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3 years ago

QUESTION 6

Physics

2 answers:

Daniel [21]3 years ago

5 0

Question 7 is false

mr Goodwill [35]3 years ago

4 0

Answer: 1. look past external characteristics and get to know more about people

2. False

Explanation:

1. The learning the past external characteristics and get to know about people who belong to diverse culture and their contribution for the overall society, region, state and nation is likely to bring the spirit of brotherhood. This will raise the feeling of unity and will help in adapting diverse culture.

2. The students of college are likely to get stress due to increase in complexity in studies. Some students also participate in extra-curricular activities along with studies which makes their schedule hectic and tiring this bring stress to their lives.

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In the 1887 experiment by Michelson and Morley, the length of each interferometer arm was 11m. The experimental limit on the mea

Vikentia [17]

For the answer to the question above,
we can get the number of fringes by dividing (delta t) by the period of the light (Which is λ/c).

fringe = (delta t) / (λ/c)

We can find (delta t) with the equation:

delta t = [v^2(L1+L2)]/c^3

Derivation of this formula can be found in your physics text book. From here we find (delta t):

600,000^2 x (11+11) / [(3x10^8)^3] = 2.93x10^-13

2.93x10^-13/ (589x10^-9 / 3x10^8) = 149 fringes

This answer is correct but may seem large. That is because of your point of reference with the ether which is usually at rest with respect to the sun, making v = 3km/s.

4 0

3 years ago

What is the heat needed to raise the temperature of 24.7 kg silver from 14.0 degrees celsius to 30.0 degrees celsius? specific h

Citrus2011 [14]

The amount of heat needed to increase the temperature of a substance by $\Delta T$ is given by
$Q=m C_s \Delta T$
where
m is the mass of the substance
$C_s$ is its specific heat capacity
$\Delta T$ is the increase of temperature

The sample of silver of our problem has a mass of $m=24.7 kg$ . Its specific heat capacity is $C_s = 236 J/g^{\circ}C$ and the increase in temperature is
$\Delta T=30.0^{\circ}-14.0^{\circ}C=16.0^{\circ}C$

Therefore, the amount of heat needed is
$Q=mC_s \Delta T=(24.7 kg)(236 J/g^{\circ}C)(16.0^{\circ}C)=9.32 \cdot 10^4 J$

8 0

4 years ago

A size-5 soccer ball of diameter 22.6 cm and mass 426 g rolls up a hill without slipping, reaching a maximum height of 5.00 m ab

Drupady [299]

Answer

According the conservation of energy

$\dfrac{1}{2}mv_i^2+\dfrac{1}{2}I\omega_i^2+0 = mgh + 0$

I for ball = $\dfrac{2}{3}mr^2$

$\dfrac{1}{2}mv_i^2+\dfrac{1}{2} \dfrac{2}{3}mr^2\omega_i^2= mgh$

$\omega_i = \dfrac{v_i}{r}$

$v_i^2+\dfrac{2}{3}r^2(\dfrac{v_i}{r})^2 = 2gh$

$v_i^2+\dfrac{2}{3}v_i^2 = 2gh$

$v_i^2+[1+\dfrac{2}{3}]=2gh$

$v_i^2\dfrac{5}{3}=2gh$

$v_i=\sqrt{\dfrac{6gh}{5}}=\sqrt{\dfrac{6\times 9.8\times 5}{5}}$

$v_i = 7.67\ m/s$

a) $\omega_i = \dfrac{v_i}{R}$

$\omega_i = \dfrac{7.67}{0.113}$

$\omega_i =67.87\ rad/s$

b) $K_{rot} = \dfrac{1}{2}\dfrac{2}{3}mR^2\omega_i^2$

$K_{rot} = \dfrac{1}{3}\times 0.426\times 0.112^2\times 67.87^2$

$K_{rot} = 8.205\ J$

4 0

3 years ago

Which property of sound waves decreases as the square of the distance from the source increases?

Vitek1552 [10]

Answer:

Intensity

Explanation:

The intensity of a sound wave is equal to the ratio between to the power emitted by the source divided by the area of the spherical surface through which the wave propagates:

$I=\frac{P}{4\pi r^2}$

where

P is the power

$4\pi r^2$ is the area of the spherical surface

r is the distance from the source

As we see from the formula, the intensity is inversely proportional to the square of the distance from the source:

$I\propto \frac{1}{r^2}$

so, intensity is the correct answer.

3 0

3 years ago

You are a geologist studying the electrical field on a given region. You find that the electric field is vertical and perpendicu

Xelga [282]

Answer:

charge Qint = 7.17 10⁻⁴ C

Explanation:

For this problem we must use Gauss's law

F = ∫ E. dA = Qint / εₙ

let's form a Gaussian surface that is parallel to the surface, for example, a Cube. As the field is vertical and perpendicular to the surface, the field lines and the area vector are parallel whereby the scalar product is reduced to an ordinary product.

Φ = E A = Qint / ε₀

A = 1 km² (1000 m / 1km)² = 1 10⁶ m²

We can calculate the charge

Qint = E A ε₀

Qint = 81 1 10⁶ 8.85 10⁻¹²

Qint = 7.17 10⁻⁴ C

5 0

3 years ago