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liubo4ka [24]
2 years ago
14

Which rule is used to determine the direction of the magnetic field and force produced by a moving electrical charge?

Physics
1 answer:
densk [106]2 years ago
7 0

Answer:

The right hand rule

Explanation:

The thumb indicating the direction of the moving charge, and the fingers indicating the direction of the magnetic field vectors in the right hand rule.

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A reversible Carnot engine has an efficiency of 30% and it operates between a heat source maintained at T 1 , and a refrigerator
bixtya [17]

To solve this problem we will apply the concepts related to the thermal efficiency given in an engine of the Carnot cycle. Here we know that efficiency is given under the equation

\eta = 1 - \frac{T_2}{T_1}

Where,

T_2 = Temperature of Cold Body

T_1 =Temperature of Hot Body

\eta= Efficiency

According to the statement our values are:

\eta = 0.3

T_2 = 210K

Replacing we have that

0.3 = 1- \frac{210}{T_1}

\frac{210}{T_1} = 0.7

T_1 = \frac{210}{0.7}

T_1 = 300K

Therefore the temperature of the heat source is 300K

6 0
3 years ago
What does it mean if an object is electrically uncharged?
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The object has been golaced in water
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Through the process of blank
maksim [4K]

Answer:

Subduction, Trench, Mantle

Explanation:

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The rectangular coordinates of a point are (5.00, y) and the polar coordinates of
Otrada [13]

Answer:

Explanation:

Polar coordinates formula

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6 0
2 years ago
One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has
Mars2501 [29]

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE)  = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = \sqrt{6gL}

Hence, v0 = 6.86 m/s

4 0
3 years ago
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