What units would you use to measure the amount of liquid inside a can of soda?

If the star Sirius emits 23 times more energy than the Sun, why does the Sun appear brighter in the sky?

Ganezh [65]

Answer:

As b ∝ (L/r²) and

the distance of the sun from the earth is 0.00001581 light years

and

the distance of the Sirius from the earth is 8.6 light years

hence,

the Sun appear brighter in the sky

Explanation:

The brightness (b) is directly proportional to the Luminosity of the star (L) and inversely proportional to the square of the distance between the star and the observer (r).

thus, mathematically,

b ∝ (L/r²)

now,

given

L for sirius is 23 times more than the sun i.e 23L

now,

the distance of the sun from the earth is 0.00001581 light years

and

the distance of the Sirius from the earth is 8.6 light years

thus,

using the the relation between conclude that the value of brightness for the Sirius comes very very low as compared to the value for brightness for the Sun.

hence, the sun appears brighter

5 0

2 years ago

Let theta denote the angular displacement of a simple pendulum oscillating in a vertical plane. If the mass of the Bob is m, the

posledela

The tension has to hold the part of the weight in the direction of the string:

T = mg*cos(theta)

Theta=0, whole weight, theta=90, T=0, if the pendulum is horizontal, the string will be loose! Yeah

8 0

2 years ago

Read 2 more answers

Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150

Sever21 [200]

Answer: $2.89\approx 2.9^{\circ}C/s$

Explanation:

Given

$Power\left ( P\right )=150 MW$

mass of core $\left ( m\right )=1.60\times 10^5 kg$

Average specific heat $\left ( C\right )=0.3349 KJ/kg^{\circ}C$

And rate of increase of temperature = $\frac{\mathrm{d}T}{\mathrm{d} t}$

Now

P= $mc\frac{\mathrm{d}T}{\mathrm{d} t}$

$150\times 10^6=1.60\times 10^5\times 0.3349\times \frac{\mathrm{d}T}{\mathrm{d} t}$

Thus $\frac{\mathrm{d}T}{\mathrm{d} t}=[tex]\frac{1.60\times 10^5\times 0.3349}{150\times 10^6}$

$\frac{\mathrm{d}T}{\mathrm{d} t}=2.89\approx 2.9^{\circ}C/s$

6 0

3 years ago

Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's m

Reil [10]

Answer:

$\Delta l=0.015m$

Explanation:

We have given initial length of the steel guitar l = 1 m

Cross sectional area $A=0.5mm^2=0.5\times 10^{-6}m^2$

Young's modulus $\gamma=2\times 10^{11}Pa$

Force F = 1500 N

So stress $=\frac{force}{area}=\frac{1500}{0.5\times 10^{-6}}=3000\times 10^{-6}=3\times 10^{9}Pa$

We know that young's modulus $=\frac{stress}{strain}$

So $2\times 10^{11}=\frac{3\times 10^{9}}{strain}$

$strain=1.5\times 10^{-2}=0.015m$

Now strain $=\frac{\Delta l}{l}$

$0.015=\frac{\Delta l}{1}$

$\Delta l=0.015m$

6 0

2 years ago

Read 2 more answers

If the vertical axis and time on the horizontal axis if the speed is steadily increasing could the speef line ever become perfec

Phantasy [73]

A graph of real speed can have a section that's as steep as you want,
but it can never be a perfectly vertical section.

Any vertical line on a graph, even it it's only a tiny tiny section, means
that at that moment in time, the speed had many different values.

It also means that the speed took no time to change from one value to
another, and THAT would mean infinite acceleration.

8 0

2 years ago