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3 years ago

An energy transformation flow diagram is shown.

Physics

1 answer:

Elza [17]3 years ago

3 0

C is the answer Because chemical energy Create heat And thermal power relates to heat energy

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The electric field on the surface of an irregularly shaped conductor varies from 74.0 kN/C to 14.0 kN/C.

mrs_skeptik [129]

Answer:

(a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Explanation:

Given that,

Electric field $E_{1}=74.0\ kN/C$

Electric field $E_{2}=14.0\ kN/C$

When the radius of curvature is greatest, the electric field at the surface will be smaller.

Where the radius of curvature is greatest

(a). We need to calculate the local surface charge density at the point on the surface

Using formula of charge density

$\sigma=\epsilon_{0}E_{2}$

Put the value into the formula

$\sigma=8.85\times10^{-12}\times14\times10^{3}$

$\sigma=1.239\times10^{-7}\ C/m^2$

$\sigma=123.9\times10^{-9}\ C/m^2$

$\sigma=123.9\ nC/m^2$

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). We need to calculate the local surface charge density at the point on the surface where the radius of curvature of the surface is smallest

Using formula of charge density

$\sigma=\epsilon_{0}E_{1}$

Put the value into the formula

$\sigma=8.85\times10^{-12}\times74\times10^{3}$

$\sigma=6.549\times10^{-7}\ C/m^2$

$\sigma=654.9\times10^{-9}\ C/m^2$

$\sigma=654.9\ nC/m^2$

The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

Hence, (a). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 123.9 nC/m².

(b). The local surface charge density at the point on the surface where the radius of curvature of the surface is greatest is 654.9 nC/m².

7 0

3 years ago

Why do we warm up when we wear warm clothes? Explain the hiccup from the point of view of physics.

bazaltina [42]

Well your body would be trapping heat when your cold your blood contricts to help hold in heat so it would help keep your body tempture up and keep heat from escaping

5 0

4 years ago

Please help me! Thank you!

amid [387]

Question 1. To solve this we are going to take advantage of the fact that 1 horsepower = 745.7 Watts, so to convert watts to horsepower, we just need to multiply by the conversion factor $\frac{1horsepower}{747.5Watts}$ .
Lets convert the power of our refrigerator from Watts to horsepower:
$500Watts* \frac{1hosepower}{747.5Watts} =0.67horsepower$

We can conclude that the power rating of a 500 W refrigerator in horsepower is 0.67 horsepower.

Question 2. An electric field tends to be strongest at the ends of pointed objects; the strongest the electric field the more charge it builds up, so at the ends of pointed objects there is enough charge to strip the atoms in the air form its electrons producing plasma. That glowing of that plasma is what we call St. Elmo's fire.

Question 3.
- Even though both phenomena are static discharges, they nature is completely different. Lightning is an electric discharge: a discharge that occurs when opposite charges accumulate until the electric field becomes strong enough to allow a current to flow. St. Elmo's fire, on the other hand, is a coronal discharge: a luminous phenomena, similar to a neon tube, that occurs when a pointed object in a strong electric field creates plasma.
- Even tough their origin an behavior are different, both of them create plasma in the air.
- Lightening is extremely hot, about the same temperature as the surface of the sun, whereas St. Elmo's fire is relatively cold.

Question 4. Since the glow in the neon tube is actually a coronal discharge that creates red plasma -due to the nature of the gas, it is basically a red St. Elmo's fire inside a tube. if we replace the neon gas with plasma, we will basically create a St. Elmo's fire inside the tube; the air inside the tube will be now a mixture of oxygen and nitrogen, so it will emit a bluish glow like St. Elmo's fire.

Question 5. St. Elmo's fire only appear during thunderstorms because it is the only time when electrons accumulate on the bottoms of clouds and induce a positive charge in the ground creating a strong electric field. The coronal discharge that creates the St. Elmo fire can only happen whiting a strong local electric field, so it can only happen during thunderstorms.

Question 6. No, air is not a good conductor of electric charge; otherwise phenomena like Sr Elmo's fire or lightening will occur outside thunder storms, which is not the case. Those phenomena only occur whiting strong electric fields, and those electric fields are only present during thunder storms. If air was a good conductor of electric charge we couldn't use electricity without killing ourselves in the process.

4 0

3 years ago

A toy cannon tosses a rubber ball straight upward. A motion sensor measures the speed of the ball as it leaves the cannon. Using

mezya [45]

Answer:

With this information is not possible to calculate the mass.

Explanation:

This is a characteristic problem of energy conservation, where kinetic energy becomes potential energy. For this particular problem, we have the initial speed as input data. The moment the ball comes out of the cannon we have the maximum kinetic energy, as the ball goes up the ball will gain more potential energy as the ball loses kinetic energy, until the moment the ball reaches the maximum height. At the maximum height point, the ball will have its maximum potential energy while its kinetic energy is zero. In other words, all the kinetic energy that was, in the beginning, was transformed into potential energy.

$E_{p}=E_{k}\\m*g*h=0.5*m*v^2$

In the above equation the masses are canceled and we can determine the maximum height, by means of the initial speed.

$h=\frac{0.5*v^2}{g} [m]$

But the mass cannot be determined, since it would be necessary to know the value of the energy, in order to determine the value of the mass.

3 0

3 years ago

A mass m at the end of a spring vibrates with a frequency

Wittaler [7]

Answer:

m = 0.59 kg.

Explanation:

First, we need to find the relation between the frequency and mass on a spring.

The Hooke's law states that

$F = -kx$

And Newton's Second Law also states that

$F = ma = m\frac{d^2x}{dt^2}$

Combining two equations yields

$a = -\frac{k}{m}x$

The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.

$\omega = \sqrt{\frac{k}{m}}$

And given that ω = 2πf

the relation between frequency and mass becomes

$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ .

Let's apply this to the variables in the question.

$0.88 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\\0.60 = \frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}\\\frac{0.88}{0.60} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{m}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}}\\1.4667 = \frac{\sqrt{m+0.68}}{\sqrt{m}}\\2.15m = m + 0.68\\1.15m = 0.68\\m = 0.59~kg$

6 0

3 years ago