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3 years ago

Very easy physics maths about current and other stuffs only one question please help me

Physics

1 answer:

Reil [10]3 years ago

7 0

Answer:

The total voltage of the circuit is 18 volts.

Explanation:

We have, three identical resistors of resistance 3 ohms are connected in series in a circuit.

For a series combination, the equivalent resistance is given by the sum of individual resistances i.e.

$R_{eq}=R_1+R_2+R_3\\\\R_{eq}=3+3+3\\\\R_{eq}=9\ \Omega$

Let V is the voltage of the battery if the current in the circuit is 2 A. So,

$V=IR$

$V=2\times 9\\\\V=18\ V$

So, the total voltage of the circuit is 18 volts.

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Suppose that instead of being inclined to Earth's orbit around the Sun, the Moon’s orbit was in the same plane as Earth’s orbit

d1i1m1o1n [39]

Answer:

Explanation:

A solar eclipse occurs when the moon in between the sun and the earth. Since in reality, the moons orbit is inclined, this aligning doesnt occur often. But if the orbit of the moon were in the same plane as the earth orbit, and since the moon orbits the earth once a month, the three objects would align every month. This makes it 12 times in a year.

8 0

3 years ago

Hey scooters dragging of 520 kg walk-through forest at a constant speed of 3.5 m/s. If the scooter is applying a force of 1850 N

Simora [160]

The coefficient of friction is 0.363

Explanation:

There are two forces acting on the scooter in the horizontal direction:

- The applied force, F = 1850 N, forward

- The frictional force, $F_f$ , backward

Since the scooter is moving at constant speed, the acceleration is zero, so the net force acting on the scooter must be zero. Therefore we can write:

$F-F_f = 0\\F_f = F = 1850 N$

The frictional force can be written as

$F_f = \mu R$ (1)

where

$\mu$ is the coefficient of friction

R is the normal reaction of the ground on the scooter

For a flat horizontal surface, there is equilibrium along the vertical direction, so the normal reaction is equal to the weight:

$R = W = mg$

where

m = 520 kg is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting into (1),

$F_f = \mu mg = 1850 N$

and solving for $\mu$ ,

$\mu=\frac{F_f}{mg}=\frac{1850}{(520)(9.8)}=0.363$

Learn more about friction:

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5 0

3 years ago

50 Joules of work in 25 seconds. How much power did she use?

Lorico [155]

P(W) = E(J) / t(s)

50/25=2

3 0

3 years ago

A.in which portion of the curve,

lozanna [386]

Answer:

the answer is 4 acceleration is 0

Explanation:

this is because when parallel line is drawn through x axis the velocity will bw uniform

plz brainliest

8 0

3 years ago

Find the force on a negative charge that is placed midway between two equal positive charges. All charges have the same magnitud

MArishka [77]

Answer: The force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.

Explanation:

Let us assume that

$q_{1} = q_{2} = +q$

$q_{3} = -q$

As $q_{3}$ is the negative charge and placed midway between two equal positive charges ( $q_{1}$ and $q_{2}$ ).

Total distance between $q_{1}$ and $q_{2}$ is 2r. This means that the distance between $q_{1}$ and $q_{3}$ , $q_{2}$ and $q_{3}$ = d = r

Now, force action on charge $q_{3}$ due to $q_{1}$ is as follows.

$F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})$

where,

k = electrostatic constant = $9 \times 10^{9} Nm^{2}/C^{2}$

Substitute the values into above formula as follows.

$F_{31} = k(\frac{q_{1} \times q_{3}}{d^{2}})\\= 9 \times 10^{9} (\frac{q \times (-q)}{r^{2}})\\= - 9 \times 10^{9} (\frac{q^{2}}{r^{2}})$ ... (1)

Similarly, force acting on $q_{3}$ due to $q_{1}$ is as follows.

$F_{32} = k \frac{q_{2}q_{3}}{d^{2}}\\= -9 \times 10^{9} \frac{q^{2}}{r^{2}}\\$ ... (2)

As both the forces represented in equation (1) and (2) are same and equal in magnitude. This means that the net force acting on charge $q_{3}$ is zero.

Thus, we can conclude that the force on a negative charge that is placed midway between two equal positive charges is zero when all charges have the same magnitude.

3 0

3 years ago