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Svetradugi [14.3K]
3 years ago
15

A 17.0 resistor and a 6.0 resistor are connected in series with a battery. The potential difference across the 6.0 resistor is m

easured as 15 V. Find the potential difference across the battery.
Physics
1 answer:
tia_tia [17]3 years ago
8 0

Answer:

V= 57.5 V

Explanation:

  • If the resistors are in the linear zone of operation, the potential difference across them, must obey Ohm's law:

        V = I*R

  • For the 6.0 Ω resistor, if the potential difference across it is 15 V, we can find the current flowing through it as follows:

       I = \frac{V}{R} = \frac{15 V}{6.0 \Omega} = 2.5 A

  • In a series circuit, the current is the same at any point of it, so the current through the battery is I = 2.5 A
  • The equivalent resistance of a series circuit is just the sum of the resistances, so, in this case, we can write the following equation:

      R_{eq} = R_{1} +R_{2} = 17.0 \Omega + 6.0 \Omega = 23.0 \Omega

  • Applying Ohm's Law to the equivalent resistance, we can find the potential difference through it, that must be equal to the potential difference across the battery, as follows:

        V = I* R_{eq}  = 2.5 A * 23.0 \Omega = 57.5 V

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Calculate the angle of refraction of 30.0° light shines from water into ice. The indices of refraction for water and ice are 1.3
WITCHER [35]

Answer:

The angle of refraction Ø2 equals 62.95° ≈ 63°

Explanation:

The relationship between the angles of incidence and

refraction , with respect to light or other waves passing through two different substances or media, such as glass, water or air is given by Snell's Law.

Snell's Law states that the when light travels from one medium to another, it generally refracts.

It is given by the mathematical expression;

[SinØ1°/SinØ2°] = [n2/n1]

Cross multiplying, we have;

n1 × SinØ1° = n2 × SinØ2°

where, n is the indices of refraction of each substance

Ø is the angle between the ray and the line normal to the surface.

Given the following values;

n1 = 1.36 n2 = 1.31 Ø1 = 30° Ø = ?

n1 × SinØ1° = n1 × SinØ2°

SinØ2° = [n1 × SinØ1°]/n2

substituting the values respectively;

SinØ2° = [1.36 × Sin30°]/1.31

SinØ2° = [1.36 × 0.5]/1.31

SinØ2° = 0.68 × 1.31

SinØ2° = 0.8906

Ø2° = Sin–¹(0.8906)

Ø2° = 62.95° ≈ 63°

4 0
3 years ago
Example 2.1
tatiyna

Answer:

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Explanation:

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8 0
3 years ago
You throw a baseball (mass 0.145 kg) vertically upward. It leaves your hand moving at 12.0 m/s. Air resistance can be neglected.
Andru [333]

Answer:

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Explanation:

Hi there!

The equations of height and velocity of the ball are the following:

y = y0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

y = height at time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

v = velocity of the ball at time t.

Placing the origin at the throwing point, y0 = 0.

Let´s use the equation of velocity to obtain the time at which the velocity is 12.0 m/s / 2 = 6.00 m/s.

v = v0 + g · t

6.00 m/s = 12.0 m/s -9.81 m/s² · t

(6.00 - 12.0)m/s / -9.81 m/s² = t

t = 0.612 s

Now, let´s calculate the height of the baseball at that time:

y = y0 + v0 · t + 1/2 · g · t²     (y0 = 0)

y = 12.0 m/s · 0.612 s - 1/2 · 9.81 m/s² · (0.612 s)²

y = 5.51 m

The ball will have an upward velocity of 6 m/s at a height of 5.51 m.

Have a nice day!

4 0
4 years ago
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Bingel [31]

Answer:

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3 years ago
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<em>number of waves that pass a given point in one second is called <u>frequency..</u></em>
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