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Tems11 [23]
2 years ago
12

After the right string is cut, the meterstick swings down to where it is vertical for an instant before it swings back up in the

other direction. What is the angular speed when the meterstick is vertical?.
Physics
1 answer:
abruzzese [7]2 years ago
4 0

The Angular Speed of the Meterstick after it becomes vertical is given as:

= 5.800 rad/s. This is computed using the formula for conservation of energy.

<h3>What is Angular Speed?</h3>

Angular Speed is defined as the measure of how quick or fast the central angle of a body that is rotational is changing in relation to time. Please see the attached for the full question.

<h3>What is the solution for the Angular Speed of the phenomena described?</h3>

Recall that the formula for conservation of energy is given as:

(1/2)(<em>Iω²) = mg(L/4) where

L=  Lenght and</em>

<em>m = Mass. Please refer to the attached questions.</em>

<em />

Factoring the moment of inertia about the pivot, we solve for the angular speed using the following formula:<em>

ω = </em>\sqrt{(mgL)/2I}

= \sqrt{(6g)/7L}

Hence, the angular speed is:

ω = 8.800rad/s

Learn more about Angular speed at:
brainly.com/question/540174

<em />

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Answer:

One side is more positive, the other is more negative

Explanation:

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The temperature of air changes from 0 to 10°C while its velocity changes from zero to a final velocity, and its elevation change
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Answer:

Final velocity = 119.83 m/s

Final elevation = 731.9 m

Explanation:

We are told that temperature of air changes from 0 to 10°C

Thus;

Change in temperature; ΔT = 10 - 0 = 10°C

Also, its velocity changes from zero to a final velocity. Thus;

v1 = 0 m/s

v2 is unknown

Also, its elevation changes from zero to a final elevation.

So, z1 = 0 and z2 is unknown

Now, we want to find v2 and z2 when the internal, kinetic and potential energy are equal.

Thus Equating the formula for both kinetic and internal energy gives;

½m(v2² - v1²) = mc_v•ΔT

m will cancel out and v1 is zero to give;

v2² = 2c_v•ΔT

v2 = √(2c_v•ΔT)

Where c_v is specific heat of constant volume of air with a constant value of 718 J/Kg.K

Thus;

v2 = √(2 × 718 × 10)

v2 = √14360

v2 = 119.83 m/s

To find z2, we will equate potential energy formula to that of the internal energy.

Thus;

mg(z2 - z1) = mc_v•ΔT

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4 0
3 years ago
A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o
gulaghasi [49]

A. 0.77 A

Using the relationship:

P=\frac{V^2}{R}

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega

The two light bulbs are connected in series, so their equivalent resistance is

R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega

The two light bulbs are connected to a voltage of

V  = 240 V

So we can find the current through the two bulbs by using Ohm's law:

I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A

B. 142.3 W

The power dissipated in the first bulb is given by:

P_1=I^2 R_1

where

I = 0.77 A is the current

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Substituting numbers, we get

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C. 42.7 W

The power dissipated in the second bulb is given by:

P_2=I^2 R_2

where

I = 0.77 A is the current

R_2 = 72 \Omega is the resistance of the bulb

Substituting numbers, we get

P_2 = (0.77 A)^2 (72 \Omega)=42.7 W

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

P=\frac{E}{t}

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

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