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salantis [7]
4 years ago
7

Which one is correct and why?​

Physics
1 answer:
Gelneren [198K]4 years ago
6 0

'A' and 'C' are exactly the same circuit, except the voltmeter's terminals are flipped.

'A' is the correct way to hook everything up.

If you start at the positive terminal of the battery, and follow the flow of current through the circuit and around to the negative terminal, you're following the path where the voltage gets lower and lower and lower all the way.

So each time you come to any device in the circuit ... whether it's a resistor or a meter ... you would be hitting the positive side of it first, and then the voltage where you come out on the other side of it would be lower.  

So the left side of the resistor is more positive, and the right side is more negative.  The voltmeter is connected correctly in 'A', but it's backwards in 'C'.  If you connect the voltmeter like in 'C' and turn things on, the voltmeter will try  to go <em>down</em> from zero.  You can't read the number on it, and It's possible that the voltmeter might be damaged.  

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Three crates with various contents are pulled by a force F pull = 3615 N across a horizontal, frictionless roller‑conveyor syste
Dmitry_Shevchenko [17]

Answer:

m1=914.9kg

m2=604.9kg

m3=864.75kg

Explanation

I think we are suppose to find the mass of the crate.

The effective force that moves the body in positive x direction is 3615N

ΣFx = Σma

Then Fx=3615N

Then the masses be m1, m2 and m3

Then,

ΣF = Σ(ma)

3615=(m1+m2+m3)a

Given that a=1.516

The masses are

m1+m2+m3=, 2384.56. Equation 1

Between mass 1 and mass 2 is, F12=1387.

The effective force that pull mass 1 is 1387.

F12=m1 ×a

Therefore,

m1=F12/a

m1=1387/1.516

m1=914.9kg.

The effective force that pulls crate 1 and crate 2 is F23

F23=(m1+m2)a

Therefore

2304=(m1+m2)a

Therefore, since a=1.516

m1+m2=2304/1.516

m1+m2=1519.8kg

Since m1=914.9kg

So, m2=1519.8-m1

m2=1519.8-914.9

m2=604.9kg

Also from equation 1

m1+m2+m3=2384.56

Since m1=914.9kg and m2=604.9kg

Then, m3=2384.56-604.9-914.9

m3=864.75kg

3 0
3 years ago
A cylinder of radius r=10.0 cm and height h=20.0 cm
Olenka [21]
D I think is the correct answer
If the cylinder is slightly
7 0
2 years ago
A typical laboratory centrifuge rotates at 4000 rpm. Testtubes have to be placed into a centrifuge very carefully because ofthe
Bogdan [553]

Answer:

A)  a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²

Explanation:

Part A) The relation of the test tube is centripetal

               a_c = v² / r

the angular and linear variables are related

              v = w r

we substitute

               a_c = w² r

let's reduce the magnitudes to the SI system

              w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s

               r = 1 cm (1 m / 100 cm) = 0.10 m

let's calculate

              a_c = 418.88² 0.1

               a_c = 1.75 10⁴ m / s²

part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor

as part of rest the initial velocity is zero and on the floor the height is zero

                v² = v₀² - 2g (y- y₀)

                v² = 0 - 2 9.8 (0 + 1)

                v =√19.6

                v = -4.427 m / s

now let's look for the applied steel to stop the test tube

                v_f = v + a t

                0 = v + at

                a = -v / t

                a = 4.427 / 0.001

                a = 4.43 10³ m / s²

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The Hubble Space Telescope (HST) orbits 569 km above Earth’s surface. Given that Earth’s mass is 5.97 × 1024 kg and its radius i
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V (<span>HST’s tangential speed</span>) =  7570 m/s

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4 years ago
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