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3 years ago

Which of the following equations illustrates the law of conservation of matter?

Physics

1 answer:

Anvisha [2.4K]3 years ago

8 0

Answer:

Explanation:

the balancing is correct in the first one

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Look at the circuit diagram. Which of these components is part of the circuit?

Roman55 [17]

Answer:

b ac power source

Explanation:

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3 years ago

A car has an engine which delivers a constant power. It accelerates from rest at time t = 0, and at t = t0 its acceleration is a

olga2289 [7]

Here is the answer of the given problem above.
Use this formula: P = FV = ma*at = ma^2 t
Substitute the values, and therefore, we got m(a0)^2t = m(x)^2 (2t)
then, solve for x which is the acceleration at 2t.
The answer would be a0/sqrt(2).
Hope this answers your question. Thanks for posting.

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3 years ago

The greater the mass of an object, the greater the _____. (fill in question)

g100num [7]

Answer:

momentum

Explanation:

4 0

3 years ago

Read 2 more answers

A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor

Nata [24]

Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

the magnetic field of the Earth is 0.500 gauss

the baton is 60.1 cm in length.

the magnetic field is oriented at 14.42°

we wil get the area due to rotation of radius of baton is

$\Delta A = \frac{1}{2} \Delta \theta R^2$

The formula for the induced emf is

$E = \frac{\Delta \phi}{\Delta t}$

$\phi = \texttt {magnetic flux}$

$E=\frac{\Delta (BA) }{\Delta t}$

$=B\frac{\Delta A}{\Delta t}$

B is the magnetic field strength

substitute

$\texttt {substitute}\ \frac{1}{2} \Delta \theta R^2 \ \ for \Delta A$

$E=B\frac{(\Delta \theta R^3/2)}{\Delta t} \\\\=\frac{1}{2} BR^2\omega$

The magnetic field of the earth is oriented at 14.42

$\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5$

we plug in the values in the equation above

so, the induce EMF will be

$E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega$

$=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V$

6 0

3 years ago

Why do electrons move from the negative end of the tube to the positive end

defon

Due to attraction ... of opposite charges

3 0

3 years ago