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2 years ago

Real-world examples of power

1 answer:

3 0

By definition, power is the amount of energy consumed (or produced) in a second. (or more precisely, it is the rate of change in energy).
so anything which uses energy in a known time period can be labeled with a power rating.

an example for power could be a nuclear plant; traditional nuclear plants produce somewhat close to 1 giga watts (which means 1 giga joules in a second)

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Choose the false statement regarding resistance.

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What happens to myosin and actin as sarcomeres relax?

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The myosin heads pull on the actin, bringing them closer together

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2 years ago

Density is a physical property that relates the mass of a substance to its volume. A. Calculate the density, in g/mL , of a liqu

Angelina_Jolie [31]

Answer:

A) 0.660 g/ml

B) 1.297 ml

C) 0.272 g

Explanation:

Every substance, body or material has mass and volume, however the mass of different substances occupy different volumes. This is where density $D$ appears as a physical characteristic property of matter that establishes a relationship between the mass $m$ of a body or substance and the volume $V$ it occupies:

$D=\frac{m}{V}$ (1)

Knowing this, let's begin with the answers:

<h2 /><h2>Answer A:</h2>

Here the mass is $m=0.155g$ and th volume $V=0.000235L=0.235mL$

Solving (1) with these values:

$D=\frac{0.155g}{0.235mL}$ (2)

$D=0.660g/mL$ (3)

<h2>Answer B:</h2>

In this case the mass of a sample is $m=4.71g$ and its density is $D=3.63g/mL$ .

Isolating $V$ from (1):

$V=\frac{m}{D}$ (4)

$V=\frac{4.71g}{3.63g/mL}$ (5)

$V=1.297mL$ (5)

<h2>Answer C:</h2>

In this case the volume of a sample is $V=0.293mL$ and its density is $D=0.930g/mL$ .

Isolating $m$ from (1):

$m=D.V$ (6)

$m=(0.930g/mL)(0.293mL)$ (7)

$m=0.272g$ (8)

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2 years ago

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A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

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2 years ago

Use the equation for magnetic force on a moving charge to derive the equation for magnetic force on a current carrying wire. Sho

max2010maxim [7]

Answer:

The formula comes from Lorentz force law which includes both the electric and magnetic field. If the electric field is zero, the force law for just the magnetic field is <u>F=q(ν×B</u>) . Here, F is force and is a vector because the force acts in a direction. q is the charge of the particle. v is velocity and is a vector because the particle is moving in some direction. B is the magnetic flux density.

We can derive an expression for the magnetic force on a current by taking a sum of the magnetic forces on individual charges. (The forces add because they are in the same direction.) The force on an individual charge moving at the drift velocity vd. Since the magnitude of B is constant at every line element of the loop (circle) and it dot product with the line element is B dl everywhere, therefore

∮B dl=μ0 I

B ∮dl=μ0 I

B 2πr=μ0 I

B=μ02πr Id=μ0/4π I dl×rr3

Since, r can be written as r=(rcosθ,rsinθ,z) and dl as dl=(dl,0,0) And now, if we take the cross product we would get

dl×r=−z dlj^+rsinθk^

and therefore the magnitude of dB is equal to

dB=μ0/4π I |dl×r|/r3=μ0/4π I z2+r2sin2θ−−−−−−−−−−√dl/r3

Thus, magnetic field is depending on r,θ,z.

Learn more about Force here-

brainly.com/question/2855467

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1 year ago