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Vinil7 [7]
3 years ago
8

Charge q1 is placed a distance r0 from charge q2 . What happens to the magnitude of the force on q1 due to q2 if the distance be

tween them is reduced to r0/4 ?
Physics
1 answer:
alexdok [17]3 years ago
8 0

Answer:

The magnitude of the force between the two charges will be increased.

Explanation:

The force between the two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Hence, if distance between charges is halved (charges remaining kept constant), the force between the two charges is quadrupled

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As speed increases, how does the potential, kinetic, and total energy levels change?
Anastaziya [24]
Potential equals kenecric at the bottom so potential would also increas
4 0
3 years ago
A sheet of metal is 2mm wide 10cm tall and 15cm long. it was 4g. what is the density? <br>​
Hoochie [10]

Answer:

Ro = 133 [kg/m³]

Explanation:

In order to solve this problem, we must apply the definition of density, which is defined as the relationship between mass and volume.

Ro = m/V

where:

m = mass [kg]

V = volume [m³]

We will convert the units of length to meters and the mass to kilograms.

L = 15 [cm] = 0.15 [m]

t = 2 [mm] = 0.002 [m]

w = 10 [cm] = 0.1 [m]

Now we can find the volume.

V = 0.15*0.002*0.1\\V = 0.00003 [m^{3} ]

And the mass m = 4 [gramm] = 0.004 [kg]

Ro = 0.004/0.00003\\Ro = 133 [kg/m^{3}]

3 0
2 years ago
A horizontal line labeled B has an arrow labeled A strike it from right and above and then another arrow D emerges from the stri
patriot [66]

Answer:

c is the actual answer.

Explanation:

7 0
2 years ago
Read 2 more answers
What quantity is the rate of change of velocity? Displacement Acceleration Final velocity
MariettaO [177]

Answer:

Acceleration

Explanation:

The quantity of the rate of change of velocity is termed the acceleration of the body.

Acceleration is the rate of change of velocity with time;

  A  = \frac{v - u}{t}  

A is the acceleration

v is the final velocity

u is the initial velocity

t is the time taken

 

7 0
2 years ago
A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A pie
liubo4ka [24]

Answer:

Explanation:

Capacitance of the capacitor = 13.5μF

Voltage across plate is 24V

Dielectric constant k=3.55.

a. Energy in capacitor is given by

E=1/2CV^2

We want to calculate energy without the dielectric substance

Given that C=13.5 μF and V=24V

The capacitance give is with dielectric so we need to remove it

C=kCo

Co=C/k

Then the Co=13.5μF/3.55

Co=3.803μF

Then

E=(1/2)×3.803×10^-6×24^2

E=1.1×10^-3J

E=1.1mJ

b. Energy in capacitor is given by

E=1/2CV^2

The capacitance given is with a dielectric, so we are going to apply it direct.

Given that C=13.5 μF and V=24V

Then

E=(1/2)×13.5×10^-6×24^2

E=3.89×10^-3J

E=3.9mJ

c. The energy without dielectric is 1.1mJ and the energy with dielectric is 3.9mJ

The energy increase when the dielectric material is added

d. Dielectrics in capacitors serve three purposes: to keep the conducting plates from coming in contact, allowing for smaller plate separations and therefore higher capacitances;

Therefore, Since dielectric allow higher capacitance, and energy of a capacitor is directly proportional to the capacitance, then the higher the capacitance the higher the energy.

6 0
3 years ago
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