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3 years ago

When the resistance in a circuit remains constant, how are the voltage and current related?

Physics

2 answers:

vagabundo [1.1K]3 years ago

8 0

Explanation:

The current doubles when the voltage doubles because they are directly proportional.

stich3 [128]3 years ago

4 0

Answer:

The current doubles when the voltage doubles because they are directly proportional.

Explanation:

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Two bicycle tires are set rolling with the same initial speed of 4.00 m/s along a long, straight road, and the distance each tra

77julia77 [94]

Answer:

The coefficient of rolling friction for the tire under low pressure is 0.0342.

Explanation:

Two bicycle tires are set rolling with the same initial speed of 4.00 m/s

Final speed of both the bicycle, speed is reduced by half is measured, v = 2 m/s.

Here,

$u_kmg = ma\\\\a=\mu g$

Using third equation of motion as :

$v^2-u^2=2as\\\\v^2-u^2=2\mu gs\\\\\mu =\dfrac{v^2-u^2}{2gs}\\\\\mu =\dfrac{4^2-2^2}{2\times 9.8\times 17.9}\\\\\mu=0.0342$

So, the coefficient of rolling friction for the tire under low pressure is 0.0342.

5 0

3 years ago

When the displacement of a mass on a spring is 1/2a the half of the amplitude, what fraction of the mechanical energy is kinetic

USPshnik [31]

Total energy is a spring:
$E = \frac{1}{2} kx^2 + \frac{1}{2} mv^2 = \frac{1}{2} ka^2$

At x = 0.5a:
$\frac{1}{2} k \frac{a}{2} ^2 + \frac{1}{2} mv^2 = \frac{1}{2} ka^2 \\ \frac{1}{2} mv^2 = \frac{1}{2} ka^2 - \frac{1}{8} ka^2 = \frac{3}{8} ka^2$

The ration:
$\frac{ \frac{3}{8}ka^2 }{ \frac{1}{2} ka^2} = \frac{3}{4}$

5 0

4 years ago

Read 2 more answers

Design a voltage divider to provide the following approximate voltages with respect to ground using a 30 V source: 8.18 V, 14.7

yarga [219]

Answer:

R₁ = 14.7 10³ Ω , R₂ = 8.18 10³ Ω , R₃ = 1.72 10³ Ω , R₄ = 5.4 10³ Ω 1/8 W resistor

Explanation:

For this exercise we must use a series circuit since the sum of the voltage on each resin is equal to the source voltage (V = 30 V)

Therefore we build a circuit with 4 resistors in series, in such a way that

V = i R

let the voltage

1st resistance

V = i R

R₁ = V / i

R₁ = 14.7 / 1 10⁻³

R₁ = 14.7 10³ Ω

power is

P = V i

P = 14.7 1 10⁻³

P = 14.7 10⁻³ W = 0.0147 W

a resistance of ⅛ W is indicated

2nd resistance

R₂ = 8.18 / 1 10⁻³

R₂ = 8.18 10³ Ω

Power

P = 8.18 1 10⁻³

P = 0.00818W

a 1/8 W resistor

3rd resistance

this resistance is calculated in such a way that

V₁ + V₂ + V₃ = 24.6

V₃ = 24.6 - V₁ -V₂

V₃ = 24.6 - 14.7 - 8.18

V₃ = 1.72 V

R₃ = 1.72 / 1 10⁻³

R₃ = 1.72 10³ Ω

power

P = Vi

P = 1.72 10⁻³

P = 0.00172 W

a resistance of ⅛ W

To obtain the voltage of 24.6 we use this three resistors together

4th resistance

The value of this resistance is calculated so that the sum of all the voltages reaches the source voltage

30 = V₁ + V₂ + V₃ + V₄

V₄ = 30 - V₁ -V₂ -V₃

V₄ = 30 -14.7 - 8.18 - 1.72

V₄ = 5.4 V

R₄ = 5.4 / 1 10⁻³

R₄ = 5.4 10³ Ω

Power

P = V i

P = 5.4 10⁻³

P = 0.0054 W

⅛ W resistance

The values of these resistance are commercially

Let's check the consumption of the circuit

R_total = R₁ + R₂ + R₃ + R₄

R_total = (14.7 + 8.18 + 1.72 + 5.4) 10³

R_total = 30 10³

the current circulating in the circuit is

i = V / R_total

i = 30/30 10³

i = 1 10⁻³ A

therefore it is within the order requirement.

for connections see attached diagram

8 0

3 years ago

Find the mass of a sample that has a dersity of 2 g/mL and a volume of 10 mL

In-s [12.5K]

Answer:

mass = volume x density.

Explanation:

2 x 10 = 20g

4 0

3 years ago

A mass is hung from a spring and set in motion so that it oscillates continually up and down. The velocity v of the weight at ti

otez555 [7]

To solve the problem it is necessary to identify the equation in the manner given above.

This equation corresponds to the displacement of a body under the principle of simple harmonic movement.

Where,

$\xi = Acos(\omega t +\phi)$

PART A) Our equation corresponds to

$y = -5cos(4\pi t)$

Therefore the value of omega is equivalent to that of

$\omega = 4\pi$

From the definition we know that the period as a function of angular velocity is equivalent to

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{4\pi}$

$T = \frac{1}{2}$

This same point is the equivalent of the maximum point of the speed that the body can reach, since the internal expression of the $cos\theta$ Is equivalent to . So the maximum speed that the body can reach is,