All categories

4 years ago

When one thing in a _____ circuit stops working, the whole circuit stops working

Physics

1 answer:

vodka [1.7K]4 years ago

4 0

Answer:

When one thing in a <u>electric </u>circuit stops working, the whole circuit stops working

Explanation:

You might be interested in

A 80 g particle is moving to the left at 22 m/s . How much net work must be done on the particle to cause it to move to the righ

AysviL [449]

Answer:

W= 38.4 J

Explanation:

Given that

m = 80 g= 0.08 kg

Initial speed ,u= 22 m/s

Final speed ,v= 38 m/s

The change in the kinetic energy of the particle

$\Delta KE=\dfrac{1}{2}m(v^2-u^2)$

$\Delta KE=\dfrac{1}{2}\times 0.08\times (38^2-22^2)\ J$

ΔKE= 38.4 J

We know that

Work done by all the forces =Change in the kinetic energy

That is why net work done = 38.4 J.

W= 38.4 J

Therefore the answer will be 38.4 J.

8 0

3 years ago

An object of mass m is dropped from height h above a planet of mass M and radius R .

Margarita [4]

Answer:

$v = \sqrt{2GM(\frac{1}{R+h)}-\frac{1}{R})}$

Explanation:

The initial mechanical energy of the object, when it is located at height h above the the planet, is just gravitational potential energy:

$E=U=\frac{GMm}{(R+h)}$

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the object

R is the radius of the planet

h is the altitude of the object

When the object hits the ground, its mechanical energy will sum of potential energy and kinetic energy:

$E=\frac{GMm}{R}+\frac{1}{2}mv^2$

where

v is the speed of the object at the ground

Since the mechanical energy is conserved, we can write

$\frac{GMm}{R}+\frac{1}{2}mv^2=\frac{GMm}{R+h}$

and solving for v, we find

$\frac{GMm}{R}+\frac{1}{2}mv^2=\frac{GMm}{R+h}\\v = \sqrt{2GM(\frac{1}{R+h)}-\frac{1}{R})}$

3 0

3 years ago

A ball is thrown vertically into the air and when it returns after an interval of 2 seconds, it is caught. Which one of the foll

baherus [9]

Answer:

Last statement option: "The acceleration after it leaves the hand is 10 m/s/s downwards."

Explanation:

At every instant of its motion, the ball is under the effects of the acceleration due to gravity (assumed to be 10 m/s^2). This is true at whatever altitude the ball is. The acceleration due to gravity is always pointing down (not up).

In the absence of air resistance, the motion is described kinematically by a parabola with the branches pointing down as a function of time (motion under constant acceleration), with the vertex indicating the maximum altitude the ball reaches. Both branches (representing motion upwards and downwards) are equidistant from the vertex, so the time going up equals the time coming down.

Therefore, the only statement option that is correct is the last one: "The acceleration after it leaves the hand is 10 m/s/s downwards."

8 0

4 years ago

How is resistance affected as more branches are added to a parallel circuit?

Olenka [21]

Answer:

The total resistance decreases

Explanation:

The total resistance of a parallel circuit is given by:

$\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}+...$

where R1, R2, etc. are the individual resistances.

We notice that if we add one more resistor in parallel, we add another positive term $\frac{1}{R}$ to the sum, therefore the term $\frac{1}{R_T}$ increases. However, this term is the reciprocal of the total resistance: therefore, this means that the total resistance will decrease.