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Contact [7]
3 years ago
10

Tessa wonders how many different ways she can light a lightbulb. Circle all the ways you think will work.

Physics
2 answers:
monitta3 years ago
5 0
E,f,h I think good luck
adell [148]3 years ago
4 0
It’s just E because ethe positiv and negative current are supposed to flow thorough the bulb in opppsote sides at a equel level.In some them negerive/postive is absent and some of them are connected to the same side
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A small charged sphere is attached to a thread and placed in an electric field. The other end of the thread is anchored so that
Katyanochek1 [597]

To find the magnitude and direction of the electric field, let us find the horizontal and vertical components of the field separately, then we will use those values to calculate the total magnitude and direction.

The tension in the thread is 6.57×10⁻²N and the thread is aligned horizontally, so the tension force is directed entirely horizontally. The sphere is in static equilibrium, therefore the horizontal component of the electrostatic force acting on the sphere, Fx, must act in the opposite direction of the tension and have a magnitude of 6.57×10⁻²N. We know this equation relating a charge, an electric field, and the force that the field exerts on the charge:

F = Eq

F is the electric force, E is the electric field, and q is the charge

Let us adjust the equation for only the horizontal components of the above quantities:

Fx = (Ex)(q)

Fx is the horizontal component of the electric force and Ex is the horizontal component of the electric field.

Given values:

F = 6.57×10⁻²N

q = 6.80×10³C

Plug in these values and solve for Ex:

6.57x10⁻² = Ex(6.80×10³)

<u>Ex = 9.66×10⁻⁶N/C</u>

<u />

Since the sphere is in static equilibrium, the vertical component of the electrostatic force acting on the sphere, Fy, must have the same magnitude and act in the opposite direction of the sphere's weight. If we assume the weight to act downwards, then Fy must act upward.

We know the weight of the sphere is given by:

W = mg

W is the weight, m is the mass, and g is the acceleration of objects due to earth's gravity field near its surface.

We also know this equation:

F = Eq

Let us adjust for the vertical components:

Fy = (Ey)(q)

Set Fy equal to W and we get:

(Ey)(q) = mg

Given values:

q = 6.80×10³C

m = 0.018kg

g = 9.81m/s²

Plug in the values and solve for Ey:

(Ey)(6.80×10³) = 0.018(9.81)

<u>Ey = 2.60×10⁻⁵N/C</u>

<u />

Let's now use the Pythagorean theorem to find the total magnitude of the electric field:

E = \sqrt{Ex^{2}+Ey^{2}}

E = 2.77×10⁻⁵N/C

The direction of the electric field is given by:

θ = tan⁻¹(Ey/Ex)

θ = 20.4° off the horizontal

4 0
3 years ago
A rocket engine uses fuel and oxidizer in a reaction that produces gas particles having a velocity of 1380 ms The desired thrust
bearhunter [10]

Answer:

a. 141.3 kg/s b. 5.49 m/s² c. i. 104228.9 N ii. 8.53 m/s² d. i. 97305.2 N ii. 9.84 m/s²

Explanation:

a. What must be the fuel/oxidizer consumption rate (in kg s1)?

The thrust T = Rv where R = mass consumption rate and v = velocity of rocket. Since T = 195000 N and v = 1380 m/s,

R = T/v = 195000 N/1380 m/s = 141.3 kg/s

b. If the initial weight of the rocket is 125000 N, what is its initial acceleration?

We also know that thrust T - W = ma since the rocket has to move against gravity. where M = mass of rocket = W/g = 125000 N/9.8m/s² = 12755.1 kg, W = weight of rocket = 125000 N, a = acceleration of rocket and T = thrust = 195000 N.

So, T - W = Ma

195000 N - 125000 N = (12755.1 kg)a

70000 N = ma

a = 70000 N/12755.1 kg = 5.49 m/s²

c. What are the weight and acceleration of the rocket at t 15.0 s after ignition?

We know that the loss in mass ΔM = mass consumption rate × time = Rt. Since R = 141.3 kg/s and t = 15 s,

ΔM = 141.3 kg/s × 15 = 2119.5 kg

The new mass is thus M = M - ΔM = 12755.1 kg - 2119.5 kg = 10635.6 kg

i.The weight after 15 seconds is thus W' = M'g = 10635.6 kg × 9.8m/s² = 104228.9 N

ii. Since T - W' = M'a. where M' is our new mass and a our new acceleration,

a = (T - W')/M'

= (195000 N - 104228.9 N)/10635.6 kg

= 90771.1 N/10635.6 kg

= 8.53 m/s²

d. What are the weight and acceleration of the rocket at 20.0 s after ignition?

We know that the loss in mass ΔM" = mass consumption rate × time = Rt. Since R = 141.3 kg/s and t = 20 s,

ΔM" = 141.3 kg/s × 20 = 2826 kg

The new mass is thus M" = M - ΔM" = 12755.1 kg - 2826 kg = 9929.1 kg

i. The weight after 20 seconds is thus W" = M"g = 9929.1 kg × 9.8m/s² = 97305.2 N

ii. Since T - W" = M"a. where M" is our new mass and a our new acceleration,

a = (T - W")/M"

= (195000 N - 97305.2 N)/9929.1 kg

= 97694.8 N/9929.1 kg

= 9.84 m/s²

4 0
3 years ago
A jet on an aircraft carrier can be launched from rest to 40 m/s in 2 seconds. What is the acceleration of the aircraft? Show st
Alex_Xolod [135]

Answer:

a=20\ m/s^2

Explanation:

Given that,

Initial speed, u = 0

Final speed, v = 40 m/s

Time, t = 2 s

We need to find the acceleration of the aircraft. We know that, acceleration is equal to the rate of change of velocity. So,

a=\dfrac{v-u}{t}\\\\a=\dfrac{40-0}{2}\\\\=20\ m/s^2

So, the acceleration of the aircraft is 20\ m/s^2.

5 0
2 years ago
How are the electric field lines around a positive charge affected when a second positive charge is near it? The field lines com
timofeeve [1]
The answer would be D hope it helps and sorry if it is wrong.  :)
3 0
3 years ago
Read 2 more answers
A lorry of mass 4000 kg is travelling at a speed of 4 m/s.
bulgar [2K]

Answer:

KE = 32,000J

v = 8m/s

Explanation:

KE = .5*m*v²

KE = .5*4000kg*(4m/s)²

KE = 32,000J

32,000J = .5*1000kg*v²

v² = 64

v = 8m/s

7 0
3 years ago
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