Y - yo = Vo*t - g * (t^2) / 2
Vo = - 9.0 m/s
t = 0.50 s
=> y - yo = -9.0 m/s * 0.5 s - 9.8 m/s^2 * (0.5s)^2 / 2 = - 4.5m - 1.225m = - 5.725 m.
Answer: option c) - 5.7
A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).
A rocket ship has several engines and thrusters. We can divide its initial movement into 2 parts:
- From t = 0 min to t = 2.0 min, the SRB and the main engines act together and the speed goes from 0 m/s (rest) to 1341 m/s.
- From t = 2.0 min to t = 8.5 min, the main engines alone accelerate the ship form 1341 m/s to 7600 m/s.
We want to know the acceleration in the first part (first 2.0 minutes). We need to consider that:
- The speed increases from 0 m/s to 1341 m/s.
- The time elpased is 2.0 min.
- 1 min = 60 s.
The acceleration of the ship during the first 2.0 minutes is:
A rocket ship is accelerated by the SRB and the main engines for 2.0 minutes and the main engines for 8.5 minutes after the launch. The acceleration of the ship during the first 2.0 minutes is 11 m/s² (D).
Learn more: brainly.com/question/16274121
<span>Density is 3.4x10^18 kg/m^3
Dime weighs 1.5x10^12 pounds
The definition of density is simply mass per volume. So let's divide the mass of the neutron star by its volume. First, we need to determine the volume. Assuming the neutron star is a sphere, the volume will be 4/3 pi r^3, so
4/3 pi 1.9x10^3
= 4/3 pi 6.859x10^3 m^3
= 2.873x10^10 m^3
Now divide the mass by the volume
9.9x10^28 kg / 2.873x10^10 m^3 = 3.44588x10^18 kg/m^3
Since we only have 2 significant digits in our data, round to 2 significant digits, giving 3.4x10^18 kg/m^3
Now to figure out how much the dime weighs, just multiply by the volume of the dime.
3.4x10^18 kg/m^3 * 2.0x10^-7 m^3 = 6.8x10^11 kg
And to convert from kg to lbs, multiply by 2.20462, so
6.8x10^11 kg * 2.20462 lb/kg = 1.5x10^12 lb</span>
Answer:
The required diagram is shown in the figure. When an object is placed in front of the convex lens, i.e., between 2F
1
and F
1
, its image is formed beyond 2F
2
on the other side of the lens. The image is real, inverted and enlarged.
solution
Explanation:
Half-life is the time taken for a radioactive material to decay to half its original composition:
Original mass = 48g
Half- life = 2hr
After four half lives;
Initially: 48g
First halving 24
Second halving 12
Third halving 6
Fourth halving 3
After second half life, we would have 12g
At fourth halving, we would have 3g
