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3 years ago

What quantities are related by Ohm's law? Check all that apply. voltage conductivity current resistance insulation

Physics

2 answers:

IrinaK [193]3 years ago

8 0

Answer:

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░░░░░░░▀███▀█░▄░░ Over Brainly

Explanation:

Free_Kalibri [48]3 years ago

7 0

Answer: Current, resistance and voltage are the quantities which are related by Ohm's law.

Explanation:

A law which states that electric current is directly proportional to voltage and inversely proportional to resistance is called Ohm's law.

Mathematically, it is represented as follows.

$I = \frac{V}{R}$

where,

I = current

V = voltage

R = resistance

This means that the quantities related by Ohm's law include current, voltage and resistance.

Thus, we can conclude that current, resistance and voltage are the quantities which are related by Ohm's law.

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Five moles of an ideal monatomic gas with an initial temperature of 121 ∘C expand and, in the process, absorb an amount of heat

liq [111]

Answer:

T₂ ≈ 107.85∘C

Explanation:

The question didn't state if the volume is constant or not as such, we can apply the first law of thermodynamic

From the first law of thermodynamic,

ΔU = Q - W

where ΔU = Internal Energy, Q = Quantity of heat absorbed, W = Amount of work done.

Q = 1200 J and W = 2020 J

∴ ΔU = 1200 -2020 = -820 J.

Using the ideal gas equation,

ΔU = 3/2nRΔT...................................equation 1

where n = number of moles, R = Molar gas constant, ΔT = Change in temperature = (T₂ - T₁).

Modifying equation 1,

ΔU = 3/2nR(T₂ -T₁)...............................equation 2.

making T₂ the subject of the relation in equation 2,

T₂ = {2/3(ΔU)/nR}+T₁........................ equation 3

where T₁=121∘C, R= 8.314 J / mol, n=5 moles, ΔU=-820 J

Substituting these values into equation 3,

∴ T₂ ={ 2/3(-820)/(5×8.314)}+121

T₂ = {2×(-820)/ (3×5×8.314)}+121

T₂={-1640/124.71}+ 121

T₂ = {-13.151} + 121

∴T₂ = 121 - 13.151 = 107. 849∘C

T₂ ≈ 107.85∘C

3 0

3 years ago

What did bohr contribute to modern atomic theory

ddd [48]

Bohr’s model showed that electron orbits had distinct radii.

5 0

3 years ago

Read 2 more answers

If you speed up from the rest to 12m/s in 3 seconds, what is your acceleration?

azamat

I believe that your acceleration would be 4 m/s

7 0

3 years ago

When reading a digital volt-ohmmeter (DVOM), you have a reading of 2168 mV, which is the same as:__________

levacccp [35]

Answer:

D, both A and B

Explanation:

2168 mV is the SI unit for potential difference and the Voltmeter.

The primary unit is Volt, represented as V. Due to the fact that there can be a much higher reading, or an even much more smaller one, comes the need for variants of the same unit.

10^-3 is called milli and represented as m

10^3 is called killo and represented as k

10^-6 is called micro and represented as µ

10^6 is called mega and represented as M

and even much higher variants of up to 10^12 and 10^-12

As we can see from the aforementioned example, 10^-3 is milli and represented as m

And our question gave us the unit in mV, which stands for millivolts.

Also, if we look at option B, it states, 2.168 volts. This 2.168 volts is also the same thing as A. Take a look at it this way, I said mV is 10^-3, right?

So, 2168*10^-3 is also 2168/100 which is 2.168. The only difference here is, once we make this conversion from mV, we have to drop the milli tag, because we have already made a conversion, and thus, leave it as V.

2168 mV = 2.168V

Hence why we picked option D, Both A & B as the right one

4 0

3 years ago

Physics Kinematics question

just olya [345]

An interesting problem, and thanks to the precise heading you put for the question.

We will assume zero air resistance.
We further assume that the angle with vertical is t=53.13 degrees, corresponding to sin(t)=0.8, and therefore cos(t)=0.6.

Given:
angle with vertical, t = 53.13 degrees
sin(t)=0.8; cos(t)=0.6;
air-borne time, T = 20 seconds
initial height, y0 = 800 m

Assume g = -9.81 m/s^2

initial velocity, v m/s (to be determined)

Solution:

(i) Determine initial velocity, v.
initial vertical velocity, vy = vsin(t)=0.8v
Using kinematics equation,
S(T)=800+(vy)T+(1/2)aT^2 ....(1)
Where S is height measured from ground.

substitute values in (1): S(20)=800+(0.8v)T+(-9.81)T^2 =>
v=((1/2)9.81(20^2)-800)/(0.8(20))=72.625 m/s for T=20 s

(ii) maximum height attained by the bomb
Differentiate (1) with respect to T, and equate to zero to find maximum
dS/dt=(vy)+aT=0 =>
Tmax=-(vy)/a = -0.8*72.625/(-9.81)= 5.9225 s

Maximum height,
Smax
=S(5.9225)
=800+(0.8*122.625)*(5.9225)+(1/2)(-9.81)(5.9225^2)
= 972.0494 m

(iii) Horizontal distance travelled by the bomb while air-borne
Horizontal velocity = vx = vcos(t) = 0.6v = 43.575 m/s
Horizontal distace travelled, Sx = (vx)T = 43.575*20 = 871.5 m

(iv) Velocity of the bomb when it strikes ground
vertical velocity with respect to time
V(T) =vy+aT...................(2)
Substitute values, vy=58.1 m/s, a=-9.81 m/s^2
V(T) = 58.130 + (-9.81)T =>
V(20)=58.130-(9.81)(20) = -138.1 m/s (vertical velocity at strike)

vx = 43.575 m/s (horizontal at strike)
resultant velocity = sqrt(43.575^2+(-138.1)^2) = 144.812 m/s (magnitude)
in direction theta = atan(43.575,138.1)
= 17.5 degrees with the vertical, downward and forward. (direction)

4 0

3 years ago