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3 years ago

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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A physics student shoves a 0.50-kg block from the bottom of a frictionless 30.0° inclined plane. The student performs 4.0 j of w

Musya8 [376]

Answer:1.63 m

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30^{\circ}$

Amount of work done $W=4 J$

block slides a distance s along the Plane

Work done =change in Potential Energy

Increase in height of block is $s\sin \theta$

Change in Potential Energy $=mg(\sin \theta -0)$

$\Delta P.E.=0.5\times 9.8\times s\sin 30$

$4=0.5\times 9.8\times s\sin 30$

$s=\frac{4}{2.45}$

$s=1.63 m$

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3 years ago

During which type of collision do the two objects stick together?

inysia [295]

Answer:

D. Perfectly inelastic

Explanation:

Kinetic energy is lost so the two bodies stick together.

6 0

3 years ago

What are tides? the regular daily rises and falls in sea level caused by the gravitational attraction of the Moon and Sun on Ear

klemol [59]

Answer:

the regular daily rises and falls in sea level caused by the gravitational attraction of the Moon and Sun on Earth.

Explanation:

Tides can be defined as the rise and fall of water level in water bodies such as lakes and oceans due to the gravitational force of attraction exerted by the moon on earth. The side closest to the moon creates a bulge of water known as high tide. Low tides are generally experienced when a sea level is not within the bulge.

Generally, the gravitational pull of the Moon cause visible changes on planet Earth's surface.

This ultimately implies that, the pull of the Moon's gravity causes high and low tides on planet Earth's surface.

The various types of ocean tides based on the position of the Earth, Moon and the Sun are;

I. Neap tides.

II. Spring tides.

III. Low tide.

IV. High tide.

V. Brown tide.

VI. Rip tide.

VII. Red tide.

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3 years ago

A type O star is likely to appear

Ludmilka [50]

A type O star is likely to appear blue in color.

The blue type-O stars are solely thirty to fifty times a lot of large than the sun. However, O stars burn 1,000,000 times brighter and it has a very short lifespan. O stars solely last a couple of million years before they die in spectacular star explosions.

7 0

3 years ago

5) A 20.0 kg cart with no friction wheels sits on a table. A light string is attached to it and runs over a low friction pulley

ella [17]

Answer:

1) Please find attached, created with Microsoft Visio

2) The acceleration of the masses connected by the light string is 0.00735 m/s²

3) The tension in the cord is 0.147 N

4) The time it would take the block to go 1.2 m to the edge of the table is approximately 18.07 s

5) The velocity of the cart as soon as it gets to the edge of the table is 0.042 m/s

Explanation:

1) Please find attached, the required free body diagram, showing the tension, weight and frictional (zero friction) forces acting on the cart and the mass created with Microsoft Visio

2) The acceleration of the masses connected by the light string is given as follows;

F = Mass, m × Acceleration, a

The mass of the truck, M = 20.0 kg

The mass attached to the string, hanging rom the pulley, m = 0.0150 kg

The force, F acting on the system = The pulling force on the cart = The tension on the cable = The weight of the hanging mass = 0.0150 × 9.8 = 0.147 N

The pulling force acting on the cart, F = M × a

∴ F = 0.147 N = 20.0 kg × a

a = 0.147 N/(20.0 kg) = 0.00735 m/s²

The acceleration of the truck = a = 0.00735 m/s²

3) The tension in the cord = F = 0.147 N

4) The time, t, it would take the block to go 1.2 m to the edge of the table is given by the kinematic equation, s = u·t + 1/2·a·t²

Where;

s = The distance to the edge of the table = 1.2 m

u = The initial velocity = 0 m/s (The cart is assumed to be initially at rest)

a = The acceleration of the cart = 0.00735 m/s²

t = The time taken

Substituting the known values, gives;

s = u·t + 1/2·a·t²

1.2 = 0 × t + 1/2 ×0.00735 × t²

1.2 = 1/2 ×0.00735 × t²

t² = 1.2/(1/2 ×0.00735) ≈ 326.5306

t = √(1.2/(1/2 ×0.00735)) ≈ 18.07

The time it would take the block to go 1.2 m to the edge of the table = t ≈ 18.07 s

5) The velocity, v, of the cart as soon as it gets to the edge of the table is given by the kinematic equation, v² = u² + 2·a·s as follows;

v² = u² + 2·a·s

u = 0 m/s

v² = 0² + 2 × 0.00735 × 1.2 = 0.001764

v = √(0.001764) = 0.042

The velocity of the cart as soon as it gets to the edge of the table = v = 0.042 m/s.

7 0

3 years ago

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