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elena-14-01-66 [18.8K]
3 years ago
12

What will the reading of the voltmeter be at the instant the switch returns to position a if the inertia of the d'Arsonval movem

ent is negligible
Physics
1 answer:
Step2247 [10]3 years ago
3 0

Answer:

hello your question is incomplete attached below is the complete question

answer :

20.16 v

Explanation:

The reading of the voltmeter at the instant the switch returns  to position a

L = 5H

i ( current through inductor ) = 1/L ∫ V(t) d(t) + Vo

                                               = 1/5 ∫ 3*10^-3  d(t)  + 0 = 0.6 * 10^-3 t

iL ( 1.6 s ) = 0.6 * 10^-3 * 1.6 = 0.96 mA

Rm ( resistance ) = 21 * 1000 = 21 kΩ

 The reading of the voltmeter ( V )

V = IR

   = 0.96 mA * 21 k Ω  = 20.16 v

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Answer:

coefficient of static friction of the surface and the normal force

Explanation:

The coefficient of static friction of the surface and the normal force exerted on the surface given by equation F = μR

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3 years ago
In some circumstances, it is useful to look at the linear velocity of a point on the blade. The linear velocity of a point in un
mihalych1998 [28]

Answer:

v=wr

Explanation:

<u>Tangent and Angular Velocities</u>

In the uniform circular motion, an object describes the same angles in the same times. If \theta is the angle formed by the trajectory of the object in a time t, then its angular velocity is

\displaystyle w=\frac{\theta}{t}

if \theta is expressed in radians and t in seconds the units of w is rad/s. If the circular motion is uniform, the object forms an angle 2\theta in 2t, or 3\theta in 3t, etc. Thus the angular velocity is constant.

The magnitude of the tangent or linear velocity is computed as the ratio between the arc length and the time taken to travel that distance:

\displaystyle v=\frac{\theta r}{t}

Replacing the formula for w, we have

\boxed{ v=wr}

4 0
3 years ago
A mango falls fromthe top its tree passing a window which is 2.4m tall by taking 0.4s
Natasha2012 [34]

Explanation:

There are three points in time we need to consider.  At point 0, the mango begins to fall from the tree.  At point 1, the mango reaches the top of the window.  At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

6 0
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SIZIF [17.4K]

Answer:

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Explanation:

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F = ma

The equation for weight is thus

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B and C

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I got it right on my a pex test

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