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3 years ago

Which of the following shows resistors in a parallel circuit

Physics

2 answers:

Alexeev081 [22]3 years ago

7 0

The bottom drawing do

liubo4ka [24]3 years ago

5 0

Answer:

Circuit (b)

Explanation:

In parallel combination of resistors, the current flowing in each branch divides while the potential difference remains the same in entire circuit. The equivalent resistance is given by the following formula as :

$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$

While in series combination, the current remains the same while the volatge across each resistor is different. Their equivalent is given by:

$R_{eq}=R_1+R_2+R_3$

In the given figure, circuit (b) shows parallel combination of resistors. Also, the resistors are connected in two different paths in circuit (b).

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3 years ago

An object is placed at 1.5 m from a convex lens with a focal length of 1.0 m.

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3 years ago

Three point charges are arranged on a line. Charge q3 = 5 nC and is at the origin. Charge q2 = - 3 nC and is at x = 4 cm. Charge

Taya2010 [7]

Answer:

q₁ = + 1.25 nC

Explanation:

Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃= 2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

$\frac{k*q_{1}*q_3 }{(d_{13})^{2} } = \frac{k*q_{2}*q_3 }{(d_{23})^{2} }$

We divide by (k * q3) on both sides of the equation

$\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }$

$q_{1} = \frac{q_{2}*(d_{13})^{2} }{(d_{23} )^{2} }$

$q_{1} = \frac{5*(2)^{2} }{(4 )^{2} }$

q₁ = + 1.25 nC

3 0

3 years ago

A projectile has an initial horizontal velocity of 34.0 M/s at the edge of a roof top. Find the horizontal and vertical componen

Sveta_85 [38]

Answer:

$v_x=34 m/s$

$v_y=53.9\ m/s$

Explanation:

<u>Horizontal Launch</u>

When an object is thrown horizontally with a speed v from a height h, it describes a curved path ruled by gravity until it eventually hits the ground.

The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:

vx=v

The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:

$v_y=g.t$

The horizontal component of the velocity is always the same:

$v_x=34 m/s$

The vertical component at t=5.5 s is:

$v_y=9.8*5.5=53.9$

$v_y=53.9\ m/s$

8 0

3 years ago

a) What is the average useful power output of a person who does6.00×10^6Jof useful work in 8.00 h? (b) Working at this rate, how

NARA [144]

Answer:

208.33 W

141.26626 seconds

Explanation:

E = Energy = $6\times 10^6\ J$

t = Time taken = 8 h

m = Mass = 2000 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Height of platform = 1.5 m

Power is obtained when we divide energy by time

$P=\frac{E}{t}\\\Rightarrow P=\frac{6\times 10^6}{8\times 60\times 60}\\\Rightarrow P=208.33\ W$

The average useful power output of the person is 208.33 W

The energy in the next part would be the potential energy

The time taken would be

$t=\frac{E}{P}\\\Rightarrow t=\frac{mgh}{208.33}\\\Rightarrow t=\frac{2000\times 9.81\times 1.5}{208.33}\\\Rightarrow t=141.26626\ s$

The time taken to lift the load is 141.26626 seconds

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4 years ago