Answer:
This is how I figured it out:
- 215.5 rounded to one significant figure is 200
- 101.02555 rounded to one significant figure is 100.
- 200 + 100 = 300.
Hope this helps!
Explanation:
<span>Our equation 1 would be
m*v=M*V1+m*V2
v=V1+V2
v-V1=V2
the equation 2 would look like this
</span>V^2=V1^2+V2^2
V^2-V1^2=V2^2
(V-V1)*(V+V1)=V2^2Dividing with the 1
V+V1=V2
Machine cycle. The four steps which the CPU carries out for each machine language instruction: fetch, decode<span>, execute, and store. hope that helped</span>
<span>22.5 newtons.
First, let's determine how much energy the stone had at the moment of impact. Kinetic energy is expressed as:
E = 0.5mv^2
where
E = Energy
m = mass
v = velocity
Substituting known values and solving gives:
E = 0.5 3.06 kg (7 m/s)^2
E = 1.53 kg 49 m^2/s^2
E = 74.97 kg*m^2/s^2
Now ignoring air resistance, how much energy should the rock have had?
We have a 3.06 kg moving over a distance of 10.0 m under a force of 9.8 m/s^2. So
3.06 kg * 10.0 m * 9.8 m/s^2 = 299.88 kg*m^2/s^2
So without air friction, we would have had 299.88 Joules of energy, but due to air friction we only have 74.97 Joules. The loss of energy is
299.88 J - 74.97 J = 224.91 J
So we can claim that 224.91 Joules of work was performed over a distance of 10 meters. So let's do the division.
224.91 J / 10 m
= 224.91 kg*m^2/s^2 / 10 m
= 22.491 kg*m/s^2
= 22.491 N
Rounding to 3 significant figures gives an average force of 22.5 newtons.</span>
Answer:
Time period, T = 403.78 years
Explanation:
It is given that,
Orbital distance, 
Mass of the Earth, 
Mass of the planet, 
Let T is the orbital period of this planet. The Kepler's third law of motion gives the relation between the orbital period and the orbital distance.
or
T = 403.78 years
So, the orbital period of this planet is 404 years. Hence, this is the required solution.
