Answer:
20 m
Explanation:
Given:
v₀ = 15 m/s
v = -25 m/s
a = -10 m/s²
Find: Δy
v² = v₀² + 2aΔy
(-25 m/s)² = (15 m/s)² + 2 (-10 m/s²) Δy
Δy = 20 m
I’ll say c Bc it make more since to find the travel distance
Answer:
Minimum work = 5060 J
Explanation:
Given:
Mass of the bucket (m) = 20.0 kg
Initial speed of the bucket (u) = 0 m/s
Final speed of the bucket (v) = 4.0 m/s
Displacement of the bucket (h) = 25.0 m
Let 'W' be the work done by the worker in lifting the bucket.
So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.
Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

Therefore, the work done by the worker in lifting the bucket is given as:

Now, plug in the values given and solve for 'W'. This gives,

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.
Answer:
a) 
b) 
c) 
Explanation:
Given masses:


Velocity of mass 1, 
Velocity of mass 2, 
a)
Initial momentum:



b)
magnitude of initial momentum:


From the conservation of momentum:



is the magnitude of final velocity.
Direction of final velocity will be in the direction of momentum:




c)
Vertical component of final velocity:


<span>6.20 m/s^2
The rocket is being accelerated towards the earth by gravity which has a value of 9.8 m/s^2. Given the total mass of the rocket, the gravitational drag will be
9.8 m/s^2 * 5.00 x 10^5 kg = 4.9 x 10^6 kg m/s^2 = 4.9 x 10^6 N
Add in the atmospheric drag and you get
4.90 x 10^6 N + 4.50 x 10^6 N = 9.4 x 10^6 N
Now subtract that total drag from the thrust available.
1.250 x 10^7 - 9.4 x 10^6 = 12.50 x 10^6 - 9.4 x 10^6 = 3.10 x 10^6 N
So we have an effective thrust of 3.10 x 10^6 N working against a mass of 5.00 x 10^5 kg. We also have N which is (kg m)/s^2 and kg. The unit we wish to end up with is m/s^2 so that indicates we need to divide the thrust by the mass. So
3.10 x 10^6 (kg m)/s^2 / 5.00 x 10^5 kg = 0.62 x 10^1 m/s^2 = 6.2 m/s^2
Since we have only 3 significant figures in our data, the answer is 6.20 m/s^2</span>