A cheetah runs at a speed of 27.6 m/s. If the

Which of the following is a likely life cycle of a star?

melomori [17]

Answer:

D.

Explanation:

But this just happen for big stars, like more than 20x the Sun mass.

Shortly: A nebula is a cloud of gas and dust, the material starts to be acummuleted and became a protostar (is like a big planet, almost a star). With enought mass this is a star, burn hydrogen and transform it in Helium.

This occurs in Main Sequence, is about almost all the life time of a star. Then starts the lack of hydrogen. Gravity compress everything, pressure goes up and heat all. Too much energy, Helium get burned and the star grews fast, became a Red Giant. Time pass and the fuel is over, no more making fusion, gravity compress the star, too much strenght, colapses, neutron star.

If it have pretty mass, ok. If have more than like 2x Sun mass, became a blackhole.

7 0

3 years ago

One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

morpeh [17]

Answer:

63.750KeV

Explanation:

We are given that

Initial velocity of second electron, $u_2=0$

Radius, $r_1=0$

$r_2=2.3 cm=\frac{2.3}{100}=0.023 m$

1 m=100 cm

Magnetic field,B=0.0370 T

We have to determine the energy of the incident electron.

Mass of electron, $m=9.1\times 10^{-31} kg$

Charge on an electron, $q=-1.6\times 10^{-19} C$

Velocity, $v=\frac{Bqr}{m}$

Using the formula

Speed of electron, $v_1=\frac{Bqr_1}{m}=\frac{0.0370\times 1.6\times 10^{-19}\times 0}{9.1\times 10^{-31}}=0$

Speed of second electron, $v_2=\frac{0.0370\times 1.6\times 10^{-19}\times 0.023}{9.1\times 10^{-31}}$

$v_2=1.5\times 10^8 m/s$

Kinetic energy of incident electron= $\frac{1}{2}mv^2_1+\frac{1}{2}mv^2_2$

Kinetic energy of incident electron= $0+\frac{1}{2}(9.1\times 10^{-31})(1.5\times 10^8)^2=1.02\times 10^{-14} J$

Kinetic energy of incident electron= $\frac{1.02\times 10^{-14}}{1.6\times 10^{-19}}=63750eV=\frac{63750}{1000}=63.750KeV$

1KeV=1000eV

3 0

3 years ago

A sled of mass 10 kg slides along the ice. it has an initial speed of 2 m/s but stops because of friction. How much work is done

NeX [460]

Answer: The correct answer is option B.

Explanation:

Mass of the sled = 10 kg

Initial speed of the sled = 2 m/s

Kinetic energy of the sled = $\frac{1}{2}mv^2$

$\frac{1}{2}\times 10 kg\times (2 m/s)^2=20 Joules$

Work done by the sled = 20 joules

The work done by the friction will be in opposite direction and equal to the magnitude of the work done of the sled that - 20 J.

Hence, correct answer is option B.

6 0

3 years ago

Read 2 more answers

Why all traveller prefer to travel easily in the morning and night and not when the sun is hot

eduard

Answer:

Travelers prefer to travel in morning because arthat time the temperature ️ is not too hot and they can easily travel here and there without any problem. and they also prefer to travel in night because moderate temperature ....

4 0

3 years ago

A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 degrees below th

zhannawk [14.2K]

Answer: $V_{f}=2.96m/s$

Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight $w$ of the block has an x-component and y-component:

$w_{x}=wsin(\theta)$ (1)

$w_{y}=wcos(\theta)$ (2)

As well as the Normal Force $N$ :

$N_{x}=Nsin(\theta)$ (3)

$N_{y}=Ncos(\theta)$ (4)

In addition, we know $N=w$ , then $\sum F_{y}=0$

In the X-component:

$\sum F_{x}=m.a$

$m.a=w_{x}$ (5)

Substituting (1) in (5):

$wsin(\theta)=m.a$ (6)

In addition, we know $w=m.g$ , where $m$ is the mass of the block and $g$ the gravity acceleration, which is equal to $9.8m/{s}^{2}$

So:

$m.g.sin(\theta)=m.a$ (7)

$a=g.sin(\theta)$ (8)

$a=5.88m/{s}^{2}$ (9) >>>>This is the acceleration of the block

On the other hand, we have the following equation that expresses a <u>relation between</u> the distance $d$ with the acceleration $a$ and time $t$ :