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Thepotemich [5.8K]
3 years ago
6

Give reason.b} String roller is an example of wheel and axel .Why?​

Physics
1 answer:
Leokris [45]3 years ago
8 0

Explanation:

string roller is called an example of wheel and axel because The thin rod which needs to be turned is called the axle and the wider object fixed to the axle, on which we apply force is called the wheel.

hence the we apply force on the roller and the string gets tight.

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Which of the following usually occurs with a short circuit? a. All parts of the circuit will begin to carry higher amounts of cu
fgiga [73]

Answer:

Hi,

The correct answer option is; D. Most of the current will flow through one part of the circuit.

Explanation:

A short circuit is a low resistance path in an electric connection between two conductors supplying current in a circuit.

It happens when excess amounts of current flow in the power source through a 'short path'.

Short circuits occur at very high temperatures which is of course caused by the heat produced during dissipation.

An example of application of short circuit is arc welding, where heating is achieved through short circuit.

6 0
3 years ago
A force of 30 N stretches a very light ideal spring 0.73 m from equilibrium. What is the force constant (spring constant) of the
pantera1 [17]

Answer:

Explanation:

F = k*n

F = 30N

k = ???

N = 0.73 M

F = k* N

30N = k * 0.73

k = 30N / 0.73

k = 41.1

5 0
3 years ago
An Na ion has a charge of +1 and an O ion has a charge of –2. How many Na ions does one O ion need to balance charges
Aleonysh [2.5K]
Because Na⁺ ion lacks an electron and O²⁻ has two extra electrons extra, to balance the charge, we need 2 Na⁺ ions.
 
All ions, atoms and molecules want to get to the minimum energy state, and that state is when the ion, atom, or molecule is neutral, that's why all of them want to balance their charges. 
7 0
3 years ago
Read 2 more answers
Bob, of mass m, drops from a tree limb at the same time that Esther, also of mass m, begins her descent down a frictionless slid
galben [10]

Answer:

Explanation:

They have the same kinetic energy

5 0
3 years ago
At locations A and B, the electric potential has the values VA = 1.83 V and VB = 5.17 V, respectively. A proton released from re
densk [106]

Answer:

a. It starts at point B.

vp = 2.53*10⁴ m/s

a. it starts at point A.

ve= 1.08*10⁶ m/s

Explanation:

a)  As the proton is a positive charge, when released from rest, it will be accelerated due to the potential difference, from the higher potential to the lower one, so it is at the point B when released.

Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:

ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)

⇒ -( e* (VA-VB) ) = \frac{1}{2}*mp*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{1.67e-27kg} } = 2.53e4 m/s

⇒ vp = 2.53*10⁴ m/s

b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:

First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.

Second, as its charge is (-e) the change in electric potential energy had been negative also:

ΔUe = -e*ΔV = -e* (VB-VA)

In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:

-( (-e)* (VB-VA) ) = \frac{1}{2}*me*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{9.1e-31kg} } = 1.08e6 m/s

⇒ ve = 1.08*10⁶ m/s

4 0
4 years ago
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