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3 years ago

Which of these letters is the symbol for current in equations A: c B: i C: r D: t

1 answer:

4 0

Answer: [B]: The letter, "<em /> I " ; for current; in units of "Amps" .
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Savanna regions developed during the Triassic period. true or false

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0.10-kilogram model rocket’s engine is designed to deliver an impulse of 6.0 newton-seconds. If the rocket engine burns for 0.75

UkoKoshka [18]

Answer:

8.0 N

Explanation:

Force: This can be defined as the mass of a body and its acceleration. The S.I unit of Force is Newton (N).

Mathematically, Fore is expressed as

F = ma ........................... equation 1

Where F = force, m = mass, a = acceleration.

and

I = mΔv

Δv = I/m ............................ Equation 2

Where I = impulse, m = mass, Δv = change in velocity

Given: I = 6.0 Newton-seconds, m = 0.1 kilogram.

Substituting into equation 2

Δv = 6.0/0.1

Δv = 60 m/s.

But

a = Δv/t

where t = time = 0.75 seconds.

a = 60/0.75

a = 80 m/s²

Substitute the values of a and m into equation 1.

F = 0.1(80)

F = 8.0 N.

Thus the average force produced = 8.0 N

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3 years ago

A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li

Kitty [74]

Answer:

$E_{net} = 6.44 \times 10^5 N/C$

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

$E = \frac{2k \lambda}{r}$

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

$E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}$

now plug in all data

$\lambda_1 = 4.68 \muC/m$

$\lambda_2 = 2.48 \mu C/m$

$r = 0.200 m$

now we have

$E_{net} = \frac{2k}{r}(4.68 + 2.48)$

$E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})$

$E_{net} = 6.44 \times 10^5 N/C$

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3 years ago

Calculate a rate of cooling down of air from 80 C to 5C Show calculation. Give an answer in cubic meters per minute and cfm.

antoniya [11.8K]

Explanation:

Given that,

Rate of cooling of air

Initial temperature= 80°C

Final temperature = 5°C

We need to calculate

Using newton's law of cooling

$\dfrac{dT}{dt}=c(T-T_{0})$

$\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})$

Where, $dT=T_{1}-T_{2}$

Here, $T =\dfrac{T_{1}+T_{2}}{2}$

$T_{0}$ = 25°C (surrounding temperature)

dt = 1 minute

$\dfrac{dT}{dt}=c(\dfrac{T_{1}+T_{2}}{2}-T_{0})$

Put the value into the formula

$\dfrac{80-5}{1}=c(\dfrac{85}{2}-25)$

$c=\dfrac{75}{17.5}$

$c=4.285\ cubic\ meter/minute$

Hence, This is the required answer.

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3 years ago

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Answer:

btw But I don't use Insta

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