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3 years ago

Manipulate p+x=r to solve for x

Physics

2 answers:

kotykmax [81]3 years ago

6 0

P+x=r
-p -p
Answer: x=r-p

Hope this helps.

Tanzania [10]3 years ago

4 0

X=r-p. Maybe I don't understand, but I am assuming that you need to isolate for X? you simply subtract p from both sides.<span />

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Given that both liquid A and B exert the same amount of pressure .What would be the height of column of liquid A if the density

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Answer:

ans 5

Explanation:

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3 years ago

If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object paral

leonid [27]

Answer:

$\theta = tan^{-1}\mu$

Explanation:

As we know that if the object is placed on the inclined plane then the force of friction on the object is counterbalanced by the component of the weight of the object along the inclined plane.

So we can say

$F_f = mgsin\theta$

now if we increase the inclination of the plane then the component of the weight weight along the inclined plane will increase and hence the friction force will also increase.

As we know that the limiting value or the maximum value of friction force at the static condition is given by

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so we have

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so we will have

$mg sin\theta = \mu (mg cos\theta)$

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3 years ago

A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If

Salsk061 [2.6K]

Complete question:

A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?

Answer:

The current in the circuit 7 ms later is 0.2499 A

Explanation:

Given;

Ideal inductor, L = 45-mH

Resistor, R = 60-Ω

Ideal voltage supply, V = 15-V

Initial current at t = 0 seconds:

I₀ = V/R

I₀ = 15/60 = 0.25 A

Time constant, is given as:

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T = (45 x 10⁻³) / (60)

T = 7.5 x 10⁻⁴ s

Change in current with respect to time, is given as;

$I(t) = I_o(1-e^{-\frac{t}{T}})$

Current in the circuit after 7 ms later:

t = 7 ms = 7 x 10⁻³ s

$I(t) = I_o(1-e^{-\frac{t}{T}})\\\\I =0.25(1-e^{-\frac{7*10^{-3}}{7.5*10^{-4}}})\\\\I = 0.25(0.9999)\\\\I = 0.2499 \ A$

Therefore, the current in the circuit 7 ms later is 0.2499 A

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3 years ago

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Answer:

Explanation:

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Answer:

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