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4 years ago

What is a bright streak of light that results when a meteoroid burns up in earth's atmosphere?

1 answer:

4 0

<span>When a meteoroid passes through Earth's atmosphere it's traveling very fast. The friction of the air makes gets the surface so hot it begins to burn or glow red</span>

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Consider a air-filled parallel-plate capacitor with plates of length 8 cm, width 5.52 cm, spaced a distance 1.99 cm apart. Now i

prohojiy [21]

Answer:

The ratio of the new potential energy to the potential energy before the insertion of the dielectric is 0.58

Explanation:

Given that,

Length of plates = 8 cm

Width = 5.52 cm

Distance = 1.99 cm

Dielectric constant = 2.6

Length = 4.4 cm

Potential = 0.8 V

We need to calculate the initial capacitance

Using formula of capacitance

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times8\times5.52\times10^{-4}}{1.99\times10^{-2}}$

$C=1.96\times10^{-12}$

We need to calculate the final capacitance

Using formula of capacitance

$C'=\dfrac{\epsilon_{0}A_{1}}{d}+\dfrac{k\epsilon_{0}A_{2}}{d}$

Put the value into the formula

$C'=(\dfrac{8.85\times10^{-12}}{1.99\times10^{-2}})((4.4\times5.52)+(3.6\times5.52)2.6)\times10^{-4}$

$C'=3.37\times10^{-12}$

We need to calculate the ratio of the new potential energy to the potential energy before the insertion of the dielectric

Using formula of energy

$\dfrac{E}{E'}=\dfrac{\dfrac{1}{2}CV^2}{\dfrac{1}{2}C'V^2}$

Put the value into the formula

$\dfrac{E}{E'}=\dfrac{1.96\times10^{-12}}{3.37\times10^{-12}}$

$\dfrac{E}{E'}=0.58$

Hence, The ratio of the new potential energy to the potential energy before the insertion of the dielectric is 0.58

4 0

3 years ago

A wheel has a rotational inertia of 16 kgm2. Over an interval of 2.0 s its angular velocity increases from 7.0 rad/s to 9.0 rad/

german

Answer:

<h2>128.61 Watts</h2>

Explanation:

Average power done by the torque is expressed as the ratio of the workdone by the toque to time.

Power = Workdone by torque/time

Workdone by the torque = $\tau \theta$ = $I\alpha * \theta$

I is the rotational inertia = 16kgm²

$\theta = angular\ displacement$

$\theta = 2 rev = 12.56 rad$

$\alpha \ is \ the\ angular\ acceleration$

To get the angular acceleration, we will use the formula;

$\alpha = \frac{\omega_f^2- \omega_i^2}{2\theta}$

$\alpha = \frac{9.0^2- 7.0^2}{2(12.54)}\\\alpha = 1.28\ rad/s^{2}$

Workdone by the torque = 16 * 1.28 * 12.56

Workdone by the torque = 257.23 Joules

Average power done by the torque = Workdone by torque/time

= 257.23/2.0

= 128.61 Watts

8 0

3 years ago

An electric heater used to boil small amounts of water consists of a 15-Ω coil that is immersed directly in the water. It operat

Allisa [31]

Answer:

t = 1120 seconds

Explanation:

Law of Conservation of Energy: It states that energy can neither be created or destroyed but it can be transformed from one form to another

The heater converts electric energy to heat energy.

According to the law of conservation of Energy,

Heat supplied by the heater = heat gained by the water.

Heat supplied by the heater = (V²/R)t, Where V = voltage, R = Resistance, t = Time.

Heat gained by the water = cm(θ₂ -θ₁)

Where c = specific heat capacity of water, m = mass of water, θ₂= final temperature of water, θ₁= initial temperature of water.

∴ (V²/R)t = cm(θ₂ -θ₁)

Making t the subject of the equation,

t = cm(θ₂ -θ₁) ×R/V²

Where c= 4200 Jkg/K, m=0.80 kg, R = 15 Ω, V = 60 V, θ₁= 20°C, θ₂ = the normal boiling point of water = 100°C

∴ t = (4200×0.8)(100 -20)×15/60²

t = (3360×80×15)/3600

t = 4032000/3600 = 1120 seconds

∴ The time required by the heater to raise the temperature of water to normal boiling point is 1120 seconds.

3 0

3 years ago

Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 3.44 kg and rotate with

Free_Kalibri [48]

Answer:

(a) 20,154.1 J

(b) 95,223.5 J

Explanation:

The expression for the moment of inertia for the uniform solid cylinder is as follows;

$I= \frac{1}{2}mr^{2}$

Here, I is the moment of inertia, r is the radius and m is the mass of the object.

The expression for the rotational kinetic energy is as follows;

$K= \frac{1}{2}I\omega ^{2}$

Here, K is the rotational kinetic energy and $\omega$ is the angular velocity.

(a)

Calculate the moment of inertia of the smaller solid cylinder.

$I= \frac{1}{2}mr^{2}$

Put m= 3.44 kg and r= 0.356 m.

$I= \frac{1}{2}(3.44)(0.356)^{2}$

$I= 0.218 kg m^{2}$

Calculate the rotational kinetic energy of the smaller cylinder.

$K= \frac{1}{2}I\omega ^{2}$

Put $I= 0.218 kg m^{2}$ and $\omega = 430 rads^{-1}$ .

$K= \frac{1}{2}(0.218)(430) ^{2}$

K= 20,154.1 J

Therefore, the rotational kinetic energy for the smaller cylinder is 20,154.1 J.

(b)

Calculate the moment of inertia of the larger solid cylinder.

$I= \frac{1}{2}mr^{2}$

Put m= 3.44 kg and r= 0.775 m.

$I= \frac{1}{2}(3.44)(0.775)^{2}$

$I= 1.03 kg m^{2}$

Calculate the rotational kinetic energy of the smaller cylinder.

$K= \frac{1}{2}I\omega ^{2}$

Put $I= 1.03 kg m^{2}$ and $\omega = 430 rads^{-1}$ .

$K= \frac{1}{2}(1.03)(430) ^{2}$

K= 95,223.5 J

Therefore, the rotational kinetic energy for the larger cylinder is 95,223.5 J.

7 0

4 years ago

Heat is transfered from the heating elements to the pot

Varvara68 [4.7K]

Hear is transferred from the heating elements to the Pot by Conductivity

8 0

4 years ago