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Fantom [35]
3 years ago
6

What is the emf produced in a 1.5 meter wire moving at a speed of 6.2 meters/second perpendicular to a magnetic field of strengt

h 3.96 × 10-3 newtons/amp·meter ? 37 volts 0.0026 volts 5.0 volts 0.37 volts 0.037 volts?
Physics
2 answers:
madam [21]3 years ago
8 0
E. 0.037 Volts. It's correct for Plato. The actual answer is around 0.0369 Volts
emmasim [6.3K]3 years ago
3 0

Answer:

0.037 volts

Explanation:

The formula for calculating the Electromagnetic field induction in a moving wire is:

Emf=B*l*v*sin(\theta )

Where:

B is the strength of the magnetic field in Webers or \frac{N*m}{A}

l is the longitude of the wire in meters

v is the speed of the wire crossing the magnetic field in meters per second (m/s)

and \theta is the angle between the magnetic field and the wire

So, substituting the values into the formula:

Emf = 3.96 * 10^{-3} W * 1.5 m * 6.2 \frac{m}{s} * sin(90)

sin(90)=1 so:

Emf = 3.96 * 10^{-3} W * 1.5 m * 6.2 \frac{m}{s} = 0.0368 volts ≅ 0.037 volts

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Question is from B to C

Answer: (b) 1.5m/s

x1=3m, x2=9m
t1=1s, t2=5s
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Time elapsed, ∆t=(5-1)s=4s

So average velocity v =∆x/∆t=6/4=1.5m/s
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2 years ago
Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
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Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

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Answer:

A) Its density will decrease

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When an object is heated, its volume increases. This is due to the fact that the particles in the medium vibrate more (if it is a solid) or they move more (if it is a liquid or a gas), therefore they tend to occupy a larger space.

At the same time, the mass of the object does not change, because the mass just represents the amount of matter contained in the object, so it does not increase/decrease at different temperatures.

The density of an object is defined as the ratio between the mass (m) and the volume (V):

d=\frac{m}{V}

We said that the mass remains unchanged while the volume increases: since the density is inversely proportional to the volume, this means that the density decreases.

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Answer:

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d₂ = 7 000 km = 7·10⁶ м

<em>Trial  3</em>

M₃ = 3·10²² kg

d₃ = 7 000 km = 7·10⁶ м

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F - ?

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F₁ = G·m·M₁ / d₁² = m·6.67·10⁻¹¹·6·10²² / (3.5·10⁶)² = 0.37·m  (N)

F₂ = G·m·M₂ / d₂² = m·6.67·10⁻¹¹·6·10²² / (7·10⁶)² = 0.08·m  (N)

F₃ = G·m·M₃ / d₃² = m·6.67·10⁻¹¹·3·10²² / (7·10⁶)² = 0.04·m  (N)

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