All categories

3 years ago

Are most main sequence stars larger or smaller than the sun?

1 answer:

3 0

Answer: Main sequence stars fuse hydrogen atoms to form helium atoms in their cores. About 90 percent of the stars in the universe, including the sun, are main sequence stars. These stars can range from about a tenth of the mass of the sun to up to 200 times as massive.

You might be interested in

As an object is undergoing free fall motion.As it falls the objects?,speed increases,acceleration increases,both of these,or non

Alex_Xolod [135]

The speed will increase, as there is acceleration, while the acceleration will remain constant, as gravity is constant.

4 0

3 years ago

During the contraction of the heart, 65 cm3 blood is ejected from the left ventricle into the aorta with a velocity of approxima

vova2212 [387]

Answer:

The value $F = 0.1396 \ N$

Explanation:

From the question we are told that

The volume blood ejected is $V = 65 \ cm^3 = 65*10^{-6} \ m^3$

The velocity of the blood ejected is $v = 103 \ cm/s = \frac{103}{100} = 1.03 \ m/s$

The density of blood is $\rho = 1060 \ kg/m^3$

The heart beat is $R = 59 \ bpm(beats \ per \ minute) = \frac{59}{60}= 0.9833\ bps$

The average force exerted by the blood on the wall of the aorta is mathematically represented as

$F = 2 * \rho * V * R * v$

=> $F = 2 * 1060 * 65*10^{-6} * 0.9833 * 1.03$

=> $F = 0.1396 \ N$

8 0

3 years ago

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 55.0 cm. She finds the pendulum m

forsale [732]

Answer:

Acceleration due to gravity will be $g=5.718m/sec^2$

Explanation:

We have given length of pendulum l = 55 cm = 0.55 m

It is given that pendulum completed 100 swings in 145 sec

So time taken by pendulum for 1 swing $=\frac{145}{100}=1.45sec$

We have to find the acceleration due to gravity at that point

We know that time period of pendulum;um is given by

$T=2\pi \sqrt{\frac{l}{g}}$

So $1.45=2\times 3.14\times \sqrt{\frac{0.55}{g}}$

$\sqrt{\frac{0.55}{g}}=0.230$

Squaring both side

${\frac{0.3025}{g}}=0.0529$

$g=5.718m/sec^2$

So acceleration due to gravity will be $g=5.718m/sec^2$

3 0

3 years ago

Newly discovered planet has twice the mass and three times the radius of the earth. What is the free-fall acceleration at its su

skad [1K]

Answer:

$g_n=\dfrac{2}{9}g$

Explanation:

M = Mass of Earth

G = Gravitational constant

R = Radius of Earth

The acceleration due to gravity on Earth is

$g=\dfrac{GM}{R^2}$

On new planet

$g_n=\dfrac{G2M}{(3R)^2}\\\Rightarrow g_n=\dfrac{2GM}{9R^2}$

Dividing the two equations we get

$\dfrac{g_n}{g}=\dfrac{\dfrac{2GM}{9R^2}}{\dfrac{GM}{R^2}}\\\Rightarrow \dfrac{g_n}{g}=\dfrac{2}{9}\\\Rightarrow g_n=\dfrac{2}{9}g$

The acceleration due to gravity on the other planet is $g_n=\dfrac{2}{9}g$

4 0

3 years ago

Read 2 more answers

The radius of Saturn is about 10 times the radius of Venus and the mass is about 100 times that of Venus. How much larger is the

lora16 [44]

let the mass of Venus is M then mass of Saturn is 100 M

similarly if the radius of Venus is R then the radius of Saturn is 10 R

now the force of gravity on a man of mass "m" at the surface of Venus is given by

$F_1 = \frac{GMm}{R^2}$

now similarly the gravitational force on the man if he is at the surface of Saturn

$F_2 = \frac{G*100M*m}{(10R)^2}$

$F_2 = \frac{GMm}{R^2}$

so here if we divide the two forces

$\frac{F_1}{F_2} = 1$

so here we can say

F1 = F2

so on both planets the gravitational force will be same

7 0

3 years ago