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3 years ago

11

Based on the number on the fuse, what do you think is the maximum current allowed in this circuit?

1 answer:

Inessa [10]3 years ago

3 0

Answer:

0.3 A

Explanation:

Build the circuit attached. Make sure the switch is turned off as you build the circuit. Set the Selected battery voltage to 10 volts.

A. Based on the number on the fuse, what do you think is the maximum current allowed in this circuit?

Solution:

A fuse is a safety device which is used in circuits for preventing very high currents which can cause overloading or start a fire. A fuse is made up of a low resistance thin piece of metal such that the flow of high current can cause the metal to melt leading to an open circuit. A fuse lets current flow up to a maximum value before it melts and the circuit will be broken.

From the image attached, since the maximum rating of the fuse is 0.3 A, hence the maximum current that can flow through the circuit is 0.3 A.

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Marysya12 [62]

The car's final velocity is 11.6 m/s in the opposite direction.

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Albert's laboratory is filled with a constant uniform magnetic field pointing straight up. Albert throws some charges into this

guajiro [1.7K]

Answer:

$\vec{F}=qB(v_y \hat{i} - v_x\hat{j})$

Explanation:

The force excerted by a magnetic field on a charged particle is given by the Lorentz force:

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Lets consider the z-direction of our coordinate system the same direction of the magnetic field, that is:

$\vec{B} = B \hat{k}$

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$\vec{v} = v_x\hat{i} + v_y \hat{j} + v_z \hat{k}$

Therefore, since k×k = 0

$\vec{v} \times \vec{B} = (v_x\hat{i} + v_y \hat{j} + v_z \hat{k}) \times B\hat{k}\\\vec{v} \times \vec{B} = (Bv_x \hat{i} \times\hat{k}) + (Bv_y \hat{j}\times\hat{k})$

And since i, j , k are a rigth hand system:

i × j = k

j × k = i

k × i = j --> i × k= -j

$\vec{v} \times \vec{B} = Bv_x (-\hat{j}) + Bv_y \hat{i} = Bv_y \hat{i} - Bv_x\hat{j}$

Threfore, if the particle has charge q and velocity v = (vx,vy,vz), the magnetic force it will feel will be

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3 years ago

Please help. All the information is in the image.

kondaur [170]

The tiger's position at time <em>t</em> is given by

<em>x</em> (horizontal) = (4.5 m/s) <em>t</em>

<em>y</em> (vertical) = 7.5 m - 1/2 <em>gt</em> ²

Solve <em>y</em> = 0 for <em>t</em> to find the time it takes for the tiger to reach the ground :

0 = 7.5 m - 1/2 (9.8 m/s²) <em>t</em> ²

===> <em>t</em> = √(2 (7.5 m) / (9.8 m/s²)) ≈ 1.2 s

Evaluate <em>x</em> at this time :

<em>x</em> = (4.5 m/s) (1.2 s) ≈ 5.6 m

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3 years ago

Two children are riding on the edge of a merry-go-round that has a mass of 100.kg and radius of 1.60m and is rotating at 20.0rpm

Gre4nikov [31]

Here since both children and merry go round is our system and there is no torque acting on this system

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nata0808 [166]

Answer:

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