The free-body diagram of your question is; 2 downward forces (253 kg mass of barbell & 133 kg body mass of Kurlovich) acting together on a point supported by 2 upward forces as normal forces exerted by Kurlovich's feet.
Solving the normal forces exerted by 2 feet :
Summation of Forces Vertical = 0
2 Dowwnard Forces = 2 Upward Forces (2F)
253 + 133 = 2F
2F = 386 Kgs
F = 386 / 2
F = 193 Kgs (Normal Force Exerted by Each Foot)
To calculate the normal force exerted on each of Vladimir Kurlovich's feet while lifting the barbell, we need to consider the weights of Kurlovich and the barbell. The normal force is equal to the sum of the two weights.
To calculate the normal force exerted on each of Vladimir Kurlovich's feet while lifting the barbell, we need to consider the weight of Kurlovich and the weight of the barbell. The normal force is equal to the sum of Kurlovich's weight and the weight of the barbell, as it balances the gravitational force pulling them downwards. Kurlovich's weight can be calculated using his mass and the acceleration due to gravity, which is approximately 9.8 m/s².
Step 1:
Calculate Kurlovich's weight:
Mass = 133 kg
Acceleration due to gravity = 9.8 m/s²
Weight = Mass × Acceleration due to gravity
Weight = 133 kg × 9.8 m/s²
Step 2:
Calculate the weight of the barbell:
Mass = 253 kg
Acceleration due to gravity = 9.8 m/s²
Weight = Mass × Acceleration due to gravity
Weight = 253 kg × 9.8 m/s²
Step 3:
Calculate the normal force exerted on each foot:
Normal force = Kurlovich's weight + Barbell's weight
#SPJ2
Answer:
+5.7 m/s
Explanation:
According to the law of conservation of momentum is that the momentum before the collision is equal to the momentum after the collision. In an equation form it would look like this:
M₁V₁+M₂V₂ = M₁V₁'+M₂V₂'
Where:
M₁ = mass of object 1 (kg)
V₁ = velocity of object 1 before the collision (m/s)
V₁' = Final velocity of object 1 after the collision (m/s)
M₂ = mass of object 2 (kg)
V₂ = velocity of object 2 before the collision (m/s)
V₂' = Final velocity of object 2 after the collision (m/s)
According to your problem you have the following given:
M₁ = 5 g = 0.005kg
V₁ = 3 m/s
V₁' = -5m/s (It bounced off so it is going the other direction)
M₂ = 6g = 0.006kg
V₂ = -1 m/s (It is coming from the opposite direction of the 3-ball)
V₂' = ?
So we plug in what we know and solve for what we don't know.
Answer:
+5.7 m/s
Explanation:
Just did the assignment :)
It has no fixed volume or shape.
B.
It has a fixed volume and varied shape.
C.
It has high energy and expands to fill the container.
D.
It has a fixed volume and shape.