All categories

3 years ago

Which of the following is one of the three main fuels that a star uses for fusion

Physics

2 answers:

Colt1911 [192]3 years ago

8 0

Hydrogen, helium, and carbon.

Elanso [62]3 years ago

8 0

Answer:

OXYGEN

Explanation:

from odyssey

You might be interested in

A fixed-geometry supersonic inlet starts at a Mach number of 3. After starting, the cruise Mach number is 2, and the operating s

Hunter-Best [27]

Answer:

Explanation: see attachment below

6 0

3 years ago

A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit

Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

$M=8.7\cdot 10^{13}kg$ is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

$F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N$

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

$m\frac{v^2}{r}=G\frac{Mm}{r^2}$

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s$

3 0

3 years ago

It is easier to climb up a slanted slope than a vertical slope

V125BC [204]

IT IS EASIER TO CLIMB A SLANTED SLOPE

3 0

3 years ago

Read 2 more answers

A man 2 m tall walks horizontally at a constant rate of 1 m/s toward the base of a tower 23 m tall. When the man is 10 m from th

Evgen [1.6K]

Answer:

$\dfrac{d\theta}{dt}=0.038\ rad/s$

Explanation:

Given that

$\dfrac{dx}{dt}= -1\ m/s$

From the diagram

$tan\theta=\dfrac{21}{x}$

By differentiating with time t

$sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}$

When x= 10 m

$tan\theta=\dfrac{21}{10}$

θ = 64.53°

Now by putting the value in equation

$sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}$

$sec^264.53^{\circ} \dfrac{d\theta}{dt}=-\dfrac{21}{10^2}\times (-1)$

$\dfrac{d\theta}{dt}=0.038\ rad/s$

Therefore rate of change in the angle is 0.038\ rad/s

8 0

3 years ago

A block of mass M is suspended from two identical springs of negligible mass, spring constant k, and unstretched length L. First

34kurt

Answer:

The answer is (C)X1=2X2

Explanation:

8 0

3 years ago