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algol [13]
3 years ago
15

The constant of proportionality between charge and voltage is the:________.

Physics
1 answer:
Brut [27]3 years ago
4 0

Answer:

The constant of proportionality between charge and voltage is the capacitance.

Explanation:

The capacitance (C), measured in farads, is the ratio of charge to voltage, that is:

C = \frac{Q}{V}

Where:

Q - Charge, measured in coulombs.

V - Voltage, measured in volts.

If charge is clear, it is easy to concluded that capacitante is a constant of direct proportionality.

Hence, the correct answer is B.

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Is distance traveled by a moving train qualitative discrete continuous or na?
butalik [34]
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8 0
3 years ago
A block with mass 0.5 kg is forced against a horizontal spring of negligible mass, compressing the spring a distance of 0.2 m. W
lianna [129]

Answer:

So coefficient of kinetic friction will be equal to 0.4081

Explanation:

We have given mass of the block m = 0.5 kg

The spring is compressed by length x = 0.2 m

Spring constant of the sprig k = 100 N/m

Blocks moves a horizontal distance of s = 1 m

Work done in stretching the spring is equal to W=\frac{1}{2}kx^2=\frac{1}{2}\times 100\times 0.2^2=2J

This energy will be equal to kinetic energy of the block

And this kinetic energy must be equal to work done by the frictional force

So \mu mg\times s=2

\mu\times  0.5\times 9.8\times 1=2

\mu =0.4081

So coefficient of kinetic friction will be equal to 0.4081

5 0
3 years ago
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8 0
3 years ago
A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of thebuilding. Ignore air resis
Bumek [7]

Answer:

a)3.5s

b)28.57m/S

c)34.33m/S

d)44.66m/S

Explanation:

Hello!

we will solve this exercise numeral by numeral

a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration

Y= VoT+0.5gt^{2}

where

Vo = Initial speed =0

T = time

g=gravity=9.81m/s^2

y = height=60m

solving for time

Y=0.5gt^2\\t=\sqrt{\frac{Y}{0.5g} } \\t=\frac{60}{0.5(9.81)} \\

T=3.5s

b)The horizontal speed remains constant since there is no horizontal acceleration. with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed

V=\frac{x}{t} =\frac{100}{3.5}=28.57m/s

c)

to find the final vertical velocity we use the equations for motion with constant velocity as follows

Vf=Vo+g.t    

Vf=0+(9.81 )(3.5)=34.335m/S          

d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares

V=\sqrt{Vx^2+Vy^2} =\sqrt{34.33^2+28.57^2} =44.67m/S

7 0
3 years ago
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