What is an input force?

At a certain place, Earth's magnetic field has magnitude B =0.703 gauss and is inclined downward at an angle of 75.4° to the hor

Irina18 [472]

Answer:

The charge flows in coulombs is

$dq=1.843x10^{-5}C$

Explanation:

The current magnitude of current is given by the resistance and the induced Emf as:

$I=N*\frac{dF}{Rdt}$

$\frac{dq}{dt}=\frac{dF}{Rdt}=dq=N*\frac{dF}{R}$

$dq=\frac{N*\beta*A*(Cos(\alpha_f)-Cos(\alpha_i}{R}$

$N=1300$ , $\beta=0.703$ , $A=\pi*r^2=\pi*0.10^2=0.01\pi m^2$ , $R=99.4+202=301.4$ Ω

$\alpha_f=14.6$ , $\alpha_i=165.4$

Replacing :

$dq=\frac{1300*0.703x10^{-4}*0.01\pi*(0.9667-(-0.9667))}{202+99.4}$

$dq=1.843x10^{-5}C$

5 0

3 years ago

Describe the water cycle process starting from an afternoon thunderstorm. There are many different variations that could happen.

spin [16.1K]

Explanation:

The water cycle basically involves five steps:

evaporation and transpiration ⇄
condensation, ⇄
precipitation, ⇄
runoff, ⇄
infiltration ⇄

So when a thunderstorm occurs it helps in completing the precipitation process by enabling the release of water vapor stored up in the atmosphere to fall on the ground as rain.

After this, the water runoffs to the surface of the ground, on plants, into rocks, rivers, and lakes.

Next, the Infiltration process enables the water on the ground surface to enter the soil some of which becomes groundwater.

The cycle begins again as the evaporation and transpiration process begins, where the groundwater as a result of heat from the sun is taken back into the atmosphere, while water in plants by means of transpiration goes back into the atmosphere.

It then condenses and falls back as precipitation again.

3 0

3 years ago

Did i draw the tangent line correctly?

Jet001 [13]

Answer:

yes i belevie so

Explanation:

8 0

3 years ago

Match each term with the appropriate definition

tatiyna

Answer:

opaque = 4

malleable = 3

ductile = 2

lustrous (or whatever the bottom word is) = 1

6 0

3 years ago

A kite 40 ft above the ground moves horizontally at a constant speed of 10 ft/s, with a child, holding the ball of kite string,

Lorico [155]

Answer:

v = 27.28 m /s, θ = 63.9º

Explanation:

For this exercise we can approximate the movement to a projectile launch, let's analyze the situation.

* We must find the horizontal speed, for this we will find the descent time and the horizontal distance

* We look for the vertical speed

At the highest point the speed is horizontal

Let's find the time it takes for the kite to reach the ground

y = y₀ + v_{oy} t - ½ g t²

0 =y₀ + 0 -1/2 gt²

t = $\sqrt{ \frac{2y_o}{g} }$

t = √(2 40/32)

t = 2.5 s

to find the horizontal velocity we must know the horizontal distance, let's use trigonometry

sin θ = y / l

θ = sin⁻¹1 y / l

θ = sin⁻¹ 40/50

θ = 53.1º

therefore the horizontal distance is

x = l cos 53.1

x = 50cos 53.1

x = 30 m

let's use the equation

x = v₀ₓ t

v₀ₓ = x / t

v₀ₓ = 30 / 2.5

v₀ₓ = 12 m / s

we look for the vertical component of the velocity

v_y = v_{oy} - g t

v_y = 0 - g t

v_y = - 9.8 2.5

v_y = -24.5 m / s

the negative sign indicates that the speed is directed downwards, because it is the arrival point, as they indicate that there is no friction, the exit speed is the same, worse with the opposite sign

We already have the two components of the velocity, let's use the Pythagorean theorem to find the modulus

v = $\sqrt{v_x^2 + v_y^2}$

v = $\sqrt{12^2 + 24.5^2}$

v = 27.28 m /s

we use trigonometry for the angle

tan θ = v_y / vₓ

θ = tan⁻¹ v_y / vₓ

θ = tan⁻¹ 24.5 / 12

θ = 63.9º

4 0

3 years ago