Which of these have an elliptical orbit around the sun?

Two uniformly charged conducting plates are parallel to each other. They each have area A. Plate #1 has a positive charge Q whil

Marrrta [24]

Answer:5

Explanation:

Given

First Plate has a charge of +Q and area A

Second Plate has a charge of -3 Q and area A

We Know electric Field due to sheet charge is given by

$E=\frac{Q}{2A\epsilon }$

Where Q=charge over the Plate

A=Area of plate

$\epsilon$ = Permittivity of free space

Electric Field Due to Positive charge will always be away from it while for negative charge it is towards it.

Net Electric Field at a point between between the Plates is the superimposition of two electric with direction

$E_{net}=\frac{Q}{2A\epsilon }+\frac{3Q}{2A\epsilon }$

$E_{net}=\frac{2Q}{A\epsilon }$

Net electric Field is towards the negative charged plate

5 0

4 years ago

A car goes from rest to a speed of 90 km/h in 10 seconds . What is the car's acceleration in m/s²

zloy xaker [14]

First you must convert Km/hr to m/s. 90 km/hr equals 25m/s (this can be done through a conversion table by plugging in the conversion values). Then you need to see what was given:
vi (initial velocity)= 0m/s
vf (final velocity= 25m/s (90km/hr)
t (time)= 10s

Next you should find an equation that requires only the values you know and gives you the value you're looking for. Sometimes that requires two equations to be used, but in this case you only need one. The best equation for this would be a=(vf-vi)/t. Finally, plug in your values (a=(25-0)/10) to get your answer which would be 2.5m/s^2. Hope this helped!

7 0

3 years ago

a supertanker traveling at 7.2 m/s decelerates to a halt in 12 min. Calculate the magnitud of its average decelaration in meters

Nataliya [291]

Explanation:

Acceleration is change in velocity over change in time.

a = Δv / Δt

a = (0 m/s − 7.2 m/s) / (12 min × 60 s/min)

a = -0.01 m/s²

8 0

3 years ago

When operated on a household 110.0-V line, typical hair dryers draw about 1650 W of power. We can model the current as a long st

Dimas [21]

Answer:

Current = 15 A

Resistance = 7.33 ohm

Magnetic field = 1.62 x 10^-4 Tesla

Explanation:

V = 110 V, P = 1650 W, r = 1.85 cm,

(a) Let i be the current

P = V x i

i = 1650 / 110 = 15 A

(b) Let R be the resistance

V = i R

R = 110 / 15 = 7.33 Ohm

(c) Let B be the magnetic field

B = μ0 / 4π x 2i / r

B = 10^-7 x 2 x 15 / 0.0185 = 1.62 x 10^-4 Tesla

5 0

3 years ago

A proton has a speed of 3.50 Ã 105 m/s when at a point where the potential is +100 V. Later, itâs at a point where the potential

ruslelena [56]

Answer:

(a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is $1.022\times10^{-16}\ J$

(c). The work done on the proton is $-8\times10^{-18}\ J$ .

Explanation:

Given that,

Speed $v= 3.50\times10^{5}\ m/s$

Initial potential V=100 V

Final potential = 150 V

(a). We need to calculate the change in the protons electric potential

Potential energy of the proton is

$U=qV=eV$

Using conservation of energy

$K_{i}+U_{i}=K_{f}+U_{f}$

$\dfrac{1}{2}mv_{i}^2+eV_{i}=\dfrac{1}{2}mv_{f}^2+eV_{f}$

$]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e(V_{f}-V_{i})$

$\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e\Delta V$

$\Delta V=\dfrac{m(v_{i}^2-v_{f}^2)}{2e}$

Put the value into the formula

$\Delta V=\dfrac{1.67\times10^{-27}(3.50\times10^{5}-0)^2}{2\times1.6\times10^{-19}}$

$\Delta V=639.2=0.639\ kV$

(b). We need to calculate the change in the potential energy of the proton

Using formula of potential energy

$\Delta U=q\Delta V$

Put the value into the formula

$\Delta U=1.6\times10^{-19}\times639.2$

$\Delta U=1.022\times10^{-16}\ J$

(c). We need to calculate the work done on the proton

Using formula of work done

$\Delta U=-W$

$W=q(V_{2}-V_{1})$

$W=-1.6\times10^{-19}(150-100)$

$W=-8\times10^{-18}\ J$

Hence, (a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is $1.022\times10^{-16}\ J$

(c). The work done on the proton is $-8\times10^{-18}\ J$ .

5 0

3 years ago