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IrinaVladis [17]
3 years ago
8

How much resistance is required to limit the current to 1.5 mA if the potential drop across the resistor is 6V

Physics
1 answer:
Whitepunk [10]3 years ago
3 0

R = V/I

R = 6v / 0.0015 A

R = 4,000 ohms

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Write the products for the following neutralization reactions
madam [21]

Answer:

answer 3

Explanation:

5 0
2 years ago
How much work is done in holding a 10 newton sack of potatoes while waiting in line at the grocery store for 3 minutes.​
hodyreva [135]

Answer:

Zero

Explanation:

W = F × s

F = 10 N,

t = 3min = 180sec

s = 0( no change in postion)

W = 10 ×0

W = 0

4 0
3 years ago
An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p
igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

Explanation:

Given:

Distance from the wire to the field point r = 2.83 \times 10^{-2} m

Speed of electron v = 35.5 \%c

Current I = 17.7 A

For finding the acceleration,

First find the magnetic field due to wire,

  B = \frac{\mu _{o}I }{2\pi r }

Where \mu_{o} = 4\pi   \times 10^{-7}

  B = \frac{4\pi \times 10^{-7}  \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }

  B = 12.50 \times 10^{-5} T

The magnetic force exerted on the electron passing through straight wire,

  F = qvB  

  F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}

  F = 21.3 \times 10^{-16} N

From the newton's second law

  F = ma

Where m = mass of electron = 9.1 \times 10^{-31} kg

So acceleration is given by,

   a = \frac{F}{m}

   a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }

   a = 2.34 \times 10^{15} \frac{m}{s^{2} }

Therefore, the magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

7 0
2 years ago
If the frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum speed of the oscillat
Maksim231197 [3]

Answer:

If the frequency of the motion of a simple harmonic oscillator is doubled , then maximum speed of the oscillator changes by the factor 2

Explanation:

We know that in a simple harmonic oscillator the maximum speed is given by

    v_{max} = Aw

  Here A is amplitude which is constant , so from above equation we see that maximum speed is directly proportional to w\\ of the oscillation .

  Since  w = 2 \pi f

      v_{max}^{|}/v_{max} = 2f/f = 2

  Where v_{max}^{|} is the maximum speed when frequency is doubled .

6 0
3 years ago
Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of
vredina [299]

Answer:

The angular speed after 6s  is \omega = 1466.67s^{-1}.

Explanation:

The equation

I\alpha  = Fd

relates the moment of inertia I of a rigid body, and its angular acceleration \alpha, with the force applied F at a distance d from the axis of rotation.

In our case, the force applied is F = 22N, at a distance d = 6cm =0.06m, to a ring with the moment of inertia of I =mr^2; therefore, the angular acceleration is

$\alpha =\frac{Fd}{I} $

$\alpha =\frac{22N*0.06m}{(1.5kg)*(0.06)^2} $

\alpha  = 244.44\: s^{-2}

Therefore, the angular speed \omega which is

\omega  = \alpha t

after 6 seconds is

\omega = 244.44$\: s^{-2}* 6s

\boxed{\omega = 1466.67s^{-1}}

7 0
3 years ago
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