PHYSICS COLLEGE

A 79-turn, 16.035-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 43 degrees away from vertical increases from 0.997 T to 6.683 T in 56.691 s. Determine the emf induced in the coil.

Answers

Answer 1

Answer:

Answer:

The induced emf is 0.0888 V.

Explanation:

Given that,

Number of turns = 79

Diameter = 16.035 cm

Angle = 43

Change in magnetic field $\Delta B=(6.683-0.997)= 5.686\ T$

Time = 56.691 s

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=(NA\Delta B\cos\theta)/(\Delta T)$

Where, N = number of turns

A = area

B = magnetic field

Put the value into the formula

$\epsilon=(79*\pi*(8.0175*10^(-2))^2*5.686*\cos43)/(56.691)$

$\epsilon =0.0888\ V$

Hence, The induced emf is 0.0888 V.

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Trevor and Nick are taxi drivers. Trevor drives a taxi using diesel oil while Nick drives a taxi using LPG. Whose taxi will cause higher levels of air pollution and why?

Answers

Answer:

Trevor's taxi would cause higher levels of air pollution

Explanation:

Trevor's taxi use diesel oil.

Diesel is less cleaner than LPG.

Compared to automotive pollution from petrol and diesel, pollutants from LPG-driven cars include lower amounts of petroleum hydrocarbons, nitrogen oxides , sulphur oxides, ozone contamination and particulate matter.

Nitrogen (N2) undergoes an internally reversible process from 6 bar, 247°C during which pν1.2 = constant. The initial volume is 0.1 m3 and the work for the process is 50 kJ. Assuming ideal gas behavior, and neglecting kinetic and potential energy effects, determine heat transfer, in kJ, and the entropy change, in kJ/K.

Answers

Answer:

$Q=25\ kJ$

$\Delta s= 0.2885 J.K^(-1)$

Explanation:

Given:

Initial pressure of nitrogen, $P_1=6* 10^5\ Pa$

initial temperature, $T_1=247+273=520\ K$

polytropic index, $n=1.2$

initial volume, $V_1=0.1\ m^3$

work done in the process, $W=50000\ J$

For heat interaction during the polytropic process we have:

$Q=W[(\gamma -n)/(\gamma-1) ]$

$Q=50*[(1.4-1.2)/(1.4-1) ]$

$Q=25\ kJ$

For ideal gas we have the Gas Law:

$P_1.V_1=m.R.T_1$

$6* 10^5* 0.1=m.R* 520$

$m.R=115.385\ J.K^(-1)$

For work we have the relation:

$W=m.R.((T_1-T_2))/((n-1))$

putting respective values

$50000=115.385* ((520-T_2))/((1.2-1))$

$T_2=433.33\ K$

We know entropy change:

$\Delta s=(dQ)/(dT)$

$\Delta s=(25)/(520-433.33)$

$\Delta s= 0.2885 J.K^(-1)$

Final answer:

The given question involves the thermodynamic process of an internally reversible process of nitrogen gas (N2) at specific pressure and temperature with a constant value of pν1.2. However, the question does not provide enough information to calculate the heat transfer and entropy change accurately.

Explanation:

The given question involves the thermodynamic process of an internally reversible process of nitrogen gas (N2) at specific pressure and temperature with a constant value of pν1.2. In order to determine the heat transfer and entropy change, we need to use the first and second laws of thermodynamics. However, the question does not provide enough information to calculate the heat transfer and entropy change accurately.

Learn more about thermodynamic process here:

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3. A particle of charge +7.5 µC is released from rest at the point x = 60 cm on an x-axis. The particle begins to move due to the presence of a charge ???? that remains fixed at the origin. What is the kinetic energy of the particle at the instant it has moved 40 cm if a) ???? = +20 µC and b) ???? = −20 µC?

Answers

Answer:

HSBC keen vs kg get it yyyyyuuy

Explanation:

hgccccxfcffgbbbbbbbbbbghhyhhhgdghcjyddhhyfdghhhfdgbxbbndgnncvbhcxgnjffccggshgdggjhddh

nnnbvvvvvggfxrugdfutdfjhyfggigftffghhjjhhjyhrdffddfvvvvvvvvvvvbbbbbbbbbvvcxccghhyhhhjjjhjnnnnnnnnnnnnnbhbfgjgfhhccccccvvjjfdbngxvncnccbnxcvbchvxxghfdgvvhhihbvhbbhhvxcgbbbcxzxvbjhcxvvbnnxvnn

Can a car moving with a negative velocity moves faster than a car moving with a positive velocity? explain.

Answers

Answer:

Yes.

Explanation:

This is because "negative velocity" just means it is in the negative in relation to the point of 0. Negative velocity doesn't equal a decrease in velocity. For example lets say you were parked next to a cone (this cone represents zero) if you accelerate forwards then that would be positive acceleration. If you were to accelerate backwards, this would be in the negative direction, aka negative velocity.

SUMMARY:

A negative velocity means that the object which has the negative velocity is moving in the opposite direction of an object moving at a positive velocity. This is a question of frame of reference. The possibility for the velocity is what makes it different to the speed. Speed is only positive.

Final answer:

In physics, a negative velocity can be faster than a positive one when considering speed alone, but not when considering motion direction. For instance, if a car is moving faster but in an opposite direction, it will have a higher speed but a negative velocity.

Explanation:

In physics, the term velocity represents both the speed of an object and its direction of motion. A negative velocity simply means that the object is moving in the opposite direction of the reference point. So, a car moving with a negative velocity can 'move faster' than a car moving with positive velocity if you're considering its speed alone.

Let's assume you have Car A moving at a speed of 40 km/hr in the eastern direction (positive velocity) and Car B moving at a speed of 60 km/hr in the western direction (negative velocity). Even though Car B is described as having a negative velocity, it is moving faster than Car A in terms of speed.

However, remember that in physics, direction matters when considering velocity. So, if you compare their velocities without ignoring the direction, Car A is moving faster to the east than Car B is to the west, even if Car B has higher speed.

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A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out. As the man reaches the horizontal section, he grabs a 20.0-kg backpack and skis off a 2.00-m-high ledge. At what horizontal distance from the edge of the ledge does the man land (the man starts at rest)?

Answers

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

$KE_(initial) + PE_(initial) = KE_(final) + PE_(final)$

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

$0 + mgH = (1)/(2)mv^(2) + 0$

$2gH = v^(2)$

$v = √(2* 9.8* 5) = 9.89\ m/s$

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

$mv = (m + m')v'$

$65.0* 9.89 = (65.0 + 20.0)v'$

$v' = 7.56\ m/s$

Now, time taken for the fall:

$h = (1)/(2)gt^(2)$

$t = \sqrt{(2h)/(g)}$

$t = \sqrt{(2* 2)/(9.8) = 0.638\ s$

Now, the horizontal distance is given by:

x = v't = $7.56* 0.638 = 4.823\ m$

Answer

given,

mass of the man = 65 kg

height = 5 m

mass of the back pack = 20 kg

skis off to 2.00 m high ledge

horizontal distance =

speed of the person before they grab back pack is equal to potential and kinetic energy

$mgh= (1)/(2)mv^2$

$v = √(2gh)$

$v = √(2* 9.8 * 5)$

v = 9.89 m/s

now he perform elastic collision

$v = (m_1v_1)/(m_1+m_2)$

$v = (65* 9.89)/(65+20)$

v = 7.57 m/s

time taken by the skies to fall is

$h = (1)/(2)gt^2$

$t = \sqrt{(2h)/(g)}$

$t = \sqrt{(2* 2)/(9.8)}$

t = 0.6388 s

distance

d = v x t

d = 7.57 x 0.6388

d = 4.84 m

When a hammer thrower releases her ball, she is aiming to maximize the distance from the starting ring. Assume she releases the ball at an angle of 54.6 degrees above horizontal, and the ball travels a total horizontal distance of 30.1 m. What angular velocity must she have achieved (in radians/s) at the moment of the throw, assuming the ball is 1.15 m from the axis of rotation during the spin?