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3 years ago

This circuit has five bulbs (A, B, C, D, and E). Select which bulbs would turn off if bulb C was unscrewed.

Physics

1 answer:

LiRa [457]3 years ago

4 0

Answer:

C, A and E

Explanation:

In theory you always assume that the prostive is going to the negtive and since C, A and E are in series they all will turn off will turn off

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Three point charges are on the x axis: −1 µC

Nina [5.8K]

Answer:

0.0078 N

Explanation:

The electrostatic force between two charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

The force is attractive if the two charges have opposite sign, and repulsive if the two charges have same sign.

In this problem, we have:

$q_1 = -1 \mu C = -1 \cdot 10^{-6}C$ located at $x_1=-3 m$

$q_2=+9 \mu C = +9\cdot 10^{-6}C$ located at $x_2=0 m$

$q_3 = -5 \mu C = -5\cdot 10^{-6} C$ located at $x_3=+3 m$

The force between charge 1 and charge 2 is:

$F_{12}=k\frac{q_1 q_2}{(x_2-x_1)^2}=(8.99\cdot 10^9)\frac{(1\cdot 10^{-6})(+9\cdot 10^{-6})}{3^2}=0.0090 N$

And since the two charges have opposite sign, the force is attractive, so the force on charge 1 is towards the right.

The force between charge 1 and charge 3 is:

$F_{13}=k\frac{q_1 q_3}{(x_3-x_1)^2}=(8.99\cdot 10^9)\frac{(1\cdot 10^{-6})(5\cdot 10^{-6})}{6^2}=0.0012 N$

And since the two charges have same sign, the force is repulsive, so the force on charge 1 is towards the left.

Therefore, the net force on charge 1 is:

$F=F_{12}-F_{13}=0.0090-0.0012 = 0.0078 N$

towards the right.

5 0

3 years ago

A 782-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli

Vadim26 [7]

Answer:

a) v = 5.59x10³ m/s

b) T = 4 h

c) F = 1.92x10³ N

Explanation:

a) We can find the satellite's orbital speed by equating the centripetal force and the gravitation force as follows:

$F_{c} = F_{G}$

$\frac{mv^{2}}{r + h} = \frac{GMm}{(r + h)^{2}}$

$v = \sqrt{\frac{gr^{2}}{r+h}$

Where:

g is the gravity = 9.81 m/s²

r: is the Earth's radius = 6371 km

h: is the satellite's height = r = 6371 km

$v = \sqrt{\frac{gr^{2}}{2r}} = \sqrt{\frac{gr}{2}} = \sqrt{\frac{9.81 m/s^{2}*6.371 \cdot 10^{6} m}{2}} = 5.59 \cdot 10^{3} m/s$

b) The period of its revolution is:

$T = \frac{2\pi}{\omega} = \frac{2\pi (r + h)}{v} = \frac{2\pi (2*6.371 \cdot 10^{6} m)}{5.59 \cdot 10^{3} m/s} = 14322.07 s = 4 h$

c) The gravitational force acting on it is given by:

$F = \frac{GMm}{(r + h)^{2}}$

Where:

M is the Earth's mass = 5.97x10²⁴ kg

m is the satellite's mass = 782 kg

G is the gravitational constant = 6.67x10⁻¹¹ Nm²kg⁻²

$F = \frac{GMm}{(r + h)^{2}} = \frac{6.67 \cdot 10^{-11} Nm^{2}kg^{-2}*5.97 \cdot 10^{24} kg*782 kg}{(2*6.371 \cdot 10^{6} m)^{2}} = 1.92 \cdot 10^{3} N$

I hope it helps you!

3 0

4 years ago

Which of the following are electromagnetic waves? sound waves water waves ground waves light waves`

Paraphin [41]

Hello! Light waves are an example of electromagnetic waves. I hope this helps!

5 0

3 years ago

To reach its intended destination on time a cruise ship needs to travel north at 70 mph. However, it is travelling in the Gulf o

bogdanovich [222]

Answer:

speed is 81.03 mph

direction is N 3.58 W

Explanation:

given data

travel north = 70 mph

Stream current = 12 mph

direction = S 25° E

result due north = 70 mph

to find out

speed and direction

solution

we will get component of resultant that is

v cosθ and v sinθ

( 12cos295 , 12 sin295 ) at ( 0, 70)

as that we can say

v sinθ + 12sin295 = 70 ....................1

v cosθ + 12 cos295 = 0 ......................2

vcosθ = -5.0714

vsinθ = 80.8756

now by ratio

cosθ /sinθ = -5.0714/ 80.8756

cot θ = -0.0627

θ = 93.58

so direction is N 3.58 W

and

we know

vcosθ = - 12cos295

v = - 12cos295 / cos(93.58)

v = 81.03 mph

so speed is 81.03 mph

4 0

3 years ago

Explain how magnets have an attractive force and repulsive force

shusha [124]

The force between two parallel wires carrying currents in the same direction is attractive. It is repulsive if the currents are in opposite directions.

7 0

3 years ago