Given:
A cylindrical container closed of both end has a radius of 7cm and height of 6cm.
Explanation:
A.) Find the total surface area of the container.
- A = 2πrh + 2πr²
- A = 2(3.14)(7)(6) + 2(3.14)(7 × 7)
- A = 263.76 + 307.72
- A = 571.48
B.) Find the volume of the container.
- V = πr²h
- V = (3.14)(7×7)(6)
- V = 923.16
Not sure huhuness.
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Answer:
9517.2 lbm
Explanation:
Electricity consumption = 14000 kWh/year
Fuel consumption = 900 gal/year
Amount of CO₂ produced per gallon = 26.4 lbm/gal
Amount of CO₂ produced per kWh = 1.54 lbm/kWh
Amount of CO₂ produced in one year

Reduction would be

The reduction in the amount of CO₂ produced is 9517.2 lbm
a) 32.3 N
The force of gravity (also called weight) on an object is given by
W = mg
where
m is the mass of the object
g is the acceleration of gravity
For the ball in the problem,
m = 3.3 kg
g = 9.8 m/s^2
Substituting, we find the force of gravity on the ball:

b) 48.3 N
The force applied

The ball is kicked with this force, so we can assume that the kick is horizontal.
This means that the applied force and the weight are perpendicular to each other. Therefore, we can find the net force by using Pythagorean's theorem:

And substituting
W = 32.3 N
Fapp = 36 N
We find

c) 
The ball's acceleration can be found by using Newton's second law, which states that
F = ma
where
F is the net force on an object
m is its mass
a is its acceleration
For the ball in this problem,
m = 3.3 kg
F = 48.3 N
Solving the equation for a, we find

Answer:
F = 100 Newtons
Explanation:
F = ?
m = 0.04kg
u = 0m/s ==> u is just an abbreviation for initial velocity, it is conventional.
s = 50m ==> s is just an abbreviation for distance, it is conventional.
v = 500m/s ==> v is just an abbreviation for final velocity, it is conventional.

Then F = ma = 0.04 x 2500 = 100N
Answer:
the magnitude of the average contact force exerted on the leg is 3466.98 N
Explanation:
Given the data in the question;
Initial velocity of hand v₀ = 5.25 m/s
final velocity of hand v = 0 m/s
time interval t = 2.65 ms = 0.00265 s
mass of hand m = 1.75 kg
We calculate force on the hand F
using equation for impulse in momentum
F
× t = m( v - v₀ )
we substitute
F
× 0.00265 = 1.75( 0 - 5.25 )
F
× 0.00265 = 1.75( - 5.25 )
F
× 0.00265 = -9.1875
F
= -9.1875 / 0.00265
F
= -3466.98 N
Next we determine force on the leg F
Using Newton's third law of motion
for every action, there is an equal opposite reaction;
so, F
= - F
we substitute
F
= - ( -3466.98 N )
F
= 3466.98 N
Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N