Answer:
The increase in gravitational potential energy is the same in both cases
Explanation:
It is easier to climb a mountain in a zigzag way rather than climbing on a straight line but since the distance is the same ( vertical height ) , mass and gravity is the same. Hence the increase in gravitational potential energy is the same in both cases.
gravitational potential energy = mgh ( same in both cases )
m = mass
g = acceleration due to gravity
h = distance ( vertical height )
Answer:
i Believe the correct answer is "An open circuit being changed into a closed circuit"
Explanation:
Answer:
2 times
Explanation:
1st time on Feb. 20, 1962 at age 41
2nd time on Oct. 29, 1998 at age 77
Answer:
The dog catches up with the man 6.1714m later.
Explanation:
The first thing to take into account is the speed formula. It is , where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is
Now, the distance equation for the man would be:
The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.
For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:
Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.
The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.
That value is used in the man's distance equation.
Finally, the dog catches up with the man 6.1714m later.
Answer:
A) At point 1, local acceleration = 0.5 m/s²
At point 2, local acceleration = 1.0 m/s²
B) Average Eulerian convective acceleration over the two points in the cross section shown = 0.5 m/s²
This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.
Explanation:
Local acceleration at those points is the instantaneous acceleration at those points and it is given as
a = dv/dt
At point 1, v₁ = 0.5 t
a₁ =dv₁/dt = 0.5 m/s²
At point 2, v₂ = 1.0 t
a₂ = dv₂/dt = 1.0 m/s²
b) Average Eulerian convective acceleration over the two points in the cross section shown = (change of velocity between the two points)/time
Change of velocity between the two points = v₂ - v₁ = 1.0t - 0.5t = 0.5 t
Time = t
Average acceleration = 0.5t/t = 0.5 m/s²
This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.