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Virty [35]
4 years ago
12

The capacitance of A conductor is affected by the presence of A second conductor that is uncharged and isolated electrically. Wh

y?
Physics
1 answer:
Komok [63]4 years ago
5 0
What happens is the potential value of the conductor decreases due to the presence of second conductor
as the capacitance is given by C = q/v
the value of v deceases as v-v1
thus the new capacitance is = C' = q/v-v1 thus the lowering of v increases the capacitance
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Simple Circuit and Ohm's Law Check-for-Understanding
Vanyuwa [196]

Answer:

current, only

Explanation:

current:I

voltage:U

resistance:R

formula: I=U/R

Increasing the battery cause the increasing in the voltage. Resistance does not normally change. And the current would increase.

4 0
3 years ago
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What is meant by an acceleration of negative 2metre per second square​
soldi70 [24.7K]

means that a body is in motion, and its velocity is measured in meters per second. And, that velocity is increasing by two meters per second, every second.

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3 years ago
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An electric dipole is formed from ±1.00nC charges spaced 3.00 mm apart. The dipole is at the origin, oriented along the x-axis.
Ronch [10]

Answer:

Value of electric field along the axis and equitorial axis  E=31.25\ N/c and E = 15.625\ N/c respectively.

Explanation:

Given :

Distance between charges , d = 3 \ mm =\dfrac{3}{1000}\ m=3\times 10^{-3}\ m.

Magnitude of charges , q=1\ nC = 10^{-9}\ C.

Dipole moment , p=qL=10^{-9}\times 3\times 10^{-3}=3\times 10^{-12} \ C\ m.

Case A) (x,y) = (12.0 cm, 0 cm) :

Electric field of dipole in its axis ,

E=\dfrac{2kp}{r^3}

Putting all values and r=12\times 10^{-2}\ m.

We get , E=31.25\ N/c.

Case B) (x,y) = (0 cm, 12.0 cm) :

Electric field of dipole on equitorial axis ,

E = \dfrac{kp}{r^3}

Putting all values and r=12\times 10^{-2}\ m.

We get , E = 15.625\ N/c.

Hence , this is the required solution.

7 0
3 years ago
1pt A cannon fires a 5-kg ball horizontally from a
Klio2033 [76]

Answer: Both cannonballs will hit the ground at the same time.

Explanation:

Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.

then the acceleration equation is only on the vertical axis, and can be written as:

a(t) = -(9.8 m/s^2)

Now, to get the vertical velocity equation, we need to integrate over time.

v(t) = -(9.8 m/s^2)*t + v0

Where v0 is the initial velocity of the object in the vertical axis.

if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s

and:

v(t) = -(9.8 m/s^2)*t

Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)

And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.

You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)

7 0
3 years ago
Assuming that Albertine's mass is 60.0 kg , for what value of μk, the coefficient of kinetic friction between the chair and the
makvit [3.9K]

Answer:

\mu_k=0.101

Explanation:

It is given that,

Mass of Albertine, m = 60 kg

It can be assumed, the spring constant of the spring, k = 95 N/m

Compression in the spring, x = 5 m

A glass sits 19.8 m from her outstretched foot, h = 19.8 m

When she just reach the glass without knocking it over, a force of friction will also act on it. Using the conservation of energy for the spring mass system such that,

\dfrac{1}{2}kx^2=\mu_k mgh

\mu_k=\dfrac{kx^2}{2mgh}

\mu_k=\dfrac{95\times (5)^2}{2\times 60\times 9.8\times 19.8}

\mu_k=0.101

So, the coefficient of kinetic friction between the chair and the waxed floor is 0.101. Hence, this is the required solution.

3 0
3 years ago
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