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When comparing two circuits, you note that circuit one has twice the resistance and double the voltage of circuit two. This mean

pochemuha

This means that there is same current flow in both the circuit, or the circuit one has twice the power of circuit two.

According to ohm's law, the resistance is given as

I=V/R

Since the circuit one has twice the voltage, and resistance

I1=I2

5 0

3 years ago

Read 2 more answers

An old-fashioned LP record rotates at 33 1/3 RPM

Ksenya-84 [330]

Answer:

Part a) $\frac{5}{9}\ \frac{rev}{sec}$

Part b) $\frac{9}{5}\ \frac{sec}{rev}$

Explanation:

Part a) what is its frequency, in rev/s

we have that

An old-fashioned LP record rotates at 33 1/3 RPM

so

$33\frac{1}{3}\ \frac{rev}{min}$

Convert mixed number to an improper fraction

$33\frac{1}{3}\ \frac{rev}{min}=\frac{33*3+1}{3}=\frac{100}{3}\ \frac{rev}{min}$

Remember that

$1\ min=60\ sec$

Convert rev/min to rev/sec

$\frac{100}{3}\ \frac{rev}{min}=\frac{100}{3}(\frac{1}{60})=\frac{100}{180}\ \frac{rev}{sec}$

Simplify

$\frac{5}{9}\ \frac{rev}{sec}$

Part b) what is it period, in seconds

we know that

The period is the reciprocal of the frequency

therefore

the frequency is

$\frac{9}{5}\ \frac{sec}{rev}$

4 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/

Katen [24]

<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

$y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}$ (1)

Where:

$y$ is the instrument's final position

$y_{o}=0$ is the instrument's initial position

$V_{o}=15m/s$ is the instrument's initial velocity

$t$ is the time the parabolic movement lasts

$g=2.5\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of planet X.

As we know $y_{o}=0$ and $y=0$ when the object hits the ground, equation (1) is rewritten as:

$0=V_{o}.t-\frac{1}{2}g.t^{2}$ (2)

Finding $t$ :

$0=t(V_{o}-\frac{1}{2}g.t^{2})$ (3)

$t=\frac{2V_{o}}{g}$ (4)

$t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}$ (5)

Finally:

$t=12s$

3 0

3 years ago

Q7:<br> A 4 kg toy is lifted off the ground and falls at 3 m/s. What is the toy's energy?

alexandr1967 [171]

Answer:

The toy's energy is 18 J.

Explanation:

We have, a 4 kg toy is lifted off the ground and falls at 3 m/s. It is required to find toy's energy.

The toy will have kinetic energy due to its motion. The energy is given by :

$E=\dfrac{1}{2}mv^2\\\\E=\dfrac{1}{2}\times 4\times 3^2\\\\E=18\ J$

So, the toy's energy is 18 J.

7 0

3 years ago