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3 years ago

5

A meter stick is held vertically with one end on the floor and is then allowed to fall. Find the speed of the other end just bef

ore it hits the floor, assuming that the end on the floor does not slip. (Hint: Consider the stick to be a thin rod and use the conservation of energy principle.) Number Enter your answer in accordance to the question statement Units Choose the answer from the menu in accordance to the question statement

1 answer:

stich3 [128]3 years ago

7 0

Answer:

v = 7.67 m/s for L= 1m

Explanation:

Let's use the conservation of mechanical energy, at the highest point and the lowest point

Initial. Vertical ruler

Em₀ = mg h

Final. Just before touching the floor

$Em_{f}$ = K = ½ I w²

Em₀ = $Em_{f}$

m g h = ½ I w²

The moment of inertia of a ruler that turns on one end is

I = 1/3 m L²

Let's replace

m g h = ½ (1/3 m L²) w²2

g h = 1/6 L² w²

They ask for the speed of the end so the height h is equal to the length of the ruler

g L = 1/6 L² w²

The linear and angular variables are related

v = w r

w = v / r

In this case the point of interest a in strangers r = L

g L = 1/6 L² v² / L²

v = √ 6 g L

Let's calculate

Assume that the length of the meter is L = 1 m

v = √ (6 9.8 1)

v = 7.67 m/s

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What is meant by an electrical current

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Answer:an electrical Current can be defined as the free flow of electrons through a circuit

Explanation:

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3 years ago

A lab cart with a mass of 15 kg is moving with constant velocity, v, along a straight horizontal track. A student drops a 2 kg m

lbvjy [14]

The equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ (option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.

The horizontal momentum is given by:

$p_{i} = p_{f}$

$m_{1}v_{1}_{i} + m_{2}v_{2}_{i} = m_{1}v_{1}_{f} + m_{2}v_{2}_{f}$

Where:

m₁: is the mass of the lab cart = 15 kg

m₂: is the <em>mass </em>of the object dropped = 2 kg

$v_{1}_{i}$ : is the initial velocity of the<em> lab cart </em>

$v_{2}_{i}$ : is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)

$v_{1}_{f}$ : is the final velocity of the<em> lab cart </em>

$v_{2}_{f}$ : is the <em>final velocity</em> of the <em>object </em>

Then, the horizontal momentum is:

$15v_{1}_{i} + 2*0 = 15v_{1}_{f} + 2v_{2}_{f}$

When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:

$15v_{1}_{i} + 2*0 = v_{f}(15 + 2)$

Therefore, the equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ represents the horizontal momentum (option 3).

Learn more about linear momentum here:

brainly.com/question/2141713?referrer=searchResults

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2 years ago

A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the c

puteri [66]

Correct question:

A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?

Answer:

the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

Explanation:

Given;

length of solenoid, L= 0.35 m

diameter of the solenoid, d = 0.04 m

current through the solenoid, I = 5.0 A

magnetic field in the center of the solenoid, 2.8 x 10⁻² T

The number of turns per meter for the solenoid is calculated as follows;

$B =\mu_o I(\frac{N}{L} )\\\\B = \mu_o I(n)\\\\n = \frac{B}{\mu_o I} \\\\n = \frac{2.8 \times 10^{-2}}{4 \pi \times 10^{-7} \times 5.0} \\\\n = 4.5 \times 10^3 \ turns/m$

Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

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2 years ago

Why do temperatures on the moons surface vary ncyb more than temperatures on earth

ollegr [7]

Answer:

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2 years ago

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lorasvet [3.4K]

Answer:

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Explanation:

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So we first find the equivalent resistance for the two resistors in parallel:

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By knowing this, we can estimate the total current through the circuit,:

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and therefore, we can estimate the voltage drop (V3) in R3 uisng Ohm's law:

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So now we know that the potential drop across the parellel resistors must be:

10 V - 4.28 V = 5.72 V

and with this info, we can calculate the current through R1 using Ohm's Law:

$I_1=\frac{V_1}{R_1} =\frac{5.72}{100} =0.0572\,\,amps$

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3 years ago