Find the speed of light in carbon tetrachlorideethyl alcohol. The refraction index is 1.461 using 3 x 10^8 m/s as the speed of light in vacuum. Answer in units of m/s.

Answers

Answer 1

Answer:

2.05 x 10^8 m /s

Explanation:

c = 3 x 10^8 m/s

μ = c / v

where, μ is the refractive index, c be the velocity of light in air and v be the velocity of light in the medium.

μ = 1.461

1.461 = 3 x 10^8 / v

v = 3 x 10^8 / 1.461

v = 2.05 x 10^8 m /s


Related Questions

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x 5 5.00 cm and released from rest at t 5 0. (a) What is the force constant of the spring? (b) What are the angular frequency v, the frequency, and the period of the motion? (c) What is the total energy of the system?

Answers

Answer:

a) 250 N/m

b) 22.4 rad/s , 3.6 Hz , 0.28 sec

c) 0.3125 J

Explanation:

a)

F = force applied on the spring = 7.50 N

x = stretch of the spring from relaxed length when force "F" is applied = 3 cm = 0.03 m

k = spring constant of the spring

Since the force applied causes the spring to stretch

F = k x

7.50 = k (0.03)

k = 250 N/m

b)

m = mass of the particle attached to the spring = 0.500 kg

Angular frequency of motion is given as

[tex]w = \sqrt{\frac{k}{m}}[/tex]

[tex]w = \sqrt{\frac{250}{0.5}}[/tex]

[tex]w [/tex] = 22.4 rad/s

[tex]f[/tex] = frequency

Angular frequency is also given as

[tex]w [/tex] = 2 π [tex]f[/tex]

22.4 = 2 (3.14) f

[tex]f[/tex]  = 3.6 Hz

[tex]T[/tex] = Time period

Time period is given as

[tex]T = \frac{1}{f}[/tex]

[tex]T = \frac{1}{3.6}[/tex]

[tex]T[/tex] = 0.28 sec

c)

A = amplitude of motion = 5 cm = 0.05 m

Total energy of the spring-block system is given as

U = (0.5) k A²

U = (0.5) (250) (0.05)²

U = 0.3125 J

(a) The force constant of the spring is 250 N/m.

(b) The angular frequency of the mass oscillation is 22.36 rad/s, frequency is 3.56 Hz and the period is 0.28 s.

(c)  the total energy of the system is 0.31 J.

Force constant of the spring

The force constant of the spring can be determined by applying Hooke's law as follows;

F = kx

k = F/x

k = (7.5)/0.03)

k = 250 N/m

Angular frequency

The angular frequency of the mass oscillation is calculated as follows;

[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{250}{0.5} }\\\\\omega = 22.36 \ rad/s[/tex]

Angular frequency

The angular frequency is calculated as follows;

ω = 2πf

f = ω/2π

f = (22.36)/2π

f = 3.56 Hz

Period of the oscillation

The period of the oscillation is calculated as follows;

T = 1/f

T = 1/3.56

T = 0.28 s

Total energy of the system

The total energy of the system is calculated as follows;

U = ¹/₂kA²

U = ¹/₂ x 250 x (0.05)²

U = 0.31 J

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Problem 1: A positively charged particle Q1 = +35 nC is held fixed at the origin. A second charge Q2 of mass m = 8.5 micrograms is floating a distance d = 25 cm above charge Q1. The net force on Q2 is equal to zero. You may assume this system is close to the surface of the Earth. A) What is the sign of charge on Q2? B) Calculate the magnitude of Q2 in units of nanocoulombs.

Answers

Answer:This problem has been solved! See the answer. Two charged particles, Q 1 = + 5.10 nC and Q 2 = -3.40 nC, are separated by 40.0 cm. (a) What is the ... has not been graded yet. (b) What is the electric potential at a point midway between the charged particles? (Note: Assume a reference level of potential V = 0 at r = infinity .)

Explanation:

Final answer:

The sign of charge Q2 is positive, and its magnitude can be calculated by setting the gravitational force equal to the electric force due to Q1 and solving for Q2.

Explanation:

To determine the sign and magnitude of the second charge Q2, we use the concept that the net force acting on it must be zero. The forces that act on Q2 include the gravitational force and the electric force due to Q1. Since Q1 is positive and the net force is zero, Q2 must also be positive for the electric force (Coulomb force) to balance the gravitational pull downwards.

The gravitational force can be calculated using Newtons's law of gravitation F = mg, where g is the acceleration due to gravity (9.8 m/s2). The electric force is given by Coulomb's law F = k|Q1Q2|/d2, where k is Coulomb's constant (8.99 x 109 Nm2/C2).

Setting the magnitude of the gravitational force equal to the magnitude of the electric force gives the equation mg = k|Q1Q2|/d2. Solving for Q2, and converting the mass of Q2 from micrograms to kilograms, we calculate the magnitude of charge Q2.

A golf ball was thrown with a speed of 25 m/sec at an angle of 65° above the horizontal. What is the horizontal distance traveled by the ball before it hits the ground?

Answers

Answer:

The horizontal distance traveled by the ball before it hits the ground is 48.85 meters.

Explanation:

It is given that,

Speed of golf ball, v = 25 m/s

Angle above horizontal or angle of projection, θ = 65°

We need to find the distance travelled by the ball before it hots the ground or in other words we need to find the range. It is given by R.

[tex]R=\dfrac{v^2\ sin2\theta}{g}[/tex]

[tex]R=\dfrac{(25\ m/s)^2\ sin2(65)}{9.8\ m/s^2}[/tex]

R = 48.85 m

So, the distance travelled by the ball before it hots the ground is 48.85 meters. Hence, this is the required solution.

A conducting bar slides without friction on two parallel horizontal rails that are 50 cm apart and connected by a wire at one end. The resistance of the bar and the rails is constant and equal to 0.10 Ω. A uniform magnetic field is perpendicular to the plane of the rails. A 0.080-N force parallel to the rails is required to keep the bar moving at a constant speed of 0.50 m/s. What is the magnitude of the magnetic field?

Answers

Answer:

0.25 T

Explanation:

F = Force required to keep the bar moving = 0.080 N

B = magnitude of magnetic field = ?

L = length of the bar = 50 cm = 0.50 m

v = speed of the bar = 0.50 m/s

R = resistance of the bar =0.10 Ω

Force is given as

[tex]F = \frac{B^{2}L^{2}v}{R}[/tex]

[tex]0.08 = \frac{B^{2}(0.50)^{2}(0.50)}{0.10}[/tex]

B = 0.25 T

Final answer:

To find the magnitude of the magnetic field, use the equation F = qvBsinθ and Ohm's Law to solve for q. Then substitute q back into the equation to find the magnetic field.

Explanation:

To find the magnitude of the magnetic field, we can use the equation F = qvBsinθ. In this case, the force (F) is given as 0.080 N, the charge (q) is not given but is not needed to find the magnetic field, the velocity (v) is given as 0.50 m/s, and sinθ is 1 because the angle between the velocity and the magnetic field is 90°. Rearranging the equation to solve for B, we get B = F / (qv sinθ). Substituting the given values, we have B = 0.080 N / (q * 0.50 m/s * 1). The only unknown variable left is the charge (q).

To calculate the charge, we can use the given resistance and Ohm's Law. Ohm's Law states that V = IR, where V is the voltage, I is the current, and R is the resistance. Rearranging the equation to solve for I, we get I = V / R. The voltage across the resistor can be calculated using the formula V = Fd, where d is the distance between the rails, which is given as 50 cm (or 0.5 m). Substituting the given values, we have V = 0.080 N * 0.5 m. Now, we can substitute the calculated current (I) into the equation B = 0.080 N / (q * 0.50 m/s * 1) and solve for q. Finally, we can substitute the calculated charge (q) back into the equation B = 0.080 N / (q * 0.50 m/s * 1) to find the magnitude of the magnetic field.

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Two man pull on a cart with a mass of 90 kg. if 1 man pulls 140 N to the right and the other man pulls 195 N to the left then find the magnitude and direction of the acceleration for the cart

Answers

Answer:

magnitude : 0.6 m/s²

Direction : Left

Explanation:

m = mass of the cart = 90 kg

Taking force in right direction as positive and force in left direction as negative

F₁ = Force applied by man 1 = 140 N

F₂ = Force applied by other man = - 195 N

a = acceleration of the cart

Force equation for the motion of the cart is given as

F₁ + F₂ = ma

140 + (- 195) = 90 a

a = - 0.6 m/s²

magnitude of acceleration is 0.6 m/s²

The negative sign indicates the direction of acceleration towards left

Final answer:

The magnitude of the cart's acceleration is approximately 0.611 m/s², and the direction is to the left.

Explanation:

When calculating the acceleration of the cart, we need to consider the net force acting on it and its mass. The net force is found by subtracting the smaller force from the larger force, taking into account their directions. With forces of 140 N to the right and 195 N to the left, the net force is the difference, which is 195 N - 140 N = 55 N, directed to the left since the larger force is in that direction. Using Newton's second law, acceleration (a) is the net force (Fnet) divided by the mass (m).

To find the magnitude of the acceleration, calculate a = Fnet / m. So, a = 55 N / 90 kg which equals approximately 0.611 m/s2. Since the larger force was to the left, the direction of the acceleration is also to the left.

A solenoid of length 0.700 m having a circular cross-section of radius 5.00 cm stores 6.00 µJ of energy when a 0.400-A current runs through it. What is the winding density of the solenoid? (μ 0 = 4π × 10-7 T · m/A

Answers

Answer:

104

Explanation:

U = Energy stored in the solenoid = 6.00 μJ = 6.00 x 10⁻⁶ J

i = current flowing through the solenoid = 0.4 A

L = inductance of the solenoid

Energy stored in the solenoid is given as

U = (0.5) L i²

6.00 x 10⁻⁶ = (0.5) L (0.4)²

L = 75 x 10⁻⁶

Inductance is given as

l = length of the solenoid = 0.7 m

N = number of turns

r = radius = 5.00 cm = 0.05 m

Area of cross-section is given as

A = πr²

A = (3.14) (0.05)²

A = 0.00785 m²

Inductance is given as

[tex]L=\frac{\mu _{o}N^{2}A}{l}[/tex]

[tex]75\times 10^{-6}=\frac{(12.56\times 10^{-7})N^{2}(0.00785)}{0.7}[/tex]

N = 73

Winding density is given as

density = n = [tex]\frac{N}{l}[/tex]

n = [tex]\frac{73}{0.7}[/tex]

n = 104

A mass of 2.82 kg is hung from a spring, causing the spring to stretch 0.331 m. If a second mass of 3.09 kg is now added to the spring, how far with it now stretch with both masses hanging?? State your answer to the correct number of significant digits and include the proper units.

Answers

Answer:

0.694 m

Explanation:

Case 1 : When only mass of 2.82 kg is hanged from spring

m = mass hanged from the spring = 2.82 kg

x = stretch caused in the spring = 0.331 m

k = spring constant

Using equilibrium of force in vertical direction

Spring force = weight of the mass

k x = m g

k (0.331) = (2.82) (9.8)

k = 83.5 N/m

Case 2 : When both masses are hanged from spring

m = mass hanged from the spring = 3.09 + 2.82 = 5.91 kg

x = stretch caused in the spring = ?

k = spring constant = 83.5 N/m

Using equilibrium of force in vertical direction

Spring force = weight of the mass

k x = m g

(83.5) x = (5.91) (9.8)

x = 0.694 m

Polarizing Filters In the fifth lab (the ters and measured the resulting intensity. Initially vertically polarized light, with an intensity Io, is sent through two (ideal) polarizing sheets, whose transmission axes are 45° and 90° to the vertical. Will the intensity of the light be greater or less than if with the polarizers), you shined light through polarizhng there were only polarizer at 90° to the yertical? Explain. Note that Malus Law one says I2= I cos2 e

Answers

Answer:

So the intensity of the light be smaller in second case with the polarizers at 90 degree with the vertical

Explanation:

When two polarizers are arranged at 45 degree and 90 degree in series

so here the intensity of light coming out of the polarizers is given as

[tex]I_1 = I_ocos^45[/tex]

[tex]I_1 = \frac{I_0}{2}[/tex]

now for second polarizer we have

[tex]I_2 = I_1 cos^2(90 - 45)[/tex]

[tex]I_2 = (\frac{I_0}{2})\frac{1}{2}[/tex]

[tex]I_2 = \frac{I_0}{4}[/tex]

now in other case when two polarizers are inclined 90 degree to the vertical

[tex]I = I_ocos^290 = 0[/tex]

so final intensity in second case will be ZERO

So the intensity of the light be smaller in second case with the polarizers at 90 degree with the vertical

A 64.0 kg skater moving initially at 2.81 m/s on rough horizontal ice comes to rest uniformly in 3.93 s due to friction from the ice. What force does friction exert on the skater?

Answers

Answer:

Force of friction, F = 45.76 N

Explanation:

t is given that,

Mass of the skater, m = 64 kg

Initial velocity of the skater, u = 2.81 m/s

Finally it comes to rest, v = 0

Time, t = 3.93 s

We need to find the force of friction. According to seconds law of motion as :

F = m × a

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

[tex]F=64\ kg\times \dfrac{0-2.81\ m/s}{3.93\ s}[/tex]

F = −45.76 N

So, the frictional force exerting on the skater is 45.76 N. Hence, this is the required solution.

What quantities determine the resistance of a piece of material? Choose all that apply.
The length of the piece of material
The cross-sectional area of the piece of material
The type of material
The voltage across the material
The current flowing through the piece of material

Answers

Answer:

Option (a), (b) and (c)

Explanation:

The resistance of a conductor depends on the length of the conductor, area of crossection of the conductor and the nature of the conductor.

The formula for the resistance is given by

R = ρ x l / A

Where, ρ is the resistivity of the conductor, l be the length of the conductor and A be the area of crossection of the conductor.

So, It depends on the length, area and the type of material.

Final answer:

Resistance of a material is determined by its length, cross-sectional area, and the type of material; voltage and current do not determine resistance but are influenced by it.

Explanation:

The Quantities Determining the Resistance of a Material

The resistance of a piece of material is determined by several key factors, specifically:

The length of the piece of material: Resistance is directly proportional to the length. The longer the material, the higher the resistance as there will be more collisions between the charges and atoms within the material.The cross-sectional area of the piece of material: Resistance is inversely proportional to the cross-sectional area. Larger areas allow more current to flow, reducing resistance.The type of material: Different materials have different intrinsic resistivity properties, which affect their resistance.

It is important to note that the voltage across the material and the current flowing through the material are not factors that determine resistance. These two quantities are actually influenced by the resistance according to Ohm's law, which states that Voltage (V) equals Current (I) times Resistance (R), or V = IR.

Three clowns move a 345 kg crate 12.5 m to the right across a smooth floor. Take the positive horizontal and vertical directions to be right and up, respectively. Moe pushes to the right with a force of 535 N , Larry pushes to the left with 225 N , and Curly pushes straight down with 705 N . Calculate the work ???? done by each of the clowns. Assume friction is negligible.

Answers

Answer:

Moe: 6687.5 J, Larry: -2812.5 J, Curly: 0 J

Explanation:

The work done by each clown is given by:

[tex]W=Fd cos \theta[/tex]

where

F is the magnitude of the force applied

d is the displacement of the box

[tex]\theta[/tex] is the angle between the direction of the force and of the displacement

Let's apply the formula to each clown:

- Moe: F = 535 N, d = 12.5 m, [tex]\theta=0^{\circ}[/tex] (because Moe pushes to the right, and the box also moves to the right)

[tex]W=(535)(12.5)cos 0^{\circ}=6687.5 J[/tex]

- Larry: F = 225 N, d = 12.5 m, [tex]\theta=180^{\circ}[/tex] (because Larry pushes to the left, while the box moves to the right)

[tex]W=(225)(12.5)cos 180^{\circ}=-2812.5 J[/tex]

- Curly: F = 705 N, d = 12.5 m, [tex]\theta=90^{\circ}[/tex] (because Curly pushes downward, while the box moves to the right)

[tex]W=(705)(12.5)cos 90^{\circ}=0[/tex]

The work done by each of the clowns to move 345 kg crate 12.5 m to the right across a smooth floor is,

The work done by the Moe is 6687.5 J.The work done by the Larry is -2812.5 J.The work done by the Larry is -2812.5 J.

What is work done?

Work done is the force applied on a body to move it over a distance. Work done to move a body by the application of force can be given as,

[tex]W=Fd\cos\theta[/tex]

Here (F) is the magnitude of force, [tex]\theta[/tex] is the angle of force applied and (d) is the distance traveled.

The mass of the crate is 345 kg. The crate moved by the clowns is 12.5 m to the right across a smooth floor.

Moe pushes to the right with a force of 535 N-

As, the Moe applies the force to the crate on the right side and the crate move in the direction of force. Here, the angle made is equal to the 0 degrees.

Therefore, the work done by the Moe is,

[tex]W=535\times12.5\times\cos (0)\\W=6687.5\rm J[/tex]

thus, the work done by the Moe is 6687.5 J.

Larry pushes to the left with 225 N. -

As, the Larry applies the force to the crate on the left side and the crate move in the opposite direction of force. Here, the angle made is equal to the 180 degrees.

Therefore, the work done by the Larry is,

[tex]W=225\times12.5\times\cos (180)\\W=-2812.5\rm J[/tex]

thus, the work done by the Larry is -2812.5 J.

Curly pushes straight down with 705 N.-

As, the Curly applies the force to the crate on the straight down and the crate move in the perpendicular direction of force. Here, the angle made is equal to the 90 degrees.

Therefore, the work done by the Curly is,

[tex]W=705\times12.5\times\cos (90)\\W=0\rm J[/tex]

thus, the work done by the Curly is 0 J.

Thus, the work done by each of the clowns to move 345 kg crate 12.5 m to the right across a smooth floor is,

The work done by the Moe is 6687.5 J.The work done by the Larry is -2812.5 J.The work done by the Larry is -2812.5 J.

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If R is the total resistance of three resistors, connected in parallel, with resistances R1, R2, R3, then 1 R = 1 R1 + 1 R2 + 1 R3 . If the resistances are measured in ohms as R1 = 100 Ω, R2 = 25 Ω, and R3 = 10 Ω, with a possible error of 0.5% in each case, estimate the maximum error in the calculated value of R. (Round your answer to three decimal places.) Ω

Answers

Final answer:

To calculate the total resistance of three resistors connected in parallel, use the formula 1/R = 1/R1 + 1/R2 + 1/R3. The maximum error in the calculated value of R can be estimated by multiplying the sum of the errors in each resistance by the calculated value of R.

Explanation:

To find the total resistance of three resistors connected in parallel, we use the formula 1/R = 1/R1 + 1/R2 + 1/R3. Given the resistances R1 = 100 Ω, R2 = 25 Ω, and R3 = 10 Ω, we can substitute these values into the formula to calculate the total resistance R. Therefore, 1/R = 1/100 + 1/25 + 1/10 = 0.01 + 0.04 + 0.1 = 0.15. Now, to estimate the maximum error in the calculated value of R, we consider the errors in each resistance. Since each resistance has a possible error of 0.5%, we can calculate the maximum error in R by multiplying the sum of the errors in each resistance by the calculated value of R. Therefore, maximum error in R = 0.005 * 0.15 = 0.00075 Ω.

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Final answer:

The maximum error in the total resistance R of three parallel resistors with a potential error of 0.5% in each resistor is approximately 0.667 ohms.

Explanation:

The question asks to calculate the maximum error in the calculated value of total resistance R when three resistors R1, R2, and R3 are connected in parallel, where R1 = 100 Ω, R2 = 25 Ω, and R3 = 10 Ω, each with a possible error of 0.5%. The resistors in parallel have a total resistance denoted by:

1/R = 1/R1 + 1/R2 + 1/R3

To find the maximum error in the calculated value of R, we will first calculate R and then use derivative rules to estimate the maximum error considering the errors in R1, R2, and R3.

After calculating 1/R using the given resistances:

1/R = 1/100 + 1/25 + 1/10

1/R = 0.01 + 0.04 + 0.1 = 0.15

Therefore, R = 1 / 0.15 = 6.667 Ω

We calculate the maximum possible errors in resistances as:

Error in R1 (eR1) = 0.005 * 100 Ω = 0.5 ΩError in R2 (eR2) = 0.005 * 25 Ω = 0.125 ΩError in R3 (eR3) = 0.005 * 10 Ω = 0.05 Ω

Using the formula for the propagation of errors for functions of several independent variables, we estimate the maximum error in R (eR) as:

eR ≈ | -R² * eR1/R1² | + | -R² * eR2/R2² | + | -R² * eR3/R3² |

Plugging in the values:

eR ≈ | -6.667² * 0.5/100² | + | -6.667² * 0.125/25² | + | -6.667² * 0.05/10² |

eR ≈ | -0.04446 | + | -0.17784 | + | -0.4446 | = 0.667 Ω (Approximated to three decimal places)

The estimated maximum error in the calculated value of R is therefore approximately 0.667 Ω.

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Water is moving at a velocity of 2.17 m/s through a hose with an internal diameter of 1.60 cm. The fluid velocity in this hose’s nozzle is 14.5 m/s. What is the nozzle’s inside diameter, in cm?

Answers

Final answer:

To find the nozzle's inside diameter, use the principle of continuity. Set up the equation A1V1 = A2V2 and solve for r2. Finally, calculate the nozzle's inside diameter.

Explanation:

To find the nozzle's inside diameter, we can use the principle of continuity, which states that the flow rate of a fluid is constant throughout a pipe or nozzle.

The formula for continuity is A1V1 = A2V2, where A is the cross-sectional area and V is the fluid velocity.

In this case, we can set up the equation as (pi * r12) * 2.17 = (pi * r22) * 14.5, where r1 and r2 are the radii of the hose and nozzle respectively.

By rearranging the equation and solving for r2, we find that r2 = sqrt((r12 * 2.17) / 14.5).

Since the radius is half of the diameter, the nozzle's inside diameter can be calculated as 2 * r2 in cm.

A large building shaped like a box is 40 m high with a face that is 6565 m wide. A strong wind blows directly at the face of the​ building, exerting a pressure of 170 N/m2 at the ground and increasing with height according to ​P(y)=170+​2y, where y is the height above the ground. Calculate the total force on the​ building, which is a measure of the resistance that must be included in the design of the building.

Answers

Answer:

[tex]F=55146000 N[/tex]

Explanation:

Given:

Pressure on the building = 170 N/m²

Pressure function P(y) = 170 + 2y

Consider a small height of the building as dy

the small force (dF) exerted on the height dy will be given as

dF = P(y)×(Area of the building face)

or

dF = (170 + 2y)(6565)dy

or

dF = 170×6565×dy + 2y×6565×dy

dF = 1116050dy + 13130ydy

integerating the function for the whole height of 40 m

[tex]\int\limits^{40}_0 {F} = \int _0^{40}\:\left(\:1116050dy\:+\:13130ydy\right)[/tex]

[tex]F = \int _0^{40}1116050dy+\int _0^{40}13130ydy[/tex]

[tex]F=\left[1116050y\right]^{40}_0+13130\left[\frac{y^2}{2}\right]^{40}_0[/tex]

[tex]F=44642000+10504000[/tex]

[tex]F=55146000 N[/tex] ; Total force exerted on the face of the building

Final answer:

To calculate the total force on a building due to wind pressure, integrate the pressure function over the building's height and multiply by the building's width. This calculation is vital for determining the building's required resistance to wind forces for safe and stable design.

Explanation:

The question asks to calculate the total force exerted on a large building by a wind blowing against one of its faces. The wind pressure increases with height as described by the function P(y)=170+2y, where y is the height in meters above the ground. To find the total force, we need to integrate this pressure function over the height of the building and multiply by the width of the building's face.

First, we can write the integral of the pressure over the height of the building:


Integrate [P(y) dy] from y=0 to y=40


= Integrate [(170+2y) dy] from y=0 to y=40

Performing this integration will give us the total pressure exerted over the entire height of the building. After integrating, we multiply the result by the width of the building (6565 m) to obtain the total force.

Using the integration results, the total force, F, can be expressed as:

F = (Integral result) × Width of the building

This force calculation is critical for architects and engineers as it provides an estimate of the resistance that the building must be designed to withstand due to wind forces, ensuring structural stability and safety.

A football player at practice pushes a 60 kg blocking sled across the field at a constant speed. The coefficient of kinetic friction between the grass and the sled is 0.30. How much force must he apply to the sled?

Answers

Answer:

180N

Explanation:

Using Newton's law of motion;

∑F = m x a       --------------------(i)

Where;

∑F = Resultant force

m = mass of the object (sled in this case)

a = acceleration of the sled

Calculate the resultant force;

Since the direction of motion is horizontal, the horizontal forces acting on the sled are the;

i. Applied force ([tex]F_{A}[/tex]) in one direction and;

ii. Frictional force ([tex]F_{R}[/tex]) in the other direction to oppose motion

Therefore, the resultant force ∑F is the vector sum of the two forces. i.e;

∑F = [tex]F_{A}[/tex] - [tex]F_{R}[/tex]  -----------------------(i)

Frictional force [tex]F_{R}[/tex] is the product of the coefficient of kinetic friction (μ) and weight(W) of the sled. i.e

[tex]F_{R}[/tex] = μ x W

Where;

W = mass(m) x gravity(g)

W = m x g

=> [tex]F_{R}[/tex] = μmg

Substitute [tex]F_{R}[/tex] into equation (ii)

∑F = [tex]F_{A}[/tex] - μmg

Substitute ∑F into equation (i)

[tex]F_{A}[/tex] - μmg = ma  -------------------(iii)

Since the motion is at constant speed, it means acceleration is zero (0)

Substitute a = 0 into equation (iii) to give;

[tex]F_{A}[/tex] - μmg = 0

=> [tex]F_{A}[/tex] = μmg

Substitute the values of μ = 0.3, m = 60kg and g = 10m/s² into the above equation to give;

=> [tex]F_{A}[/tex]  = 0.3 x 60 x 10

=> [tex]F_{A}[/tex] = 180N

This means that the applied force should be 180N

The amount of force the football player must apply to the sled is 176.4 Newton.

Given the following data:

Mass of sled = 60 kgCoefficient of kinetic friction = 0.30

We know that acceleration due to gravity (g) on Earth is equal to 9.8 [tex]m/s^2[/tex]

To find how much force the football player must apply to the sled:

Mathematically, the force of kinetic friction is given by the formula;

[tex]Fk = umg[/tex]

Where;

Fk represents the force of kinetic friction.μ represents the coefficient of friction.m represents the mass.g is the acceleration due to gravity.

Substituting the given parameters into the formula, we have;

[tex]Fk = 0.30\times60\times9.8[/tex]

Force, Fk = 176.4 Newton.

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A piece of iron block moves across a rough horizontal surface before coming to rest. The mass of the block is 3.3 kg, and its initial speed is 3.2 m/s. How much does the block's temperature increase, if it absorbs 74% of its initial kinetic energy as internal energy? The specific heat of iron is 452 J/(kg · °C).
°C

Answers

Answer:

Rise in temperature is given as

[tex]\Delta T = 8.4 \times 10^{-3} ^0C[/tex]

Explanation:

Initial kinetic energy of the block is given as

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

here we will have

m = 3.3 kg

v = 3.2 m/s

now we will have

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

now we will have

[tex]KE = \frac{1}{2}(3.3)(3.2)^2[/tex]

[tex]KE = 17 J[/tex]

now we know that 74% of initial kinetic energy is absorbed as internal energy of the block

so the rise in temperature of the block is given as

[tex]KE = ms\Delta T[/tex]

[tex]0.74 \times 17 J = (3.3)(452)\Delta T[/tex]

[tex]12.5 = 1491.6 \Delta T[/tex]

[tex]\Delta T = 8.4 \times 10^{-3} ^0C[/tex]

Answer:

The temperature increases 0.0084ºC

Explanation:

Please look at the solution in the attached Word file.

A place kicker applies an average force of 2400 N to a football of .040 kg. The force is applied at an angle of 20.0 degrees from the horizontal. Contact time is .010 sec. a. Find the velocity of the ball upon leaving the foot. b. Assuming no air resistance find the time to reach the goal posts 40.0 m away. c. The posts are 4.00 m high. Is the kick good? By how much?

Answers

Answer:

a)  The velocity of the ball upon leaving the foot = 600 m/s

b)  Time to reach the goal posts 40.0 m away = 0.07 seconds

c)  The kick won't e going inside goal post, it is higher by 10.34m.

Explanation:

a) Rate of change of momentum = Force

   [tex]\frac{\texttt{Final momentum - Initial momentum}}{\texttt{Time}}=\texttt{Force}\\\\\frac{0.040v-0.040\times 0}{0.010}=2400\\\\v=600m/s[/tex]

  The velocity of the ball upon leaving the foot = 600 m/s

b) Horizontal velocity = 600 cos20 = 563.82 m/s

   Horizontal displacement = 40 m

   Time

            [tex]t=\frac{40}{563.82}=0.07s[/tex]

   Time to reach the goal posts 40.0 m away = 0.07 seconds

c) Vertical velocity = 600 sin20 = 205.21 m/s

    Time to reach the goal posts 40.0 m away = 0.07 seconds

    Acceleration = -9.81m/s²

    Substituting in s = ut + 0.5at²

             s = 205.21 x 0.07 - 0.5 x 9.81 x 0.07²= 14.34 m

    Height of ball = 14.34 m

    Height of post = 4 m

    Difference in height = 14.34 - 4 = 10.34 m

    The kick won't e going inside goal post, it is higher by 10.34m.

A moon is in orbit around a planet. The​ moon's orbit has a semimajor axis of 4.3 times 10 Superscript 8 Baseline m and has an orbital period of 1.516 days. Use these data to estimate the mass of the planet.

Answers

Answer:

The mass of the planet is [tex]2.7\times10^{27}\ kg[/tex].

Explanation:

Given that,

Semi major axis [tex]a= 4.3\times10^{8}[/tex]

Orbital period T=1.516 days

Using Kepler's third law

[tex]T^2=\dfrac{4\pi^2}{GM}a^3[/tex]

[tex]M=\dfrac{4\pi^2}{GT^2}a^3[/tex]

Where, T = days

G = gravitational constant

a = semi major axis

Put the value into the formula

[tex]M=\dfrac{4\times(3.14)^2}{6.67\times10^{-11}(1.516\times24\times60\times60)^2}(4.3\times10^{8})^3[/tex]

[tex]M=2.7\times10^{27}\ kg[/tex]

Hence, The mass of the planet is [tex]2.7\times10^{27}\ kg[/tex].

A cat leaps to catch a bird. If the cat's jump was at 60.0° off the ground and its initial velocity was point of its trajectory? 0.30 m 3.44 m/s, what is the highest O 13.76 m 0.45 m 0.90 m

Answers

Answer:

The highest of its trajectory = 0.45 m

Option C is the correct answer.

Explanation:

Considering vertical motion of cat:-

Initial velocity, u =  3.44 sin60 = 2.98 m/s

Acceleration , a = -9.81 m/s²

Final velocity, v = 0 m/s

We have equation of motion v² = u² + 2as

Substituting

   v² = u² + 2as

    0² = 2.98² + 2 x -9.81 x s

    s = 0.45 m

The highest of its trajectory = 0.45 m

Option C is the correct answer.

A water pump draws about 9 A when connected to 240 V. What is the cost (with electrical energy at 13 cents per kWh) of running the pump for 16 h? (Give your answer in dollars)

Answers

Answer:

4.49 dollars

Explanation:

i = 9 A, V = 240 V, t = 16 h

Energy = V x i x t = 240 x 9 x 16 = 34560 W h = 34.56 kWh

The cost of 1 kWh is 13 cents.

Cost of 34.56 kWh = 13 x 34.56 = 449.28 cents = 449.28 / 100 = 4.49 dollars

Final answer:

The cost of running a water pump drawing 9 A at 240 V for 16 hours, with an energy cost of 13 cents per kWh, is $4.49.

Explanation:

To calculate the cost of running a water pump for 16 hours at an electrical energy cost of 13 cents per kWh, you need to follow these steps:

First, calculate the power consumed by the pump. Power (P) is given by the formula P = VI, where V is the voltage and I is the current. For a pump drawing 9 A at 240 V, the power consumed is P = 240V * 9A = 2160W or 2.16kW.Next, calculate the total energy consumed in kilowatt-hours. Energy (E) is calculated as E = Pt, where P is the power in kW and t is the time in hours. For 16 hours, the energy consumed is E = 2.16kW * 16h = 34.56kWh.Finally, calculate the cost by multiplying the total energy consumed by the cost per kilowatt-hour. The cost (C) is C = E * cost/kWh. So, the cost to run the pump for 16 hours is C = 34.56kWh * $0.13/kWh = $4.4928.

In dollars, the cost of running the pump for 16 hours is $4.49 (rounded to the nearest cent).

A camper dives from the edge of a swimming pool at water level with a speed of 8.0 m/s at an angle of 30.0° above the horizontal. (a) How long is the diver in the air? (b) How high is the diver in the air? (c)How far out in the pool does the diver land?

Answers

Answer:

(a) 0.816 s

(b) 0.816 m

(c) 5.656 m

Explanation:

u = 8 m/s, theta = 30 degree,

(a) Use the formula for time of flight

T = 2 u Sin theta / g

T = ( 2 x 8 x Sin 30 ) / 9.8 = 0.816 s

(b) Use the formula for maximum height

H = u^2 Sin^2theta / 2 g

H = ( 8 x 8 x Sin^2 30) / ( 2 x 9.8)

H = 0.816 m

(c) Use the formula for horizontal range

R = u^2 Sin 2 theta / g

R = ( 8 x 8 x Sin 60) / 9.8

R = 5.656 m

A DVD drive is spinning at 100.0 rpm. A dime (2.00 gm) is placed 3.00 cm from the center of the DVD. What must the coefficient of friction be to keep the dime on the disk?

Answers

Answer:

0.3375

Explanation:

w = angular speed of the DVD drive = 100.0 rpm = [tex]100.0 \frac{rev}{min}\frac{2\pi rad}{1 rev}\frac{1 min}{60 sec}[/tex] = [tex]10.5\frac{rad}{sec}[/tex]

m = mass of the dime = 2 g = 0.002 kg

r = radius = 3 cm = 0.03 m

μ = Coefficient of friction

The frictional force provides the necessary centripetal force to move in circle. hence

frictional force = centripetal force

μ mg = m r w²

μ g =  r w²

μ (9.8) =  (0.03) (10.5)²

μ = 0.3375

Determine explicitly which is faster, 75 miles per hour or 75 m/s? Express all your results in MKS units and explain your conclusions from the numbers, and explain using words. Show all work.

Answers

Answer:

75 m/s is faster

Explanation:

MKS units stands for meter kilogram seconds

75 miles per hour = 75 mph

1 mile = 1609.34 meters

1 hour = 60×60 = 3600 seconds

1 mph = 1609.34/3600 = 0.44704 m/s

75 mph = 75×0.44707 = 33.52792 m/s

Comparing 75 mph = 33.52792 m/s with 75 m/s it can be seen that 75 m/s is faster. Even without calculating the values you can know the answer. 75 mph means that in 1 hour the object will move 75 miles. 75 m/s means that in one second the object will cover 75 meters multiply by 3600 and you will get 270000 m/h that is 270 km/h divide it by 1.6 and you can approximately get the value in mph that will be around 168 mph which is faster than 75 mph.

A 79-kg worker at the Burj Khalifa, the tallest building in the world with its roof 828 m above the ground below, rides a series of elevators from the ground level to the roof. What is the worker's change in gravitational potential energy?

Answers

Answer:

Worker's change in gravitational potential energy = 641.69 kJ

Explanation:

Potential energy = Mass x Acceleration due to gravity x Height

PE = mgh

Mass, m = 79 kg

Acceleration due to gravity, g = 9.81 m/s²

Height, h = 828 m

Potential energy, PE = 79 x 9.81 x 828 = 641691.72 J = 641.69 kJ

Worker's change in gravitational potential energy = 641.69 kJ

A net force of 200 N acts on a 100-kg boulder, and a force of the same magnitude acts on a 100-g pebble. Is the change of the boulder’s momentum in one second greater than, less than, or equal to the change of the pebble’s momentum in the same time period?

Answers

Answer:

equal

Explanation:

According to the Newton's second law of motion, the force acting on the body is directly proportional to the rate of change of momentum.

If the force is same, then the change in momentum is also same for a given time.

The change of the boulder’s momentum in one second greater is equal to the change of the pebble’s momentum in the same time period.

What is force?

Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration.

The given data in the problem is;

F is the net force=200 N

m is the mass of boulder=100 kg

M is the mass of pebble=100 g

The force acting on the body is directly proportional to the rate of change in momentum, according to Newton's second law of motion.

For a given time, if the force is the same, the change in momentum is likewise the same.

Hence the change of the boulder’s momentum in one second greater is equal to the change of the pebble’s momentum in the same time period.

To learn more about the force refer to the link;

https://brainly.com/question/26115859

A person hums into the top of a well and finds that standing waves are established at frequencies of 140, 196, and 252 Hz. The frequency of 140 Hz is not necessarily the fundamental frequency. The speed of sound is 343 m/s. How deep is the well?

Answers

Answer:

Depth of well 3.06m

Explanation:

We know that for a pipe closed at one end the frequencies are in ratios if 1:2:3:5:7.... to the fundamental frequency

In our case the given frequencies are in the ratio of

a)[tex]\frac{140}{196}=\frac{5}{7}[/tex]

b) [tex]\frac{196}{252}=\frac{7}{9}[/tex]

Thus the fundamental frequency can be calculated as [tex]140Hz=5n[/tex]

[tex]\therefore n=\frac{140}{5}=28Hz[/tex]

Now we know that

[tex]\lambda_{1}=4l\\\\\frac{v}{f}=4l\\\\l=\frac{v}{4f}[/tex]

Applying values we get

[tex]L=3.06m[/tex]

Final answer:

By identifying 56 Hz as the fundamental frequency and using the formula for the resonance of open-ended tubes, the depth of the well is calculated to be approximately 3.06 meters using the speed of sound as 343 m/s.

Explanation:

The student has observed standing waves at frequencies of 140 Hz, 196 Hz, and 252 Hz in a well. These frequencies represent the natural resonant frequencies or harmonics of the well. To find the depth of the well, we need to consider these frequencies as harmonics of a sound wave corresponding to the lengths of the air column in the well where each harmonic creates a standing wave.

Given the multiple frequencies, there is a constant difference of 56 Hz (196 Hz - 140 Hz, 252 Hz - 196 Hz) between consecutive frequencies. This indicates that 56 Hz is the fundamental frequency of the harmonics. We can represent these frequencies as [tex]F1 = 56 Hz \(n=1\), F2 = 112 Hz \(n=2\)[/tex]ere the observed frequencies correspond to F3, F4, and F5.

To determine the length of the well (L), we use the formula for the resonance of open-ended tubes, [tex]\(L = \frac{v}{2f}\)[/tex]d of sound (343 m/s) and 'f' is the frequency (56 Hz in this case). Therefore, [tex]L = \frac{343 m/s}{2 \times 56 Hz} = 3.0625 m[/tex]ly 3.06 meters.

The Sun radiates electromagnetic energy at the rate of 3.92 × 1026 W. The mass of the Sun is 1.99 × 1030 kg. What fraction of the Sun’s mass is lost during a human lifetime of 75 years? A. 2.16 × 10–13 B. 1.42 × 10–14 C. 6.90 × 10–14 D. 5.18 × 10–12 E. 8.63 × 10

Answers

Answer:

D. 5.18 x 10⁻¹²

Explanation:

[tex]\frac{dE}{dt}[/tex] = rate at which sun radiates energy = 3.92 x 10²⁶ W

M = mass of sun = 1.99 x 10³⁰ kg

[tex]\frac{dm}{dt}[/tex] = rate at which sun's mass is lost

c = speed of light

Energy is given as

E = m c²

Taking derivative both side relative to "t"

[tex]\frac{dE}{dt}=c^{2}\frac{dm}{dt}[/tex]

[tex]3.92\times 10^{26}=(3\times 10^{8})^{2}\frac{dm}{dt}[/tex]

[tex]\frac{dm}{dt}[/tex] = 4.4 x 10⁹ kg/s

t = time interval = 75 yrs = 75 x 365 days = 75 x 365 x 24 hours = 75 x 365 x 24 x 3600 sec = 2.4 x 10⁹ sec

[tex]m[/tex] = mass lost

mass lost is given as

[tex]m = t\frac{dm}{dt}[/tex]

[tex]m = (2.4\times 10^{9})(4.4\times 10^{9})[/tex]

m = 10.56 x 10¹⁸ kg

fraction is given as

fraction = [tex]\frac{m}{M}[/tex]

fraction = [tex]\frac{10.56\times 10^{18}}{1.99\times 10^{30}}[/tex]

fraction = 5.18 x 10⁻¹²

An electrical motor spins at a constant 1975.0 rpm. If the armature radius is 7.112 cm, what is the acceleration of the edge of the rotor? O 305,200 m/s O 152.3 m/s O 15.20 m/s O 3042 m/s

Answers

Answer:

Option D is the correct answer.

Explanation:

Since value of angular acceleration is constant, the body has only centripetal acceleration.

Centripetal acceleration

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

We have radius = 7.112 cm = 0.07112 m

Frequency, f = 1975 rpm = 32.92 rps

Angular frequency, ω = 2πf = 2 x π x 32.92 = 206.82 rad/s

Substituting in centripetal acceleration equation,

              [tex]a=r\omega ^2=0.07112\times 206.82^2=3042.17m/s^2[/tex]

Option D is the correct answer.                

An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner's 2.00 T field with his fingers pointing in the direction of the field. His wedding ring has a diameter of 2.5 cm and it takes 0.45 s to move it into the field. What average current is induced in the ring in A if its resistance is 0.0100 Ω ?

Answers

The average induced current in the loop is 0.218 A.

Induced emf in the loop

The emf induced in the loop is determined by applying Faraday's law as shown below;

emf = dФ/dt

emf = BA/t

where;

A is the area

A = πr² = πd²/4

A = π x (0.025)²/4

A = 4.908 x 10⁻³ m²

emf = (2 x 4.908 x 10⁻³)/(0.45)

emf = 2.18 x  10⁻³ V

Average induced current

The average induced current in the loop is calculated as follows;

I = emf/R

I = 2.18 x  10⁻³/0.01

I = 0.218 A

Learn more about induced emf here: https://brainly.com/question/13744192

A coil has a resistance of 5Ω and an inductance of 100 mH . At a particular instant in time after a battery is connected across the coil, the current is 2 A, and is increasing at a rate of 20A/s a) What is the voltage of the battery. b) What is the time-constant of the circuit. c) What is the final value of the current. d) Plot I() Vs time.

Answers

I have seen this question before and the correct answer would be B

Hope this helped!!

Answer:

The correct answer would be B

Explanation:

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