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AleksandrR [38]

1 year ago

14

Two parallel plates of area 2.3410-3 M2 have 7.0710-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the p

lates. What is the magnitude of the electric force on q1?

1 answer:

ira [324]1 year ago

8 0

Answer:

The force exerted on the $q_1$ is $F = 2.25*10^{3} \ N$

Explanation:

From the question we are told that

The area is $A = 2.34*10^{-3} \ m^2$

The magnitude of charge placed on them is $q = 7.07 * 10^{-7} C$

The charge placed between the plate is $q_1 = 6.62 *10^{-5} C$

The electric field generated around the plate is mathematically represented as

$E = \frac{q}{A \epsilon_o}$

Substituting values

$E = \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}$

$E = 34*10^{6} \ V/m$

The force exerted the charge $q_1$ is mathematically represented as

$F = q_1 * E$

Substituting values

$F = 6.62 *10^{-5} * 34*10^{6}$

$F = 2.25*10^{3} \ N$

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1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r

DENIUS [597]

Answer:

1)

$v_{oy}=11.29\ m/s$

2)

$y=7.39\ m$

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle $\theta$ with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

Where $y_o$ is the initial height above the ground level, $v_{oy}$ is the vertical component of the initial velocity and t is the time

The y-component of the speed is

$v_y=v_{oy}-gt$

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of $v_o$

The object will reach the maximum height when $v_y=0$ . It allows us to compute the time to reach that point

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum heigh is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Solving for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Replacing the known values

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

$y=7.39\ m$

5 0

2 years ago

You are piloting a small airplane in which you want to reach a destination that is 750 km due north of your starting location. O

alexira [117]

Answer:

v_wind = 101.46 km / h , θ = 61.8

Explanation:

This is a velocity composition exercise.

Let's do the problem in parts. Let's start by knowing the speed of the plane without air.

v = d / t

v = 750 / 3.14

v = 238.85 km / h

This is the speed of the plane relative to the Earth and it does not change.

In the second part, when there is wind, the travel time is greater than when there is no wind, therefore the wind delays the plane. To be more general, suppose that the wind has two components vₓ and $v_{y}$

Let's use trigonometry to find the components of the plane's speed

cos θ = v_N / v

sin θ = v_W / v

v_N = v cos θ

v_W = v sin θ

let's calculate

V _N = 238.85 cos 22 = 221.46 km / h

v_W = -238.85 sin 22 = -89.47

the negative sign is because the plane is going west and the positive sign is the east direction.

As it indicates that the destination of the avine is towards the north, the x component of the wind must be

vₓ - v_W = 0

vₓ = v-w

vₓ = 89.47 km / h

in the direction to the East.

Now let's analyze the component of the wind in the Nort-South direction,

Indicate the travel time, let's calculate the speed that the component must have the speed of the plane

v_total = d / t

v_total = 750 / 4.32

v_total = 173.61 km / h

This is the final speed of the plane, which can be written

v_total = v_n - vy

vy = v_n - v_total

vy = 221.46 - 173.61

vy = 47.85 km

this component is directed towards the south

Let's use the Pythagorean Theorem, to find the magnitude

v_wind² = vₓ² + vy²

v_wind = √ (89.47² + 47.85²)

v_wind = 101.46 km / h

the address will then be found using trigonometry

θ = Vy / vx

θ = tan⁻¹ (vy / vx)

θ = tan⁻¹1 (47.85 / 89.47)

θ = 28.14

Therefore, the magnitude of the wind speed is 101.5 km / h and its direction is 28º south of the East, to give this value

90- θtea = 90- 28.2

θ = 61.8

East of South

5 0

1 year ago

The charges for four ions are sodium (Na) +1, calcium (Ca) +2, fluorine (F) –1, and sulfur (S) –2. Which would be an ionic bond

Ludmilka [50]

The correct answer would be B. It is the compound Na2S that forms an ionic bond from the given ion above. From the choices, it is the only compound that exist and is available. Metals cannot react with another metal as well as nonmetal to nonmetal.

3 0

2 years ago

Read 2 more answers

A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of 25.0° below the horizontal. It is held in position by a fri

liubo4ka [24]

Answer:

$T = 7.64 kN$

$F_y = 0.52 kN$ (Downwards)

$F_x = 3.23 kN$ (Towards Left)

Explanation:

As we know that beam is in equilibrium

So here we can use torque balance as well as force balance for the beam

Now by torque balance equation at the pivot we can say

$F(4.50 cos\theta) + mg(2cos\theta) = T \times 3$

As we know that

mg = 1.40 kN

F = 5 kN

so we will have

$5 kN(4.50 cos25) + 1.40 kN(2 cos25) = 3 T$

$T = 7.64 kN$

Now force balance in vertical direction

$F + mg = Tsin65 + F_y$

$5 + 1.40 = 7.64 sin65 + F_y$

$F_y = 0.52 kN$ (Downwards)

Force balance in horizontal direction

$F_x = T cos65$

$F_x = 7.64 cos65$

$F_x = 3.23 kN$ (Towards Left)

7 0

2 years ago

Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the

Stells [14]

Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

q =75*10^-6 C

A = 0.25

eo = 8.85*10^-12

Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

8 0

1 year ago