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Sphinxa [80]
3 years ago
6

(a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500 - μC charge and flies due west at a sp

eed of 660 m/s over the Earth’s south magnetic pole, where the 8.00 10 -T 5  magnetic field points straight up. What are the direction and the magnitude of the magnetic force on the plane? (b) Discuss whether the value obtained in part (a) implies this is a significant or negligible effect.
Physics
1 answer:
12345 [234]3 years ago
8 0

(a) 2.64\cdot 10^{-8} N north

We can treat the aircraft as a single point charge moving in a magnetic field. In this case, the magnetic force exerted on the plane is

F=qvB sin \theta

where

q=0.500 \mu C = 0.500\cdot 10^{-6} C is the charge on the plane

v = 660 m/s is the velocity

B=8.00\cdot 10^{-5} T is the magnitude of the magnetic field

\theta=90^{\circ} is the angle between the direction of motion of the jet and of the magnetic field

Substituting,

F=(0.5\cdot 10^{-6})(660)(8.0\cdot 10^{-5})=2.64\cdot 10^{-8} N

The direction can be found by using Fleming's left hand rule. We have:

- index finger: magnetic field direction (straight up)

- middle finger: velocity of the plane (due west)

- force: thumb --> north

(b) Not negligible

As we can see from part (a), the magnitude of the force is not really big, so the effects are negligible.

For instance, we can compare this force with the weight of a plane. If we take a Boeing 737, its mass is about 80,000 kg, so its weight is

W=mg=(80000)(9.8)=784,000 N

As we can see, this is several orders of magnitude bigger than the magnetic force calculated at point (a), so the effects of the magnetic force are negligible.

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A leaky 12-kg bucket is lifted from the ground to a height of 15 m at a constant speed with a rope that weights 0.75 kg/m. Initi
Scrat [10]

Answer:

Approximately 5.5 kJ.

Explanation:

  • The mass of water in the bucket is \rm 40\; kg at a height of 0 meters above the ground.
  • At a height of 15 meters, the mass of the water would be \rm 0\; kg.

The bucket goes up at a constant rate. Also, the water leaks at a constant rate. As a result, the mass of water in the bucket changes at a constant rate as height increases.

Find the rate at which the mass of water decreases over height:

\begin{aligned} &\text{Rate at which water leaks}\cr  &= \frac{\text{Change in mass of Water}}{\text{Change in height}} \cr &= \rm \frac{40 \; kg - 0\; kg}{15\; m} \cr &= \rm \frac{40}{15}\; kg \cdot m^{-1}\end{aligned}.

In other words, \displaystyle \rm \frac{40}{15}\; kg of water leaks out of the bucket everytime its height increases by \rm 1\; m.

Hence, the mass of water at h\; \rm m would be \displaystyle 40 - \frac{40}{15}\, h \rm kg. Write that as a function of height:

\displaystyle m_\text{water}(h) = 40 - \frac{40}{15}\, h = \frac{40}{15} \, (15 - h).

On the other hand, (40 - h)\; \rm m of rope is still in the air while the bucket is at a height of h\; \rm m. The mass of the rope that needs to be lifted at would thus be 0.75\; (15 - h)\; \rm kg.

Write that also as a function of the height of the bucket:

\displaystyle m_\text{rope}(h) = 0.75\; (15 - h).

Add the mass of the bucket:

\displaystyle m(h) = \frac{40}{15}\, (15 - h) + 0.75\; (15 - h) + 12.

Simplify this expression:

\begin{aligned} m(h) &= \frac{40}{15}\, (15 - h) + 0.75\; (40 - h) + 12 \cr &= \left(\frac{40}{15} + 0.75\right) (15 - h) + 12\end{aligned}.

The only two forces on the bucket are gravitational force and tension force from the rope. Since the bucket is moving at constant speed, the net force on it should be zero. The two forces must balance each other. In other words, they must act in opposite directions. Most importantly, the must be of equal magnitude. That is:

\begin{aligned} F(h) &= W(h) \cr &= m(h) \cdot g \cr &= 9.81 \times \left(\left(\frac{40}{15} + 0.75\right) (15 - h) + 12\right)\cr &= 9.81 \times \left(\frac{40}{15} + 0.75\right) (15 - h) + 9.81 \times 12 \cr &\approx 502.763 - 33.5175\; h + 117.72 \cr &= 620.483 - 33.5175\; h\end{aligned}.

Integrate force over distance travelled to find work done. In this case, since the bucket is moving upwards, distance travelled is equal to height.

\begin{aligned}W &= \int \limits_{0}^{15} (620.483 - 33.5175\; h) \, dh \cr &= \left[620.483\; h - \frac{1}{2}\times 33.5175\; h^2\right]^{15}_{0} \cr &= \left(620.483 \times 15 - \frac{1}{2} \times 33.5175 \times 15^2\right) - 0 \cr &\approx 5.5\times 10^{3}\; \rm J \cr &= 5.5\;\rm kJ \end{aligned}.

7 0
3 years ago
If 160 J of work was done to move a car and the distance the car moved was 40 m. what amount of force was used?
Monica [59]

Answer:

Force = 4N

Explanation:

Workdone = force × distance

Force = Workdone/distance

Force =160/40 = 4N

7 0
4 years ago
Seberkas cahaya monokromatik dijatuhkan pada kisi difraksi dengan 5000 goresan/cm menghasilkan spectrum garis terang orde kedua
Vlada [557]

Answer:

500 nm

Explanation:

In this problem, we have a diffraction pattern created by light passing through a diffraction grating.

The formula to find a maximum in the pattern produced by a diffraction grating is the following:

d sin \theta = m\lambda

where:

d is the distance between the lines in the grating

\theta is the angle at which the maximum is located

m is the order of the maximum

\lambda is the wavelength of the light used

In this problem we have:

\theta=30^{\circ} is the angle at which is located the 2nd-order bright line, which is the 2nd maximum

n = 5000 lines/cm is the number of lines per centimetre, so the distance between two lines is

d=\frac{1}{d}=\frac{1}{5000}=2\cdot 10^{-4} cm = 2\cdot 10^{-6} m

Re-arranging the equation for \lambda, we find the wavelength of the light used:

\lambda=\frac{d sin \theta}{m}=\frac{(2\cdot 10^{-6})(sin 30^{\circ})}{2}=5\cdot 10^{-7} m = 500 nm

4 0
3 years ago
Can anyone tell me which ones of these are true and which ones are false pls
Akimi4 [234]

There are two important facts to recall that will help answer this question:

1. The resistance of a segment of conducting wire is given by this equation:

R = ρL/A

ρ is the resistivity of the material making up the wire. This value is a constant that depends on the properties of the material. Resistivities for various materials can be found with a quick Google search.

L is the length of the wire.

A is the cross-sectional area of the wire.

From this equation you can tell that a wire's resistance will increase if it is made longer and/or thinner, and the resistance will decrease if it is made shorter and/or thicker. Mathematically speaking, the resistance is directly proportional to the length and inversely proportional to the cross-sectional area.

2. The other fact is that a conductor's resistance is also dependent on its temperature. Generally, as a conductor gets hotter, its resistance increases.


Let us now tackle the list of statements:

1. A shorter wire will allow electricity to move through at a higher rate than a longer wire.

According to the equation for a conductor's resistance, a shorter wire will have a smaller resistance.

Now recall that current is the movement of electric charges and Ohm's law:

V = IR

V is the applied potential difference between the ends of the wire.

I is the current.

R is the resistance.

Assuming you keep the potential difference constant, when you have a smaller resistance, you will have a larger current.

Statement 1 is correct.


2. A short, thick, cold wire is the best conductor.

According to the equation for a conductor's resistance, a shorter, thicker wire will have lesser resistance. A cold temperature will also help to keep the resistance low. A low resistance means a higher current.

Statement 2 is correct.


3. How well a material conducts current is an internal factor affecting resistance.

Statement 3 is correct, assuming the physical property in question is the material's resistivity. The resistivity is one of the factors in the equation for a conductor's resistance.

4. If you double the length of a wire, you cut the resistance in half.

According to the equation for a conductor's resistance, increasing the length of a wire increases the resistance. Statement 4 is false.


5. If you double the thickness of a wire, you cut the resistance in half.

According to the equation for a conductor's resistance, increasing the thickness of a wire decreases its resistance. Statement 5 is true.


6. Superconductors have no measurable resistance.

A superconductor by definition is able to conductor electric current with virtually no resistance. Statement 6 is true.


7. The higher the temperature of the conductor, the lower the resistance.

A conductor's resistance generally increases with temperature. Statement 7 is false.


8. The resistance in a wire with less thickness is less.

According to the equation for a conductor's resistance, making a wire thinner will increase its resistance. Statement 8 is false.


9. Thickness, length, and temperature are internal factors that affect resistance.

Thickness (cross-sectional area) and length are both factors in the equation determining a conductor's resistance. Temperature is also known as a factor that affects resistance. Statement 9 is true.


10. When a light is first switched on, the light bulb's filament has a lower resistance than after it gives off light for awhile.

A device that draws a current will generally heat up given sufficient time. This increases the device's resistance. Statement 10 is correct.

4 0
4 years ago
Read 2 more answers
An air mass that originates over land in Canada is most likely
liberstina [14]

The air mass that originates over land in Canada is most likely to be cold and dry

<u>Explanation:</u>

  • The air masses overland from a continent is usually Dry. The air masses are formed by the uneven warming and cooling of the surface of the earth by the sun gives rise to air masses.
  • Canada is affected by the five air masses they are Continental Arctic, Maritime Arctic, Maritime polar, Maritime tropical, and Continental Tropic.
  • As Canada is most likely affected by the Continental Arctic the air masses are cold and dry.

4 0
3 years ago
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