge volumetric density, σ.

σ = Q/V

Where

But to get Q/V, we need to get the total shell volume value, V.

V = 4/3 π r³

V = 4/3 π (r2³ - r1³)

V = 4/3 π (0.15³ - 0.1³)

V = 4/3 π (0.00237)

V = 0.00993m³

Now that we have the total volume, and we know the value of the total charge, we need to get the density of the shell.

σ = Q/V

σ = 15 μC / 0.00993m³

σ = 15x10⁻⁶ C / 0.00993m³

σ = 0.00151 C/m³

Now, we will calculate the volume v between r and r1, which is what we are interested in.

v = 4/3 π r³

v = 4/3 π (r³ - r1³)

v = 4/3 π (0.12³ - 0.1³)

v = 4/3 π (0.0007)

v = 0.00305m³

Finally, using this new v value and the σ value, we can get the charge q distributed  throughout the volume v.

q = vσ

q = 0.00305m³ x 0.00151 C/m³

q = 4.6x10⁻⁶ C

Now we can get the electric field at a distance r from the center of the shell

E = kq/r²

E = ((8.99 × 10⁹ N ∙ m²/C²) x (4.6x10⁻⁶ C) ) / 0.12²m

E = 2.8x10⁶N/C

Hence, the electric field at a distance r is 2.8x10⁶N/C

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Answer:

[tex]43.28\frac{lbf}{in^3}=43.28\times 27460.68\frac{kg}{m^3}=1188498.44\frac{kg}{m^3}[/tex]

Explanation:

Here the given values corresponds to the density of a substance

The density of a substance in standard SI unit is given as Kg/m³.

Given value = 43.28lbf/in³

also

we know,

1 lbf = 0.45kg

1in = 0.0254 m

thus,

[tex]\frac{1lbf}{1in^3}=\frac{0.45kg}{0.0254^3m^3}=27460.68\frac{kg}{m^3}[/tex]

therefore,

[tex]43.28\frac{lbf}{in^3}=43.28\times 27460.68\frac{kg}{m^3}=1188498.44\frac{kg}{m^3}[/tex]

Answer:

Vector angular momentum about this axis of the sphere is:

L= 3.76[tex]\hat k[/tex] kg-m²/sec

Explanation:

The formula for the moment of inertia of a sphere is:

[tex]I=\frac{2}{5}\times MR^2[/tex]

Given:

Mass of the sphere = 13.5 kg

Radius of the sphere = 0.490 m

Thus, moment of inertia :

[tex]I=\frac{2}{5}\times 13.5\times (0.490)^2 kg\ m^2[/tex]

[tex]I=1.29654 kg\ m^2[/tex]

The expression for the angular momentum is:

L=I×ω

Given:

Angular speed(ω) = 2.9 rad/s

I, above calculated = 1.29654 kgm⁻²

Thus, angular momentum is:

L= 1.29654×2.9 kg-m²/sec

L= 3.76 kg-m²/sec

Given, the sphere is turning counterclockwise about the vertical axis. Thus, the direction of the angular momentum will be on the upper side of the plane. ( [tex]+\hat k[/tex] ).

Thus, angular momentum with direction is:

L= 3.76[tex]\hat k[/tex] kg-m²/sec

Answer:

(a) 9.65 mH

(b) doubled

Explanation:

A = 6 cm^2 = 6 x 10^-4 m^2, l = 20 cm = 0.2 m, i = 2.8 A, N = 1600

(a) The formula for the self inductance of a solenoid is given by

L = μ0 N^2 A / l

L = 4 x 3.14 x 10^-7 x 1600 x 1600 x 6 x 10^-4 / 0.2

L = 9.65 x 10^-3 H

L = 9.65 mH

(b) As we observe that the inductance is proportional to the area so if the area is doubled, then inductance is also doubled.

Answer:

[tex]v_f = 20 m/s[/tex]

Explanation:

Since the hoop is rolling on the floor so its total kinetic energy is given as

[tex]KE = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2[/tex]

now for pure rolling condition we will have

[tex]v = R\omega[/tex]

also we have

[tex]I = mR^2[/tex]

now we will have

[tex]KE = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\frac{v^2}{R^2}[/tex]

[tex]KE = mv^2[/tex]

now by work energy theorem we can say

[tex]W = KE_f - KE_i[/tex]

[tex]842 J = mv_f^2 - mv_i^2[/tex]

[tex]842 = 3(v_f^2) - 3\times 11^2[/tex]

now solve for final speed

[tex]v_f = 20 m/s[/tex]