Answer:
The kinetic energy of the weight is 344.5 J
Explanation:
Given that:
Force = F = 65 newton
distance = d = 5.3 meters
We have to find change in kinetic energy ΔK.E
Now we know that, initially kinetic energy was 0 So the formula we use will be:
Work done = Change in kinetic energy
Mathematically,
W = ΔK.E
As we know W = F . d and ΔK.E = K.E(final) - K.E(initial)
So by putting values:
F . d = K.E(final) - K.E(initial)
F . d = K.E(final)
As K.E(initial) is 0 so by putting values of F and d
(65)* (5.3) = K.E(final)
344.5 J = K.E(final)
So the change in K.E will also be 344.5 J
i hope it will help you!
3-SAT ≤p TSP
If P ¹ NP, then no NP-complete problem can be solved in polynomial time.
both the statements are true.
<u>Explanation:</u>
- 3-SAT ≤p TSP due to any complete problem of NP to other problem by exits of reductions.
- If P ¹ NP, then 3-SAT ≤p 2-SAT are the polynomial time algorithm are not for 3-SAT. In P, 2-SAT is found, 3- SAT polynomial time algorithm implies the exit of reductions. 3 SAT does not have polynomial time algorithm when P≠NP.
- If P ¹ NP, then no NP-complete problem can be solved in polynomial time. because for the NP complete problem individually gets the polynomial time algorithm for the others. It may be in P for all the problems, the implication of latter is P≠NP.
Answer:
(a) BP = 11.99 KPa
(b) h = 2 m
Explanation:
(a)
Since, the fluid pressure and blood pressure balance each other. Therefore:
BP = ρgh
where,
BP = Blood Pressure
ρ = density of fluid = 1020 kg/m³
g = acceleration due to gravity = 9.8 m/s²
h = height of fluid = 1.2 m
Therefore,
BP = (1020 kg/m³)(9.8 m/s²)(1.2 m)
<u>BP = 11995.2 Pa = 11.99 KPa</u>
(b)
Again using the equation:
P = ρgh
with data:
P = Gauge Pressure = 20 KPa = 20000 Pa
ρ = density of fluid = 1020 kg/m³
g = acceleration due to gravity = 9.8 m/s²
h = height of fluid = ?
Therefore,
20000 Pa = (1020 kg/m³)(9.8 m/s²)h
<u>h = 2 m</u>
Answer:
Energy produce in one year =20.49 x 10¹⁶ J/year
Explanation:
Given that
Plant produce 6.50 × 10⁸ J/s of energy.
It produce 6.50 × 10⁸ J in 1 s.
We know that
1 year = 365 days
1 days = 24 hr
1 hr = 3600 s
1 year = 365 x 24 x 3600 s
1 year = 31536000 s
So energy produce in 1 year = 31536000 x 6.50 × 10⁸ J/year
Energy produce in one year = 204984 x 10¹² J/year
Energy produce in one year =20.49 x 10¹⁶ J/year
