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Deffense [45]
3 years ago
8

Prior Knowledge Questions (Do these BEFORE using the Gizmo.) Strings of holiday lights can be designed in one of two ways. In so

me strings of lights, each light is connected to the others along a single wire (in series). In others, each light is attached to its own wire (in parallel). Suppose a single light bulb burns out. How do you think this will affect lights that are strung along a single wire
Physics
1 answer:
IgorLugansk [536]3 years ago
8 0

Answer:

They would go out

Explanation:

This is because, in a series connection, the same current passes through each light. Since the current is the same, if one light burns out, it cuts off the rest of the other lights and thus, no current flows in the string again.

Whereas, in a parallel connection, each light is attached to its own wire and thus has a different current flowing through it than the rest of the other wires.  If one of the lights goes out, current stops flowing through it but, it doesn't affect the other lights.

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A -5.0 μC charge experiences a 11 i^ N electric force in a certain electric field. [Recall that i^ is a unit vector in the x-dir
Pachacha [2.7K]

Answer:

\vec{F}= -3.52\times 10^{-13}\hat{i}\ N

Explanation:

given,

charge = -5.0 μC

Electric force, F = 11 i^ N

force would a proton experience = ?

we know

\vec{F} = q \vec{E}

11\hat{i} = -5 \times 10^{-6} \vec{E}

\vec{E} =-2.2 \times 10^{6}\hat{i}

we know charge of proton is equal to 1.6 x 10⁻¹⁹ C

using formula

\vec{F} = q \vec{E}

\vec{F}= 1.6 \times 10^{-19}\times -2.2 \times 10^{6}\hat{i}

\vec{F}= -3.52\times 10^{-13}\hat{i}\ N

Force experienced by the photon in the same field is equal to \vec{F}= -3.52\times 10^{-13}\hat{i}\ N

3 0
3 years ago
ball is dropped from rest and hits the ground 1.1 seconds later. What height was the ball released from? ​
Aleonysh [2.5K]
Height = 1/2 * 9.8 * (1.1^2)
5 0
3 years ago
A beryllium-9 ion has a positive charge that is double the charge of a proton, and a mass of 1.50 ✕ 10−26 kg. At a particular in
seropon [69]

Answer:

Magnetic force, F = 3.52\times 10^{-13}\ N

Explanation:

Given that,

A beryllium-9 ion has a positive charge that is double the charge of a proton, q=2\times 1.6\times 10^{-19}\ C=3.2\times 10^{-19}\ C

Speed of the ion in the magnetic field, v=5\times 10^6\ m/s

Its velocity makes an angle of 61° with the direction of the magnetic field at the ion's location.

The magnitude of the field is 0.220 T.

We need to find the magnitude of the magnetic force on the ion. It is given by :

F=qvB\\\\F=3.2\times 10^{-19}\times 5\times 10^6\times 0.22\\\\F=3.52\times 10^{-13}\ N

So, the magnitude of magnetic force on the ion is 3.52\times 10^{-13}\ N.

3 0
3 years ago
Read 2 more answers
If the vertical axis and time on the horizontal axis if the speed is steadily increasing could the speef line ever become perfec
Phantasy [73]

A graph of real speed can have a section that's as steep as you want,
but it can never be a perfectly vertical section.

Any vertical line on a graph, even it it's only a tiny tiny section, means
that at that moment in time, the speed had many different values.

It also means that the speed took no time to change from one value to
another, and THAT would mean infinite acceleration.

8 0
3 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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