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slega [8]
3 years ago
10

3. A 10-centimeter diameter solid sphere made of a conducting material has 10 micro-Coulombs of charge placed upon it. What is t

he potential difference between a point on one side of the sphere to a point on the exact opposite side of the sphere
Physics
1 answer:
bazaltina [42]3 years ago
4 0

Answer:

zero

Explanation:

For a solid conducting sphere, charges are present on the surface of the sphere due to a phenomenon known as electrostatic sheilding. This affects the charge present in the body and makes it zero. However, the electrostatic potential appears to be equal to the whole present point that shows on the surface. The surface of a spherical conducting solid sphere is known as an equipotential surface. Thus, the potential difference between the two opposite points on the surface of the sphere will also be zero.

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A 48.0-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocit
Kipish [7]

Answer:

F = 47.6 N

Explanation:

  • Newton's 2nd law can be expressed as the rate of change of the total momentum, respect of time, as follows:

       F = \frac{\Delta p}{\Delta t}

  • So, in order to find the average force exerted by the skater on the wall, we can find the change in momentum due to the force exerted by the wall (which is equal and opposite to the one exerted by the skater), and  divide it by the time interval , as follows:

       F_{wall} = \frac{\Delta p}{\Delta t} =\frac{(48.0 kg*(-1.06m/s)}{1.07s} = -47.6 N

       ⇒ Fsk = 47.6 N (normal to the wall)

3 0
3 years ago
In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole.
Fiesta28 [93]

Answer:

6.0 m/s

Explanation:

According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.

Therefore, we can write:

KE_i+PE_i =KE_f+PE_f

or

\frac{1}{2}mu^2+0=\frac{1}{2}mv^2+mgh

where:

m is the mass of the athlete

u is the initial speed of the athlete (at the bottom)

0 is the initial potential energy of the athlete (at the bottom)

v = 0.80 m/s is the final speed of the athlete (at the top)

g=9.8 m/s^2 is the acceleration due to gravity

h = 1.80 m is the final height of the athlete (at the top)

Solving the equation for u, we find the initial speed at which the athlete must jump:

u=\sqrt{v^2+2gh}=\sqrt{0.80^2+2(9.8)(1.80)}=6.0 m/s

4 0
3 years ago
A spherical balloon with radius r inches has volume V(r) = 4 3 πr3. Find an expression that represents the amount of air require
DerKrebs [107]
So the question ask to formulate an expression to represent the amount of air required to inflate the balloon from a radius of R+4 and the expression would be V=16pi/3*(3r^2+12r+16). I hope you are satisfied with my answer and feel free to ask for more if you have question and further clarifications
4 0
3 years ago
________ is an arrhythmia in which there is a very fast but regular rhythm (250 beats per minute) of the atria or ventricles.
erastova [34]

Answer:

Flutter

Explanation:

Flutter is a type of arrhythmia that causes very fast and regular ryth of the atria of about 250 beats per minute.

Arrhythmia can be defined as any sort of irregularity heart rate or rhythm is also called as dysrhythmia.

Arrhythmias can be categorized as heart block, bradycardia, tachycardia, fibrillation, flutter, sick sinus syndrome, and is diagnosed by Electrocardiography.

In Flutter, the heart chambers do get sufficient time to get filled with blood completely prior to next contraction.

4 0
3 years ago
A capacitor consists of two metal surfaces separated by an insulating layer. A new capacitor has no charge on either of its surf
mihalych1998 [28]
<h2>Answer:</h2><h3>(A) the positively charged surface increases and the energy stored in the capacitor increases.</h3>

When charging a capacitor transferring charge from one surface to the other, the first surface becomes negatively charged while the second surface becomes positively charged. As you transfer the charge, the voltage of the positively charged surface increases and the energy stored in the capacitor also increases. We can solve this by the definition of <em>capacitance</em><em> </em>that is <em>a  measure of the ability of a capacitor  to store energy. </em>For any capacitor, the capacitance is  a constant defined as:

C=\frac{Q}{V_{ab}}

To maintain C constant, if Q increases V also increases.

On the other hand, the potential energy U can be expressed as:

U=\frac{Q^{2}}{2C}

In conclusion, as Q increases the potential energy also increases.

5 0
3 years ago
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