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AlexFokin [52]
2 years ago
9

Why do meteors in a meteor shower appear to come from just one point in the sky?

Physics
1 answer:
kolbaska11 [484]2 years ago
4 0

Answer:

All the meteors in a meteor shower typically come from the same comet, which is in a periodic orbit around the Sun. The dust from the comet follows the path of the parent body and as Earth intersects the debris along the comet's orbit, it plows into this material. Since the dust particles are all moving together in space before they encounter Earth, if you trace back the streak of each meteor, its path will appear to diverge from one place in the sky.

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8.4-1 Consider a magnetic field probe consisting of a flat circular loop of wire with radius 10 cm. The probe’s terminals corres
Vlad1618 [11]

Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T

hence, B_o = 1.013μT

6 0
2 years ago
An 80-kg person stands at one end of a 130-kg boat. He then walks to the other end of the boat so that the boat moves 80 cm with
a_sh-v [17]

Explanation:

We assume that length of boat is L cm. And, we suppose to have a coordinate system relative to the bottom of the lake with:

Boat's position initially from x = 0 cm to x = L cm

Let person (S) is initially at x = L cm (right hand end of boat)

and, let boat's center of mass at x = C cm

Therefore, initially center of mass of overall (system's) is at X_{1} where (taking moments about x = 0).

             (80 + 130)X_{1}

             = 130C + 80L ........... (i)  

After S walks left to the other end of the boat, the boat moves right 80 cm.

Hence,  

Boat's position is now from x = 80 cm to x = (L+ 80) cm

S's position is x = 80 (left end of boat)

Boat's center of mass is at (C + 80).

Therefore, overall (system's) final center of mass is at X_{2} where (taking moments about x = 0).

                (80 + 130)X_{2}

             = (130)(C + 80) + 80 \times 80 ............. (ii)

As no external force is acted on the system, the system's center of mass has not moved, so X_{1} = X_{2}.

This means the left hand sides of both equations (i) and (ii) are equal. So, we can equate the right hand sides giving: the equation as follows.

                130C + 80L = (130)(C+80) + 80 \times 80


                           80L = 16800 cm

                           L = 210 cm

                             = 2.1 m

Length of boat = 2.1 m

When the person is walking from one end to the other as there was no external force then center of mass of the person-boat system did not move.

6 0
2 years ago
A device that uses electricity and magnetism to create motion is called a (motor magnet generator) . In a reverse process, a dev
Tems11 [23]

the answers are a. and c.

I hope I help you out!

6 0
2 years ago
Read 2 more answers
What factors affect the ability of a substance to transfer thermal energy to heat or be heated by it's surroundings?
xeze [42]

The prime factors that affect the ability of substances to transfer the thermal energy to heat are the temperature difference between the two objects, area of cross-section, time, and distance travelled by the thermal energy.

<u>Explanation: </u>

The process of heat conduction takes place through contact between two or more objects. But this conduction depends on multiple factors that are responsible for thermal conduction. They are-

  • Temperature Difference(\Delta T) - The two objects must have a temperature difference else there will be no thermal conduction between them. The more the difference in their temperatures, the more thermal energy flows from one object to the other.
  • Area of Cross-section (A) - Larger areas of contact provide as better medium of thermal conduction.
  • Time (t) - The more time we give for the thermal conduction, the more energy is transmitted from one system to the other.
  • Distance Travelled (l) - The longer the distance, lesser the conduction. Means, the distance should be minimized in order to achieve the optimum thermal conduction between two objects.

Consider metal pot and its handle, it is being boiled for 15 m. The molecules present near the source of heat, showing fast vibration and bounce off. It actually indicates the heats of substance. That’s why, handle remains hot as heat conduction takes place.  It can be estimated by,

                                       Q=\frac{k A \Delta T t}{l}

k - Thermal conductivity of the material, measured in J/s.m.^{\circ} \mathrm{C}

4 0
2 years ago
Water is boiled at 300 kPa pressure in a pressure cooker. The cooker initially contains 3 kg of water. Once boiling started, it
Inessa [10]

Answer:

The average rate of energy transfer to the cooker is 1.80 kW.

Explanation:

Given that,

Pressure of boiled water = 300 kPa

Mass of water = 3 kg

Time = 30 min

Dryness friction of water = 0.5

Suppose, what is the average rate of energy transfer to the cooker?

We know that,

The specific enthalpy of evaporate at 300 kPa pressure

h_{f}=561.47\ kJ/kg

h_{fg}=2163.8\ kJ/kg

We need to calculate the enthalpy of water at initial state

h_{1}=h_{f}

h_{1}=561.47\ kJ/kg

We need to calculate the enthalpy of water at final state

Using formula of enthalpy

h_{2}=h_{f}+xh_{fg}

Put the value into the formula

h_{2}=561.47+0.5\times2163.8

h_{2}=1643.37\ kJ/kg

We need to calculate the rate of energy transfer to the cooker

Using formula of rate of energy

Q=\dfrac{m(h_{2}-h_{1})}{t}

Put the value into the formula

Q=\dfrac{3\times(1643.37-561.47)}{30\times60}

Q=1.80\ kW

Hence, The average rate of energy transfer to the cooker is 1.80 kW.

3 0
2 years ago
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