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lilavasa [31]
3 years ago
14

How does a resistor affect an electrical circuit?

Physics
2 answers:
Klio2033 [76]3 years ago
5 0

Answer:

D.

It reduces the rate at which electrons flow.

Vanyuwa [196]3 years ago
3 0

Answer:

D

Explanation:

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What is light emitting diode<br>​
Natali [406]

Answer:

A light emitting diode (a semiconductor diode which glows when voltage is applied.

8 0
3 years ago
How many joules of energy are required to accelerate one kilogram of mass from rest to a velocity of 0.866c?
Arlecino [84]

Answer:

the amount of energy needed is 1.8 x 10¹⁷ J.

Explanation:

Given;

mass of the object, m₀ = 1 kg

velocity of the object, v = 0.866 c

By physics convection, c is the speed of light = 3 x 10⁸ m/s

The energy needed is calculated as follows;

E = Mc²

As the object approaches the speed of light, the change in the mass of the object is given by Einstein's relativity formula;

M = \frac{M_0}{\sqrt{1- \frac{v^2}{c^2} } } \\\\  M = \frac{1}{\sqrt{1- \frac{(0.866c)^2}{c^2} } }\\\\  M = \frac{1}{\sqrt{1- \frac{0.74996c^2}{c^2} } }\\\\  M = \frac{1}{\sqrt{0.25} } \\\\ M = 2 \ kg

The energy required is calculated as;

E = 2 x (3 x 10⁸)²

E = 1.8 x 10¹⁷ J

Therefore, the amount of energy needed is 1.8 x 10¹⁷ J.

3 0
3 years ago
Which activities are you always doing when you have a warm-up activity?​
Sati [7]
A jog because it helps to get the muscles moving and your heart pumping blood around your body.

I hope this helps.
3 0
3 years ago
Read 2 more answers
A 1,600 kg car is traveling with a speed of 20 m/s. Find the net force that is required to bring the car to a halt in a distance
Pie
256N. I love physics so.... yerp
4 0
4 years ago
Read 2 more answers
A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi
Dmitriy789 [7]

Answer:

a

 A =  0.081 \  m

b

The value is  u =  0.2569 \  m/s

Explanation:

From the question we are told that

   The mass is  m  =  0.750 \ kg

   The spring constant is  k  =  17.5 \  N/m

    The instantaneous speed is  v  =  39.0 \  cm/s= 0.39 \  m/s

    The position consider is  x =  0.750A  meters from equilibrium point

   

Generally from the law of  energy conservation we have that

        The kinetic energy induced by the hammer  =  The energy stored in the spring

So

          \frac{1}{2} *  m * v^2  =  \frac{1}{2}  *  k  *  A^2

Here a is the amplitude of the subsequent oscillations

=>      A =  \sqrt{\frac{m *  v^ 2 }{ k} }

=>      A =  \sqrt{\frac{0.750 *  0.39 ^ 2 }{17.5} }

=>       A =  0.081 \  m

Generally from the law of  energy conservation we have that

The kinetic energy  by the hammer  =  The energy stored in the spring at the point considered   +   The kinetic energy at the considered point

             \frac{1}{2}  * m *  v^2 = \frac{1}{2}  * k x^2 + \frac{1}{2}  * m *  u^2

=>          \frac{1}{2}  * 0.750 *  0.39^2 = \frac{1}{2}  * 17.5* 0.750(0.081 )^2 + \frac{1}{2}  * 0.750 *  u^2

=>          u =  0.2569 \  m/s

3 0
3 years ago
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