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Lerok [7]
3 years ago
15

The following voltage drops are measured across three resistors in series: 5.5 V. 8.2 V. and 12,3 V. What is the value of the so

urce voltage to which these resistors arc connected?
Physics
1 answer:
DiKsa [7]3 years ago
7 0

Answer:

26 V

Explanation:

Given:

The three resistors 5.5 V. 8.2 V. and 12,3 V. are connected in series.

Now we have to find out the source voltage  to which these resistors arc connected ?

Solution:

As we know in series the magnitude of current is uniform but the voltage divides between the resistors supplied from source voltage.So the magnitude of source voltage is, 5.5 V + 8.2 V + 12.3 V = 26 V

Hence, the value of the source voltage to which these resistors are connected is  26 V

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A car has an initial velocity of 11.2 m /sec. the car accelerates at 10.0 m /s2 for 8.0 seconds. what is the velocity of the car
Vinil7 [7]
Initial velocity(u) = 11.2 m/s.
Final velocity(v) = ?
acceleration(a) = 10.2 m/s²

Using kinematic equation v = u + at

v = 11.2 + 10 x 8 = 11.2 + 80 = 91.2 m/s.

Therefore final velocity is 91.2 m/s.
7 0
3 years ago
A 2.0-kg projectile is fired with initial velocity components v0x = 30 m/s and v0y = 40 m/s from a point on the Earth's surface.
EleoNora [17]

(a) The kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

(b) The work done  in firing the projectile is 2,500 J.

<h3>Kinetic energy of the projectile at maximum height</h3>

The kinetic energy of the projectile when it reaches the highest point in its trajectory is calculated as follows;

K.E = ¹/₂mv₀ₓ²

where;

  • m is mass of the projectile
  • v₀ₓ is the initial horizontal component of the velocity at maximum height

<u>Note:</u> At maximum height the final vertical velocity is zero and the final horizontal velocity is equal to the initial horizontal velocity.

K.E = (0.5)(2)(30²)

K.E = 900 J

<h3>Work done in firing the projectile</h3>

Based on the principle of conservation of energy, the work done in firing the projectile is equal to the initial kinetic energy of the projectile.

W = K.E(i) = ¹/₂mv²

where;

  • v is the resultant velocity

v = √(30² + 40²)

v = 50 m/s

W = (0.5)(2)(50²)

W = 2,500 J

Thus, the kinetic energy of the projectile when it reaches the highest point in its trajectory is 900 J.

The work done  in firing the projectile is 2,500 J.

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

5 0
1 year ago
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velikii [3]
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I need help on this science
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2 years ago
A vertical spring (ignore its mass), whose spring constant is 1070 N/m, is attached to a table and is compressed 0.100 m.
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I can not solve the problem if I do not have the mass.

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3 years ago
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