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Verizon [17]
3 years ago
8

A star with an original mass of 8 to 25 solar masses will ultimately become a ?

Physics
1 answer:
BaLLatris [955]3 years ago
8 0
The answer is C. Neutron star.

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Using software, a simulation of a javelin being thrown both on Earth and on the moon is created. In both cases, the velocity and
mario62 [17]

Horizontal distance covered by a projectile is X = Vix *T

where Vix is the initial horizontal component of velocity and T is time taken by the projectile

Vix = ViCos theta

In question they said that initial velocity and angle is same on earth and moon

so Vix would remains same

now let's see about time taken T

time taken to reach the highest point

Vfy = Viy +gt

at highest point vertical velocity become zero so Vfy =0

0 = Vi Sin theta + gt

t = Vi Sintheta /g

Total time taken to land will be twice of that

On earth

Te= 2t

Te = 2Sinθ/g

on moon g is one-sixth of g(earth)

Tm = 2Sinθ/(g/6)

Tm = 6(2Sinθ/g)

Tm = 6Te

so total time taken by the projectile on moon will be six times the time taken on earth

From first equation X = Vix*T

we can see that X will also be 6 times on moon than earth

so projectile will cover 6 times distance on moon than on earth

4 0
2 years ago
Read 2 more answers
Two Earth satellites, A and B, each of mass m, are to be launched into circular orbits about Earth's center. Satellite A is to o
Pachacha [2.7K]

(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

U=-\frac{GMm}{r}

where

G is the gravitational constant

M is the Earth's mass

m is the satellite's mass

r is the distance of the satellite from the Earth's centre, which is sum of the Earth's radius (R) and the altitude of the satellite (h):

r = R + h

We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}

and so, substituting:

R=6370 km\\h_A = 5970 km\\h_B = 21200 km

We find

\frac{U_B}{U_A}=\frac{6370 km+5970 km}{6370 km+21200 km}=0.448

(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

K=\frac{1}{2}\frac{GMm}{r}

So, the ratio between the two kinetic energies is

\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}

Which is exactly identical to the ratio of the potential energies. Therefore, this ratio is also equal to 0.448.

(c) B

The total energy of a satellite is given by the sum of the potential energy and the kinetic energy:

E=U+K=-\frac{GMm}{R+h}+\frac{1}{2}\frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}

For satellite A, we have

E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J

For satellite B, we have

E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J

So, satellite B has the greater total energy (since the energy is negative).

(d) -2.57\cdot 10^8 J

The difference between the energy of the two satellites is:

E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J

4 0
3 years ago
A speedboat increases its speed from 14.5 m/s to 29.3 m/s in a distance of 172 m.
dalvyx [7]

Answer:

a. Acceleration, a = 1.88 m/s²

b. Time, t = 7.87 seconds.

Explanation:

Given the following data;

Initial velocity, U = 14.5m/s

Final velocity, V = 29.3m/s

Distance, S = 172m

a. To find the acceleration of the speedboat;

We would use the third equation of motion;

V² = U² + 2aS

Substituting into the formula

29.3² = 14.5² + 2a*172

858.49 = 210.25 + 344a

344a = 858.49 - 210.25

344a = 648.24

a = 648.24/344

Acceleration, a = 1.88 m/s²

b. To find the time;

We would use the first equation of motion;

V = U + at

29.3 = 14.5 + 1.88t

1.88t = 29.3 - 14.5

1.88t = 14.8

Time, t = 14.8/1.88

Time, t = 7.87 seconds.

6 0
3 years ago
Although wave power does not produce pollution, some people may not want to invest in it because it is _____. 100 percent renewa
notka56 [123]
In my view, correct answer should look like this: Although wave power does not produce pollution, some people may not want to invest in it because it is <span>prone to storm damage and limited to particular areas of the ocean.</span>
6 0
2 years ago
Read 2 more answers
Which student is doing work?
Vilka [71]

Answer:

d

hope this helps

Explanation:

8 0
3 years ago
Read 2 more answers
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