Does energy generation through a water generator seem to depend more on potential or kinetic energy? Explain.

the magnitude of the magnetic field at point p for a certain electromagnetic wave is 2.21. What is the magnitude of the elctic f

vesna_86 [32]

Answer:

$6.63\times 10^8\ N/C$

Explanation:

Given that,

The magnitude of magnetic field, B = 2.21

We need to find the magnitude of the electric field. Let it is E. So,

$\dfrac{E}{B}=c\\\\E=Bc$

Put all the values,

$E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C$

So, the magnitude of the electric field is equal to $6.63\times 10^8\ N/C$ .

7 0

2 years ago

A child drops a ball from a window. The ball strikes the ground in 3.0 seconds. What is the velocity of the ball the instant bef

inessss [21]

Answer:

29.396988 m/s

Explanation:

Really, it depends on where the child is when he drops the ball - e.g., which planet he is on, and his distance from the center of that planet.

I'll assume that the child is on Earth at sea level at the equator, so that his distance from the geocenter is 6378000 meters.

The acceleration, g, is found from

g = GM/r²

G = 6.6743e-11 m³ kg⁻¹ sec⁻²

M = 5.9724e+24 kg

r = 6.378e+6 m

g = 9.799086 m sec⁻²

An approximate answer is found from an equation from constant acceleration kinematics:

v = gt

t = 3.0 sec

v = 29.397259 m/s

Now, the above method is an approximation that makes the technically incorrect assumption that the acceleration of gravity is a constant throughout the entire fall. You get away with it because the drop is very short. In another situation, it might not be. So it would be nice to develop a more accurate method that does not assume constant gravitational acceleration. For that, we begin with the Vis Viva equation:

v = √[GM(2/r − 1/a)]

Here,

a = the semimajor axis of a plunge orbit, which is equal to half of the apoapsis distance of 6378000+h, where

h = the altitude from which the ball is dropped

We can (using some math) develop the following equation:

t − t₀ = √[d/(2GM)] { √(rd−r²) + d arctan √(d/r−1) }

t − t₀ = 3 sec

r = 6378000 meters

d = r + h

Using an iterative method (e.g. Newton's or Danby's), we can determine that the altitude,

h = 44.0954 meters

So,

d = 6378044.09538 meters

a = d/2 = 3189022.04769 meters

Now we can calculate that

v = 29.396988 m/s

This is the more nearly correct answer because it takes into account the variability of the gravitational acceleration during the fall.

5 0

2 years ago

An echo is heard from a cliff 3.71 s after a rifle is fired. How many feet away is the cliff

LenKa [72]

The cliff is 2042 ft away.

We know that the speed of sound in air is directly proportional to the absolute temperature.

First convert the Fahrenheit temperature to Celsius;

°C = 5/9(44.5 - 32)

°C = 6.9 °C

Applying the formula;

V1/V2 = √T1/T2

Where; V1 = velocity of sound in air at 0°C

V2 = Velocity of sound in air at 6.9 °C

1087/V2 = √273/279.9

V2= 1101 ft/s

Given that; V = 2s/t

Where s is the distance of the cliff

t is the time taken

1101 ft/s = 2s/3.71 s

s = 1101 ft/s × 3.71 s/2

s = 2042 ft

Learn more:brainly.com/question/15381147

3 0

2 years ago

A 0.4 kg soccer ball approaches a player with a velocity of 18 m/s to the east. The player strikes

Zigmanuir [339]

Answer:

16Newton sec

Explanation:

Ft = m( v + u)

= 0.4( 22 + 18)

= 0.4 x 40

= 16newton sec

7 0

2 years ago

Bob is standing at rest. Wendy runs past him at a constant speed of 3.00 m/s. At the instant that Wendy passes him, Bob starts t

77julia77 [94]

Answer:

90 meters

Explanation:

To find the distance in which Bob catches up with Wendy, you first write down the motion equations of both Bob and Wendy.

Wendy has a constant speed, then you have:

$x_1=v_1t$ (1)

Bob has an accelerated motion, then, his equation of motion is:

$x_2=v_2t+\frac{1}{2}at^2$ (2)

v1: constant speed of Wendy = 3.00 m/s

v2: initial speed of Bob = 0 m/s

a: acceleration of Bob = 0.200m/s^2

When Bob catches up with Wendy x1 = x2, then you equal both equations and solve for time t:

$x_1=x_2\\\\v_1t=\frac{1}{2}at^2\\\\t=\frac{2v_1}{a}$

you replace the values of a and v1:

$t=\frac{2(3.00m/s)}{0.200m/s^2}=30 s$

Finally, you replace this value of time either the equation for x1 or the equation for x2, and you calculate the distance:

$x_2=x_1=(3.00m/s)(30s)=90m$

hence, Bob catches up with Wendy for a distance of 90m

3 0

2 years ago