Is the speed of light faster in helium or air?

What are the main causes of the convention currents in the asthenosphere?

Tju [1.3M]

I think it is c density and temperature

8 0

3 years ago

A 5.75 mm high firefly sits on the axis of, and 11.3 cm in front of, the thin lens A, whose focal length is 5.77 cm . Behind len

weeeeeb [17]

Answer

given,

focal length of lens A = 5.77 cm

focal length of lens B= 27.9 cm

flies distance from mirror = 11.3 m

now,

Using lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q}$

$\dfrac{1}{5.77} = \dfrac{1}{11.3} + \dfrac{1}{q}$

q =11.79 cm

image of lens A is object of lens B

distance of lens = 59.9 - 11.79 = 48.11

now, Again applying lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q'}$

$\dfrac{1}{27.9} = \dfrac{1}{48.11} + \dfrac{1}{q'}$

q' =66.41 cm

hence, the image distance from the second lens is equal to q' =66.41 cm

6 0

2 years ago

A generator produces 60 A of current at 120 V. The voltage is usually stepped up to 4500 V by a transformer and transmitted thro

aalyn [17]

Answer:

The percentage power lost in the transmission line if the voltage not stepped up is 50%.

Explanation:

Given that,

Current = 60 A

Voltage = 120 V

Resistance = 1.0 ohm

We need to calculate the power

Using formula of power

$P=I\times V$

Where,I =current

V = voltage

Put the value into the formula

$P=60\times120$

$P=7200\ W$

We need to calculate the percentage power lost in the transmission line

If the voltage is not stepped up

Then, the power loss

$P'=I^2\times R$

Put the value into the formula

$P'=(60)^2\times1$

$P'=3600\ W$

The percentage power loss P''

$P''=\dfrac{P'}{P}\times100=\dfrac{3600}{7200}\times100$

$P''=50\%$

Hence, The percentage power lost in the transmission line if the voltage not stepped up is 50%.

5 0

2 years ago

Interactive Solution 9.37 presents a method for modeling this problem. Multiple-Concept Example 10 offers useful background for

viva [34]

Answer:

21.67 rad/s²

208.36538 N

Explanation:

$\omega_f$ = Final angular velocity = $\dfrac{1}{6}78=13\ rad/s$

$\omega_i$ = Initial angular velocity = 78 rad/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation

t = Time taken

r = Radius = 0.13

I = Moment of inertia = 1.25 kgm²

From equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{13-78}{3}\\\Rightarrow \alpha=-21.67\ rad/s^2$

The magnitude of the angular deceleration of the cylinder is 21.67 rad/s²

Torque is given by

$\tau=I\alpha\\\Rightarrow \tau=1.25\times -21.67\\\Rightarrow \tau=-27.0875$

Frictional force is given by

$F=\dfrac{\tau}{r}\\\Rightarrow F=\dfrac{-27.0875}{0.13}\\\Rightarrow F=-208.36538\ N$

The magnitude of the force of friction applied by the brake shoe is 208.36538 N

5 0

3 years ago

One of the basic rules of science is

adoni [48]

Geocentric means earth centered
So answer is C - the earth in no longer believed to be the center of the solar system

6 0

1 year ago