What is the equivalent resistance for a parallel circuit that has two resistors: 18.0 ohms and 23.5 ohms?

Answers

Answer 1

Answer:
With only two resistors in parallel, it's relatively easy
to calculate their combined equivalent resistance.

   R = (their product) / (their sum)

   = (18Ω x 23.5Ω) / (18Ω + 23.5Ω)

   =         (423Ω²) / (41.5Ω)

   =    10.2 Ω

Related Questions

A proton moves along the x axis according to the equation x = 38 t + 14 t2, where x is in meters and t is in seconds. Calculate (a) the average velocity of the proton during the first 3.0 s of its motion, (b) the instantaneous velocity of the proton at t = 3.0 s, and (c) the instantaneous acceleration of the proton at t = 3.0 s.
2a. Abecomesjeep starts from it thestarts from it the state of rest. If itsvelocity 60m/s and its take 5 minute: what is the acceleraution of the jeepthe distanced travelled bythe jeep:
A long jumper leaves the ground at a 30 degree angle and travels 8.50 m. What is his take off speed?
A 6.0-newton force and an 8.0-newton force act concurrently on a point. As the angle between these forces increases from 0 degrees to 90 degrees, the magnitude of their resultant…(1) decreases (2) increases (3) remains the same
Electrons in excited hydrogen atoms are in then = 3 energy level. How many different photon frequencies could be emitted as the atoms return to the ground state? (1) 1 (3) 3 (2) 2 (4) 4

If you were standing at the center of curvature in front of a concave mirror, what image would be projected?A. The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are.

B.The image would be upright, would look shorter than you, and would be closer to the mirror than you are.

C.The image would be upside down, would look shorter than you, and would be closer to the mirror than you are.

D.The image would be upright, would look as tall you, and would be at the same distance from the mirror as you are.

Answers

The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are.

Answer:

A. The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are.

Explanation:

As we know by the mirror formula

$(1)/(d_i) + (1)/(d_o) = (1)/(f)$

now we know that object is placed at distance equal to the radius of mirror

$d_o = -R$

also we know that focal length of mirror is half of the radius of mirror

$f = (R)/(2)$

now we have

$(1)/(d_i) - (1)/(R) = -(2)/(R)$

so we have

$d_i = -R$

so image will be at same position as that the position of object and it is inverted in position so correct answer will be

A. The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are.

If we assume that 60 % of the kinetic energy delivered by a 1.80-kg hammer with a speed of 7.80 m/s is transformed into heat that flows into the nail and does not flow out, what is the temperature increase of an 8.00-g aluminum nail after it is struck ten times?

Answers

Answer:

45.6°C

Explanation:

Kinetic energy = $(1)/(2)mv^2$

Use m= 1.80kg and v=7.80m/s (mass and speed of hammer).

K = 0,5*1.80kg*(7.80m/s)^2 = K=54.8J

Heat is 60% of Kinetic energy. Q = 0.6*54.8J = 32.9J

As it is stuck 10 times the total heat is 10*32.9J = Total Heat = 329J

Use the equation $Q = mC_v \Delta T$ to find change of temperature:

$\Delta T = (Q)/(mC_v)$

Q = 329J; m = 8.00g of aluminium; C_v = 0.900J/g°C (For aluminium)

$\Delta T = (329J)/(8.00g*0.900J/g°C)$

Calculating gives Change of Temperature = 45.6°C

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 3.7 m/s2 for 4.4 seconds. It then continues at a constant speed for 8.3 seconds, before applying the brakes such that the carâs speed decreases uniformly coming to rest 216.0 meters from where it started. The yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop. 1. How fast is the blue car going 1.8 seconds after it starts? 2. How fast is the blue car going 10.0 seconds after it starts? 3. How far does the blue car travel before its brakes are applied to slow down? 4. What is the acceleration of the blue car once the brakes are applied? 5. What is the total time the blue car is moving? 6. What is the acceleration of the yellow car?

Answers

1. How fast is the blue car going 1.8 seconds after it starts?

Recall this kinematic equation:

Vf = Vi + aΔt

Vf is the final velocity.

Vi is the initial velocity.

a is the acceleration.

Δt is the amount of elapsed time.

Given values:

Vi = 0 m/s (the car starts at rest)

a = 3.7 m/s² (this is the acceleration between t = 0s and t = 4.4s)

Δt = 1.8 s

Substitute the terms in the equation with the given values and solve for Vf:

Vf = 0 + 3.7×1.8

Vf = 6.66 m/s

2. How fast is the blue car going 10.0 seconds after it starts?

The car stops accelerating after t = 4.4s and continues at a constant velocity for the next 8.3 seconds. This means the car is traveling at a constant velocity between t = 4.4s and t = 12.7s. At t = 10s the car is still traveling at this constant velocity.

We must use the kinematic equation from the previous question to solve for this velocity. Use the same values except Δt = 4.4s which is the entire time interval during which the car is accelerating:

Vf = 0 + 3.7×4.4

Vf = 16.28 m/s

The constant velocity at which the car is traveling at t = 10s is 16.28 m/s

3. How far does the blue car travel before its brakes are applied to slow down?

We must break down the car's path into two parts: When it is traveling under constant acceleration and when it is traveling at constant velocity.

Traveling under constant acceleration:

Recall this kinematic equation:

d = $(Vi+Vf)/(2)$ ×Δt

d is the distance traveled.

Vi is the initial velocity.

Vf is the final velocity.

Δt is the amount of elapsed time.

Given values:

Vi = 0 m/s (the car starts at rest).

Vf = 16.28 m/s (determined from question 2).

Δt = 4.4 s

Substitute the terms in the equation with the given values and solve for d:

d = $(0+16.28)/(2)$ ×4.4

d = 35.8 m

Traveling at constant velocity:

Recall the relationship between velocity and distance:

d = vΔt

d is the distance traveled.

v is the velocity.

Δt is the amount of elapsed time.

Given values:

v = 16.28 m/s (the constant velocity from question 2).

Δt = 8.3 s (the time interval during which the car travels at constant velocity)

Substitute the terms in the equation with the given values:

d = 16.28×8.3

d = 135.1 m

Add up the distances traveled.

d = 35.8 + 135.1

d = 170.9 m

4. What is the acceleration of the blue car once the brakes are applied?

Recall this kinematic equation:

Vf²=Vi²+2ad

Vf is the final velocity.

Vi is the initial velocity.

a is the acceleration

d is the distance traveled.

Given values:

Vi = 16.28 m/s

Vf = 0 m/s

d = 216 m - 170.9 m = 45.1 m (subtracting the distance already traveled from the total path length)

Substitute the terms in the equation with the given values and solve for a:

0² = 16.28²+2a×45.1

a = -2.94 m/s²

5. What is the total time the blue car is moving?

We already know the time during which the car is traveling under constant acceleration and traveling at constant velocity. We now need to solve for the amount of time during which the car is decelerating.

Recall again:

d = $(Vi+Vf)/(2)$ ×Δt

Given values:

d = 45.1 m

Vi = 16.28 m/s (the velocity the car was traveling at before hitting the brakes).

Vf = 0 m/s (the car slows to a stop).

Substitute the terms in the equation with the given values and solve for Δt:

45.1 = $(16.28+0)/(2)$ ×Δt

Δt = 5.54s

Add up the times to get the total travel time:

t = 4.4 + 8.3 + 5.54 =

t = 18.24s

6. What is the acceleration of the yellow car?

Recall this kinematic equation:

d = ViΔt + 0.5aΔt²

d is the distance traveled.

Vi is the initial velocity.

a is the acceleration.

Δt is the amount of elapsed time.

Given values:

d = 216 m (both cars meet at 216m)

Vi = 0 m/s (the car starts at rest)

Δt = 18.24 s (take the same amount of time to reach 216m)

Substitute the terms in the equation with the given values and solve for a:

216 = 0×18.24 + 0.5a×18.24²

a = 1.3 m/s²

What is the ultimate source of electromagnetic waves?A. Vibrating atoms
B. Vibrating molecules
C. Radio sets
D. Vibrating charged particles
E. TV antennas

Answers

The ultimate source of electromagnetic waves are the radio sets. The answer is letter C. One example of an electromagnetic radiation is the visible light. And visible light can be radio waves, infrared light and X - rays. The rays of the sun are a form of visible light. It has an electromagnetic radiation of UV (ultra violet) rays. That is why the radiation at day is greater than at day due to sun’s rays.

Like magnetic poles will repel each other, while opposite magnetic poles will attract each other true or false