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rewona [7]
3 years ago
11

Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.

Physics
1 answer:
Drupady [299]3 years ago
5 0

Answer:

14.57 ohms

Explanation:

Here in the figure ,Rb & R₄are in series  & also  Rc & R₅ are in series. As they are in series , ( Rb + R₄ ) & (Rc & R₅) are in parallel . So the equivalent resistance in that branch = ( 2 + 18 ) ║ ( 3 + 12 )

                                          = 20 ║ 15

                                          = (20×15) / (20 + 15)

                                          = 8.57 ohms

Also Ra ( 6 ohm ) is in series with that branch ,. So the equivalent resistance of the whole circuit = 8.57 + 6 = 14.57 ohms.

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Write down two advantages of parallel combination ​
Simora [160]

Answer:

In parallel combination each appliance gets the full voltage.

If one appliance is switched on/of others are not affected.

The parallel circuit divide the current through the appliances.

In a parallel combination it is very easy to connect or disconnect a new appliance without affecting the working of other appliances.

pls mark me as brainlist

Explanation:

8 0
3 years ago
baseball is hit into the air at an initial speed of 37.2 m/s and an angle of 49.3 ° above the horizontal. At the same time, the
Agata [3.3K]

Answer:

The average speed of the fielder is 5.24 m/s

Explanation:

The position vector of the ball after it was hit can be calculated using the following equation:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem.

When the ball is caught, its position vector will be (see r1 in the figure):

r1 = (r1x, 0.873 m)

Then, using the equation of the position vector written above:

r1x = x0 + v0 · t · cos α

0.873 m = y0 + v0 · t · sin α + 1/2 · g · t²

Since the frame of reference is located at the point where the ball was hit, x0 and y0 = 0. Then:

r1x = v0 · t · cos α

0.873 m = v0 · t · sin α + 1/2 · g · t²

Let´s use the equation of the y-component of r1 to obtain the time of flight of the ball:

0.873 m = 37.2 m/s · t · sin 49.3° - 1/2 · 9.8 m/s² · t²

0 = -0.873 m + 37.2 m/s · t · sin 49.3° - 4.9 m/s² · t²

Solving the quadratic equation:

t = 0.03 s and t = 5.72 s.

It would be impossible to catch the ball immediately after it is hit at t = 0.03 s. Besides, the problem says that the ball was caught on its way down. Then, the time of flight of the ball is 5.72 s.

With this time, we can calculate r1x which is the horizontal distance traveled by the ball from home:

r1x = v0 · t · cos α

r1x = 37.2 m/s · 5.72 s · cos 49.3°

r1x = 1.39 × 10² m

The distance traveled by the fielder is (1.39 × 10² m - 1.09 × 10² m) 30.0 m.

The average velocity is calculated as the traveled distance over time, then:

average velocity = treveled distance / elapsed time

average velocity = 30.0 m / 5.72 s = 5.24 m/s

8 0
3 years ago
Which of these is an example of a mechanical wave
anastassius [24]
Some of the most common examples of mechanical waves are water waves, sound waves, and seismic waves. There are three types of mechanical waves: transverse waves, longitudinal waves, and surface waves.
6 0
3 years ago
Calculate the acceleration of a car if it's velocity increases from 15 m/s to 75m/s in 5 seconds. ​
Andreas93 [3]
  • initial velocity=u=15m/s
  • Final velocity=v=75m/s
  • Time=t=5s

\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \sf\longmapsto Acceleration=\dfrac{75-15}{5}

\\ \sf\longmapsto Acceleration=\dfrac{60}{5}

\\ \sf\longmapsto Acceleration=12m/s^2

7 0
3 years ago
A force of 5N and a force of 8N act to the same point and are inclined at 45degree to each other. Find the magnitude and directi
Alex_Xolod [135]
  • Magnitude: 12.1 N.
  • Direction: 17.0° to the 8 N force.
<h3>Explanation</h3>

Refer to the diagram attached (created with GeoGebra). Consider the 5 N force in two directions: parallel to the 8 N force and normal to the 8 N force.

  • \displaystyle F_{\text{1, Parallel}} = F_1 \cdot \cos{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}.
  • \displaystyle F_{\text{1, Normal}} = F_1 \cdot \sin{45^\textdegree} = \dfrac{5\sqrt{2}}{2}\;\text{N}.

The sum of forces on each direction will be the resultant force on that direction:

  • Resultant force parallel to the 8 N force: (8 + \dfrac{5\sqrt{2}}{2})\;\text{N}.
  • Resultant force normal to the 8 N force: \dfrac{5\sqrt{2}}{2}\;\text{N}.

Apply the Pythagorean Theorem to find the magnitude of the resultant force.

\displaystyle \Sigma F = \sqrt{{(8 + \dfrac{5\sqrt{2}}{2})}^2 + {(\dfrac{5\sqrt{2}}{2})}^2} = 12.1\;\text{N} (3 sig. fig.).

The size of the angle between the resultant force and the 8 N force can be found from the tangent value of the angle. Tangent of the angle:

\displaystyle \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}} = \dfrac{8 + \dfrac{5\sqrt{2}}{2}}{\dfrac{5\sqrt{2}}{2}} \approx 0.306491.

Find the size of the angle using inverse tangent:

\displaystyle \arctan{ \dfrac{\Sigma F_\text{Normal}}{\Sigma F_\text{Parallel}}} = \arctan{0.306491} = 17.0\textdegree.

In other words, the resultant force is 17.0° relative to the 8 N force.

4 0
3 years ago
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