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Orlov [11]
3 years ago
15

A resistor and a capacitor are connected in series across an ideal battery having a constant voltage across its terminals. (a) A

t the moment contact is made with the battery the voltage across the capacitor is
Physics
1 answer:
maks197457 [2]3 years ago
6 0

Answer:

(a) D.  Zero.

(b) C.  Equal to the battery's terminal voltage.

Explanation:

The question is incomplete, see the complete question for your reference and information.

A resistor and a capacitor are connected in series across an ideal battery having a constant voltage

across its terminals. At the moment contact is made with the battery

(a) the voltage across the capacitor is

A) equal to the battery's terminal voltage.

B) less than the battery's terminal voltage, but greater than zero.

C) equal to the battery's terminal voltage.

D) zero.

(b) the voltage across the resistor is

A) equal to the battery's terminal voltage.

B) less than the battery's terminal voltage, but greater than zero.

C) equal to the battery's terminal voltage.

D) zero

A RC circuit is a circuit that is composed of both resistors and capacitors connect to a source of current or voltage.

basically when a voltage source is applied to an RC circuit, the capacitor, C charges up through the resistance, R

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A solid sphere has a radius of 0.200 m and a mass of 150.0 kg. how much work is required to get the sphere rolling with an angul
Allisa [31]

Here in this case we can use work energy theorem

As per work energy theorem

Work done by all forces = Change in kinetic Energy of the object

Total kinetic energy of the solid sphere is ZERO initially as it is given at rest.

Final total kinetic energy is sum of rotational kinetic energy and translational kinetic energy

KE = \frac{1}{2}Iw^2 +\frac{1}{2} mv^2

also we know that

I = \frac{2}{5}mR^2

w= \frac{v}{R}

Now kinetic energy is given by

KE = \frac{1}{2}(\frac{2}{5}mR^2)w^2 +\frac{1}{2} m(Rw)^2

KE = \frac{1}{5}mR^2w^2 +\frac{1}{2} mR^2w^2

KE = \frac{7}{10}mR^2w^2

KE = \frac{7}{10}*150*(0.200)^2(50)^2

KE = 10500 J

Now by work energy theorem

Work done = 10500 - 0 = 10500 J

So in the above case work done on sphere is 10500 J

7 0
3 years ago
Which one of these Target games uses the underhand throw?
vesna_86 [32]

Answer:

Cornhole

Explanation:

4 0
2 years ago
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The equation for the conservation of mechanical energy is:
marusya05 [52]

Answer:

The answer is A because the equation is  KEi+PEi=KEf+PEf

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7 0
2 years ago
What change will always result in an increase in gravitational force between two objects
tekilochka [14]

The gravitational force between two object depends on their masses and on their distance.


Since the formula is


F = G\frac{m_1m_2}{d^2}


If the masses grow, the force also grows. But I'm assuming the two objects are fixed, so you can't enlarge their mass.


So, the only option remaining is to lower their distance: since it sits at the denominator, a smaller value of d results in a bigger value for F.


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3 years ago
An acorn falls from a tree and hits the ground in 0.8 s. How far did the acorn fall . Use g = 9.8 m/s^2. Round your final result
mylen [45]

The distance covered by the acorn is 3.136 m.

<u>Explanation:</u>

The time taken for the acorn to hit the ground is 0.8 s. As it is a free fall, the acorn will be completely under the influence of gravity. So the acceleration will be acceleration due to gravity.

Then using the second law of equation,

s=ut+\frac{1}{2}gt^{2}

Since the initial velocity and time is zero, then the time taken to reach the ground is stated as 0.8 s, so

   s=0+\left(\frac{1}{2} \times 9.8 \times 0.8 \times 0.8\right)=\frac{6.272}{2}=3.136 \mathrm{m}

So the distance covered by the acorn is 3.136 m.

8 0
3 years ago
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