<h3><u>Answer;</u></h3>
100 times
<h3><u>Explanation;</u></h3>
- The largest stars are 100 times the mass of the Sun.
- <u>The giant stars are about 10 to 100 times the radius of the sun</u>, which means they are 100 times brighter than the sun.
- <em><u>The largest known star in terms of mass and brightness is known as the Pistol Star. It is believed to be 100 times as massive as our Sun, and 10,000,000 times as bright.</u></em>
The maximum velocity in a banked road, ignoring friction, is given by;
v = Sqrt (Rg tan ∅), where R = Radius of the curved road = 2*1000/2 = 1000 m, g = gravitational acceleration = 9.81 m/s^2, ∅ = Angle of bank.
Substituting;
30 m/s = Sqrt (1000*9.81*tan∅)
30^2 = 1000*9.81*tan∅
tan ∅ = (30^2)/(1000*9.81) = 0.0917
∅ = tan^-1(0.0917) = 5.24°
Therefore, the road has been banked at 5.24°.
Explanation:
I think it will increase a little bit ... just image ... if the temperature is 0, the velocity will be 0 too. because the vibration of atom is so weak and the sound cant progation.
Maybe you can split up the questions. I will try to answer your first question.
1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)
2. Momentum: p = m₁v₁ + m₂v₂
m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B
Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0
Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0
3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.
4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7
5.You figure out.
<span>To find the acceleration we are given two facts to begin. The impact at 16 km/h and the dent of 6.4 cm, or 0.064 meters. In solving the problem uniform acceleration is assumed, which would mean the avg speed during the impact was 8 km/hr by taking 16/2. We know distance = rate*time (d=r*t) . So t = d / r, so 0.64/8 = 0.008hr for t. Now we can solve for acceleration by taking a = 16 / 0.008 = 2000 km/hr.</span>
