A hemispherical surface (half of a spherical surface) of radius R is located in a uniform electric field of magnitude E that is parallel to the axis of the hemisphere. What is the magnitude of the electric flux through the hemisphere surface?

Answers

Answer 1

Answer:

Answer:

π*R²*E

Explanation:

According to the definition of electric flux, it can be calculated integrating the product E*dA, across the surface.

As the electric field E is uniform and parallel to the hemisphere axis, and no charge is enclosed within it, the net flux will be zero, so, in magnitude, the flux across the opening defining the hemisphere, must be equal to the one across the surface.

The flux across the open surface can be expressed as follows:

$\int\ {E} \, dA = E*A = E*\pi *R^(2)$

As E is constant, and parallel to the surface vector dA at any point, can be taken out of the integral, which is just the area of the surface, π*R².

⇒Flux = E*π*R²

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A lead ball is dropped into a lake from a diving board 5.0 m above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 3.0 s after it is released. How deep is the lake

Answers

Answer:

$|D_(depth) |=19.697m$

Explanation:

To find Depth D of lake we must need to find the time taken to hit the water.So we use equation of simple motion as:

Δx=vit+(1/2)at²

$x_(f)-x_(i)=v_(i)t+(1/2)at^(2)\n -5.0m=(o)t+(1/2)(-9.8m/s^(2) )t^(2)\n -4.9t^(2)=-5.0\n t^(2)=5/4.9\nt=√(1.02) \nt=1.01s$

As we have find the time taken now we need to find the final velocity vf from below equation as

$v_(f)=v_(i)+at\nv_(f)=0+(-9.8m/s^(2) )(1.01s) \nv_(f)=-9.898m/s$

So the depth of lake is given by:

first we need to find total time as

t=3.0-1.01 =1.99 s

$|D_(depth) |=|vt|\n|D_(depth) |=|(-9.898m/s)(1.99s)|\n|D_(depth) |=19.697m$

Someone claps his or her hands is an example of… motion energy to sound energy sound energy to potential energy motion energy to radiant energy

Answers

motion energy to sound energy i think

A 21 kg mountain lion carries a 3kg cub in it's mouth as it jumps from rest on the ground to the top of a 2 m talk rock. It takes 1 seconds for the mountain lion to jump and reach the top. How much power did the mountain lion exert? I need help to solve for power

Answers

Answer:

The power exerted by the mountain lion is 1,472.35 W.

Explanation:

Given;

mass of mountain, m₁ = 21 kg

mass of the cub, m₂ = 3 kg

height jumped by the mountain lion, h = 2 m

time taken for the mountain lion to jump, t = 1 s

Determine the weight of the lions on the top rock;

W = F = (m₁ + m₂)g

F = (21 + 3) x 9.8

F = (24) x 9.8

F = 235.2 N

Determine the final velocity of the mountain rock as it jumped to the top;

v² = u² + 2gh

where;

u is the initial velocity = 0

h is the height jumped = 2 m

v² = 0 + 2 x 9.8 x 2

v² = 39.2

v = √39.2

v = 6.26 m/s

The power exerted by the mountain lion is calculated as;

P = Fv

P = 235.2 x 6.26

P = 1,472.35 W

Therefore, the power exerted by the mountain lion is 1,472.35 W.

A 372-g mass is attached to a spring and undergoes simple harmonic motion. Its maximum acceleration is 17.6 m/s2 , and its maximum speed is 1.75 m/s. a)Determine the angular frequency. b)Determine the amplitude. c)Determine the spring constant.

Answers

Answer with Explanation:

We are given that

Mass , m=372 g= $(372)/(1000)=0.372 Kg$

1 kg=1000g

Maximum acceleration, a= $17.6 m/s^2$

Maximum speed ,v=1.75 m/s

a.We know that

Maximum acceleration, a= $A\omega^2$

Maximum speed, v= $\omega A$

$17.6=A\omega^2$

$1.75=A\omega$

$(17.6)/(1.75)=(A\omega^2)/(A\omega)=\omega$

Angular frequency, $\omega=10.06 rad/s$

b.Substitute the value of angular frequency

$1.75=A(10.06)$

$A=(1.75)/(10.06)=0.17 m$

Hence, the amplitude=0.17 m

c.Spring constant,k= $m\omega^2$

Using the formula

$k=0.372* (10.06)^2$

Hence, the spring constant,k=37.6 N/m

A baseball player standing on a platform throws a baseball out over a level playing field. The ball is released from a point 3.50 m above the field with an initial speed of 14.3 m/s at an upward angle of 27.0 degrees above the horizontal. What horizontal distance will the ball travel before hitting the ground?

Answers

Answer:

22.1 m

Explanation:

$v_(o)$ = initial speed of ball = 14.3 m/s

$\theta$ = Angle of launch = 27°

Consider the motion of the ball along the vertical direction.

$v_(oy)$ = initial speed of ball = $v_(o) Sin\theta = 14.3 Sin27 = 6.5 ms^(-1)$

$a_(y)$ = acceleration due to gravity = - 9.8 ms⁻²

$t$ = time of travel

$y$ = vertical displacement = - 3.50 m

Using the kinematics equation that suits the above list of data, we have

$y = v_(oy) t + (0.5) a_(y) t^(2) \n- 3.50 = (6.5) t + (0.5) (- 9.8) t^(2)\n- 3.50 = (6.5) t - 4.9 t^(2)\nt = 1.74 s$

Consider the motion of the ball along the horizontal direction.

$v_(ox)$ = initial speed of ball = $v_(o) Cos\theta = 14.3 Cos27 = 12.7 ms^(-1)$

$X$ = Horizontal distance traveled

$t$ = time taken = 1.74 s

Since there is no acceleration along the horizontal direction, we have

$X = v_(ox) t\nX = (12.7)(1.74)\nX = 22.1 m$

If 2050 J of heat are added to a 150 g object its temperature increases by 15°C.(a) What is the heat capacity of this object?
(b) What is the object's specific heat?

Answers

When an object gets heated by a temperature ΔT energy needed, E = mcΔT

Here energy is given E = 2050 J

Mass of object = 150 g

Change in temperature ΔT = 15 $^0C$ = 15 K

a) Heat capacity of an object equal to the ratio of the heat added to (or removed from) an object to the resulting temperature change.

So heat capacity = E/ΔT = 2050/15 = 136.67 J/K

b) We have E = mcΔT

c = $(2050)/(150*10^(-3)*15) = 911.11 J/kgK$

So object's specific heat = 911.11 J/kgK

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