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Tom [10]
2 years ago
10

A cylindrical region of radius R contains a uniform magnetic field parallel to its axis. The field is zero outside the cylinder.

If the magnitude of the field is changing at the rate dB/dt, the electric field induced at a point 2R from the cylinder axis is:
Physics
1 answer:
PtichkaEL [24]2 years ago
3 0

The magnitude of the induced electric field is (RdB/dt)/4

The induced electric field is gotten from

-∫E.dl = dФ/dt where E = induced electric field, dl = path length vector, Ф = magnetic flux through cylindrical region = AB where A = area of magnetic flux = πR² where R = radius of cylindrical region and B = magnetic field.

So, -∫E.dl = dФ/dt

-∫E.dl = dAB/dt

-∫Edlcos0 = AdB/dt (where E.dl = Edlcos0 = Edl since E and dl are parallel to each other.)

So -∫Edl = πR²dB/dt

-E∫dl = πR²dB/dt (∫dl = 2πr since the integral is the circumference of the path)

-E(2πr) = πR²dB/dt (we integrate dl from r = 0 to 2R)

-E2π(2R - 0) = πR²dB/dt

-E4πR= πR²dB/dt

E = πR²dB/dt ÷ 4πR

E = -(RdB/dt)/4

So, the magnitude of the induced electric field is (RdB/dt)/4

Learn more about induced electric field here:

brainly.com/question/15730392

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Short Answer Questions:
Otrada [13]

According to Archimede's principle, a physical object experiences an upthrust due to a difference in pressure between upper and lower  surfaces.

<h3>What is an upthrust?</h3>

An upthrust is also referred to as buoyancy and it can be defined as an upward force which is exerted by a fluid (liquid or gas), so as to oppose the weight of a partially or fully immersed physical object that is floating in it.

Based on scientific information, a physical object experiences an upthrust when it is immersed in a fluid due to a difference in height and pressure between upper (top) and lower (bottom) surfaces.

According to Archimede's principle, there is a higher pressure at the bottom of the physical object due to height, and a lower pressure at the top of the physical object.

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4 0
2 years ago
PLEASE HELP!!!!!!!! More science questions coming soon!!
Naily [24]

Answer:

1.) A - 1,2,1,2

2.) Both forces can act without objects touching.

3.) C

Both can be modeled as waves having amplitude, frequency and wave lengths.

4.) A, B and C

Explanation:

1.) SO2 + ___ H2 ----------> _____ S + _______H2O

Looking at above equation, we have one molecule of surfur at the right hand side (RHS), and one molecule of surfur at the left hand side (LHS).

Two atoms of hydrogen and both RHS and LHS

But oxygen is not balanced. We have two atoms of oxygen at the RHS while having one at the LHS.

So let's make oxygen 2 atoms at LHS by adding 2 to water molecules and hydrogen molecules at the RHS.

SO2 + ___ 2H2 ----------> _____ S + _______2H2O

The correct answer is A - 1,2,1,2

2.) Both magnetic force and gravitational force obey inverse square law with distance. They are not directly proportional but inversely proportional to the square distance. B is the correct answer because Both forces can act without objects touching.

3.) Light waves are transverse waves which can travel through a vacuum without a medium while sound waves are longitudinal waves which cannot travel through vacuum without a medium. But both can be modeled as waves having amplitude, frequency and wave lengths.

4.) If an object is slowing down, due to conservation of energy,

The potential energy could be increasing, like a ball thrown into the air.

The kinetic energy could be lost to friction or air resistance.

The ball could be returning to its natural resting state.

7 0
3 years ago
In the Olympic shot-put event, an athlete throws the shot with an initial speed of 12.0m/s at a 40.0? angle from the horizontal.
HACTEHA [7]

A) Horizontal range: 16.34 m

B) Horizontal range: 16.38 m

C) Horizontal range: 16.34 m

D) Horizontal range: 16.07 m

E) The angle that gives the maximum range is 41.9^{\circ}

Explanation:

A)

The motion of the shot is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion is a uniformly accelerated motion, so we can use the following suvat equation to find the time of flight:

s=u_y t + \frac{1}{2}at^2 (1)

where

s = -1.80 m is the vertical displacement of the shot to reach the ground (negative = downward)

u_y = u sin \theta is the initial vertical velocity, where

u = 12.0 m/s is the initial speed

\theta=40.0^{\circ} is the angle of projection

So

u_y=(12.0)(sin 40.0^{\circ})=7.7 m/s

a=g=-9.8 m/s^2 is the acceleration due to gravity (downward)

Substituting the numbers, we get

-1.80 = 7.7t -4.9t^2\\4.9t^2-7.7t-1.80=0

which has two solutions:

t = -0.21 s (negative, we ignore it)

t = 1.778 s (this is the time of flight)

The horizontal motion is instead uniform, so the horizontal range is given by

d=u_x t

where

u_x = u cos \theta=(12.0)(cos 40^{\circ})=9.19 m/s is the horizontal velocity

t = 1.778 s is the time of flight

Solving, we find

d=(9.19)(1.778)=16.34 m

B)

In this second case,

\theta=42.5^{\circ}

So the vertical velocity is

u_y = u sin \theta = (12.0)(sin 42.5^{\circ})=8.1 m/s

So the equation for the vertical motion becomes

4.9t^2-8.1t-1.80=0

Solving for t, we find that the time of flight is

t = 1.851 s

The horizontal velocity is

u_x = u cos \theta = (12.0)(cos 42.5^{\circ})=8.85 m/s

So, the range of the shot is

d=u_x t = (8.85)(1.851)=16.38 m

C)

In this third case,

\theta=45^{\circ}

So the vertical velocity is

u_y = u sin \theta = (12.0)(sin 45^{\circ})=8.5 m/s

So the equation for the vertical motion becomes

4.9t^2-8.5t-1.80=0

Solving for t, we find that the time of flight is

t = 1.925 s

The horizontal velocity is

u_x = u cos \theta = (12.0)(cos 45^{\circ})=8.49 m/s

So, the range of the shot is

d=u_x t = (8.49)(1.925)=16.34 m

D)

In this 4th case,

\theta=47.5^{\circ}

So the vertical velocity is

u_y = u sin \theta = (12.0)(sin 47.5^{\circ})=8.8 m/s

So the equation for the vertical motion becomes

4.9t^2-8.8t-1.80=0

Solving for t, we find that the time of flight is

t = 1.981 s

The horizontal velocity is

u_x = u cos \theta = (12.0)(cos 47.5^{\circ})=8.11 m/s

So, the range of the shot is

d=u_x t = (8.11)(1.981)=16.07 m

E)

From the previous parts, we see that the maximum range is obtained when the angle of releases is \theta=42.5^{\circ}.

The actual angle of release which corresponds to the maximum range can be obtained as follows:

The equation for the vertical motion can be rewritten as

s-u sin \theta t + \frac{1}{2}gt^2=0

The solutions of this quadratic equation are

t=\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g}

This is the time of flight: so, the horizontal range is

d=u_x t = u cos \theta (\frac{u sin \theta \pm \sqrt{u^2 sin^2 \theta+2gs}}{-g})=\\=\frac{u^2}{-2g}(1+\sqrt{1+\frac{2gs}{u^2 sin^2 \theta}})sin 2\theta

It can be found that the maximum of this function is obtained when the angle is

\theta=cos^{-1}(\sqrt{\frac{2gs+u^2}{2gs+2u^2}})

Therefore in this problem, the angle which leads to the maximum range is

\theta=cos^{-1}(\sqrt{\frac{2(-9.8)(-1.80)+(12.0)^2}{2(-9.8)(-1.80)+2(12.0)^2}})=41.9^{\circ}

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

8 0
3 years ago
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