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Gennadij [26K]
3 years ago
6

Why do the stars appear like pointed objects​

Physics
1 answer:
saw5 [17]3 years ago
8 0

Answer:

because of where we are and our distance away from them, in reality they're giant balls of burning gas, like the sun. some stars are even bigger than the sun

Explanation:

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5 0
3 years ago
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Astronaut Rob leaves Earth in a spaceship at a speed of 0.960c relative to an observer on Earth. Rob's destination is a star sys
xxMikexx [17]

Answer:

A) 15.0 years

Explanation:

Due to the distance to the star system is in light-year units, we can compute the time by using:

t=\frac{d}{v}=\frac{14.4 l-y}{0.960}=15l-y

then, Rob will take to complete the trip about 15 light-years.

hope this helps!!

3 0
3 years ago
A proton with mass 1.67*10^-27kg is propelled at an initialspeed of 3.00*10^5m/s directly toward a uranium nucleus 5.00maway. Th
Tresset [83]

Answer:

Explanation:

F = 2.12 x 10⁻²⁶ / x²

Work done by electric field of nucleus

W = ∫ Fdx

= ∫2.12 x 10⁻²⁶ / x² dx

= 2.12 x 10⁻²⁶ ( - 1 / x )

= - 2.12 x 10⁻²⁶ ( 1/5 - 1 / 8 x 10⁻¹⁰ )

= - .265 x 10⁻¹⁶ J

1/ 2 x mv² = .5 x 1.67 x 10⁻²⁷ x 9 x 10¹⁰ - .265 x 10⁻¹⁶

= 7.515 x 10⁻¹⁷ - .265 x 10⁻¹⁶

=( .7515 - .265 )x 10⁻¹⁶

= .4865 x 10⁻¹⁶

.5 x 1.67 x 10⁻²⁷ x v² = .4865 x 10⁻¹⁶

v² = .5826 x 10¹¹

v² = 5.826 x 10¹⁰

v = 2.41 x 10⁵ m /s

b )

Let r be the closest distance

Potential at this point

2.12 x 10⁻²⁶ (  1 / r )

Kinetic energy

= 0

Total energy = 2.12 x 10⁻²⁶ (  1 / r )

Total energy at 5 m

= .5 x 1.67 x 10⁻²⁷ x 9 x 10¹⁰ + 0 ( potential energy at 5 m will be negligible as compared with that near the center )

= 7.515 x 10⁻¹⁷ J

So ,

2.12 x 10⁻²⁶ (  1 / r ) = 7.515 x 10⁻¹⁷

r = 2.12 x 10⁻²⁶ / 7.515 x 10⁻¹⁷

= .282 x 10⁻⁹

= 2.82 x 10⁻¹⁰ m

c ) As electric field is conservative , no dissipation of energy takes place . Hence its speed at 5m on returning back to this point will be same as

3.00 x 10⁵ m /s

7 0
4 years ago
Pls help ASAP !! Phyics
mina [271]

Answer:

Explanation:

Given

mass (m)= 20 kg

acceleration (a)= 10 m/s^2

Force (f)= m a

             = 20 * 10

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7 0
3 years ago
49. A 6.1-kg bowling ball is liſted 2.1 m to a shell. Find
stealth61 [152]

Answer:

125.5J

Explanation:

Given parameters:

Mass of the bowling ball  = 6.1kg

Height of lifting  = 2.1m

Unknown:

Increase in the ball energy  = ?

Solution:

The ball has changed position by moving it from one point to another. So, it has acquired more potential energy.

 Potential energy  = mgh

     m is the mass

     g is the gravity  

     h is the height

Now insert the given parameters and solve;

       Potential energy  = 6.1 x 9.8 x 2.1  = 125.5J

5 0
3 years ago
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