What's the thrust force of 20,000N

A closed circuit is formed by a solid conductor in contact with two parallel conducting rails that are closed on their left end,

castortr0y [4]

this is an equation that you need to solve for motional emf. motional emf=vBL, where v is velocity in meters/second, B is magnetic field in Teslas and L is length or distance the rails are apart from each other. when we plug everything into the formula given above, we get: motional emf=5m/s*0.80T*0.20m. solving all this we get 0.8 volts. pretty sure that since they are giving you the direction of the field, they want to know which way the current will flow . since the conductor is moving from left to right the area of the field is increasing which means magnetic flux is increasing as Ф(magnetic flux)=B(magnetic field)*A(area)*cosФ(little phi is the angle to the normal. in this case little fee is 0 degrees so the cosФ doesn't matter). so ↑Ф=B↑A. if magnetic flux is increasing, the induced magnetic field is in the opposite direction as the original magnetic field meaning the induced magnetic field will be out of the page. using the right hand rule which says that if the field is in to the page, the current should go clockwise and if the field is out of the page, the current is counterclockwise so that means that the current should be going counter clockwise since the induced field is going out of the screen. the top of the conducting wire will have its current go to the left and the bottom of the conducting wire will have the current go to the right.

8 0

2 years ago

During the fission reaction shown, how did the target nucleus change ?

Zarrin [17]

Answer:

A. The target nucleus split into two nuclei, each with fewer nucleons than the original.

Explanation:

5 0

3 years ago

A simple series circuit consists of a 150 Ω resistor, a 29 V battery, a switch, and a 2.1 pF parallel-plate capacitor (initially

Rufina [12.5K]

Find the electric flux and the disp at t=0.50ns
Given: 
Resistor R = 160 Ω 
Voltage ε = 22.0 V 
Capacitor C = 3.10 pF = 3.10 * 10^-12 F 
time t = 0.5 ns = 0.5 * 10^-9 s 
ε0 = 8.85 * 10^-12 
Solution: 
ELECTRIC FLUX: 
Φ = Q/ε0 
we have ε0, we need to find Q the charge 
STEP 1: FIND Q 
Q = C ε ( 1 - e^(-t/RC) ) 
Q = { 3.10 * 10^-12 } { 22.0 } { 1 - e^(- 0.5 * 10^-9 / 160 *3.10 * 10^-12 ) } 
Q = { 3.10 * 10^-12 } { 22.0 } { 1 - 0.365 } 
Q = { 3.10 * 10^-12 } { 22.0 } { 0.635 } 
Q = 43.31 * 10^-12 C 
STEP 2: WE HAVE Q AND ε0 > >>> SOLVE FOR ELECTRIC FLUX >>> 
Φ = Q/ε0 
Φ = { 43.31 * 10^-12 C } / { ε0 = 8.85 * 10^-12 } 
Φ = 4.8937 = 4.9 V.m 
DISPLACEMENT CURRENT 
we use the following equation: 
I = { ε / R } { e^(-t/RC) } 
I = { 22 / 160 } { e^(- 0.5 * 10^-9 / 160 *3.10 * 10^-12 ) } 
I = { 0.1375 } { 0.365 } 
I = 0.0502 A = 0.05 A

8 0

3 years ago

A car is parked on a steep incline, making an angle of 37.0° below the horizontal and overlooking the ocean, when its brakes fai

patriot [66]

Answer:

a) The speed of the car when it reaches the edge of the cliff is 19.4 m/s

b) The time it takes the car to reach the edge is 4.79 s

c) The velocity of the car when it lands in the ocean is 31.0 m/s at 60.2º below the horizontal

d) The total time interval the car is in motion is 6.34 s

e) The car lands 24 m from the base of the cliff.

Explanation:

Please, see the figure for a description of the situation.

a) The equation for the position of an accelerated object moving in a straight line is as follows:

x =x0 + v0 * t + 1/2 a * t²

where:

x = position of the car at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the car starts from rest and the origin of the reference system is located where the car starts moving, v0 and x0 = 0. Then, the position of the car will be:

x = 1/2 a * t²

With the data we have, we can calculate the time it takes the car to reach the edge and with that time we can calculate the velocity at that point.

46.5 m = 1/2 * 4.05 m/s² * t²

2* 46.5 m / 4.05 m/s² = t²

t = 4.79 s

The equation for velocity is as follows:

v = v0 + a* t

Where:

v = velocity

v0 = initial velocity

a = acceleration

t = time

For the car, the velocity will be

v = a * t

at the edge, the velocity will be:

v = 4.05 m/s² * 4.79 s = 19.4 m/s

b) The time interval was calculated above, using the equation of the position:

x = 1/2 a * t²

46.5 m = 1/2 * 4.05 m/s² * t²

2* 46.5 m / 4.05 m/s² = t²

t = 4.79 s

c) When the car falls, the position and velocity of the car are given by the following vectors:

r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)

v =(v0x, v0y + g * t)

Where:

r = position vector

x0 = initial horizontal position

v0x = initial horizontal velocity

t = time

y0 = initial vertical position

v0y = initial vertical velocity

g = acceleration due to gravity

v = velocity vector

First, let´s calculate the initial vertical and horizontal velocities (v0x and v0y). For this part of the problem let´s place the center of the reference system where the car starts falling.

Seeing the figure, notice that the vectors v0x and v0y form a right triangle with the vector v0. Then, using trigonometry, we can calculate the magnitude of each velocity:

cos -37.0º = v0x / v0

(the angle is negative because it was measured clockwise and is below the horizontal)

(Note that now v0 is the velocity the car has when it reaches the edge. it was calculated in a) and is 19,4 m/s)

v0x = v0 * cos -37.0 = 19.4 m/s * cos -37.0º = 15.5 m/s

sin 37.0º = v0y/v0

v0y = v0 * sin -37.0 = 19.4 m/s * sin -37.0 = - 11. 7 m/s

Now that we have v0y, we can calculate the time it takes the car to land in the ocean, using the y-component of the vector "r final" (see figure):

y = y0 + v0y * t + 1/2 * g * t²

Notice in the figure that the y-component of the vector "r final" is -30 m, then:

-30 m = y0 + v0y * t + 1/2 * g * t²

According to our reference system, y0 = 0:

-30 m = v0y * t + 1/2 g * t²

-30 m = -11.7 m/s * t - 1/2 * 9.8 m/s² * t²

0 = 30 m - 11.7 m/s * t - 4.9 m/s² * t²

Solving this quadratic equation:

t = 1.55 s ( the other value was discarded because it was negative).

Now that we have the time, we can calculate the value of the y-component of the velocity vector when the car lands:

vy = v0y + g * t

vy = - 11. 7 m/s - 9.8 m/s² * 1.55s = -26.9 m/s

The x-component of the velocity vector is constant, then, vx = v0x = 15.5 m/s (calculated above).

The velocity vector when the car lands is:

v = (15.5 m/s, -26.9 m/s)

We have to express it in magnitude and direction, so let´s find the magnitude:

$|v| = \sqrt{(15.5 m/s)^{2} + (-26.9 m/s)^{2}} = 31.0m/s$

To find the direction, let´s use trigonometry again:

sin α = vy / v

sin α = 26.9 m/s / 31.0 m/s

α = 60.2º

(notice that the angle is measured below the horizontal, then it has to be negative).

Then, the vector velocity expressed in terms of its magnitude and direction is:

vy = v * sin -60.2º

vx = v * cos -60.2º

v = (31.0 m/s cos -60.2º, 31.0 m/s sin -60.2º)

The velocity is 31.0 m/s at 60.2º below the horizontal

d) The total time the car is in motion is the sum of the falling and rolling time. This times where calculated above.

total time = falling time + rolling time

total time = 1,55 s + 4.79 s = 6.34 s

e) Using the equation for the position vector, we have to find "r final 1" (see figure):

r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)

Notice that the y-component is 0 ( figure)

we have already calculated the falling time and the v0x. The initial position x0 is 0. Then.

r final 1 = ( v0x * t, 0)

r final 1 = (15.5 m/s * 1.55 s, 0)

r final 1 = (24.0 m, 0)

The car lands 24 m from the base of the cliff.

PHEW!, it was a very complete problem :)

5 0

2 years ago

Two forces, F1 = -3000 N and F2 = +1500 N acts on a plane horizontally. The craft is moving with a constant velocity of -850 m/s

jonny [76]

Because the net force must be zero, we conclude that the magnitude of the force is 1500 newtons, and the direction is in the positive axis.

<h3>What is the magnitude and direction of the third force?</h3>

By Newton's laws, we know that if the net force applied to an object is different than zero, then the object is accelerated.

In this case, we know that the object moves with constant velocity, so there is no acceleration, meaning that the net force is equal to zero.

Then we must have:

F1 + F2 + F3 = 0N

Replacing F1 and F2 we get:

-3000 N + 1500N + F3 = 0

F3 = 3000N - 1500 N = 1500N

Then the magnitude of the force is 1500 newtons, and the direction is in the positive axis.

If you want to learn more about Newton's laws, you can read:

brainly.com/question/10454047

8 0

2 years ago