Complete Question
An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.
What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.
Answer:
The torque is 
Explanation:
From the question we are told that
The mass of the steel ball is 
The length of arm is 
The mass of the arm is 
Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as
=> 
=> 
Generally the magnitude of torque about the athlete shoulder is mathematically represented as
=> 
=>
Pushing a broke down car, even done by more than one person, is difficult especially if the distance to be covered is quite far. A car is heavy and it requires a lot of force to start the car moving. This is because the inertia of the car to remain at rest is great. Additionally, the force applied in pushing the car must be greater than the frictional force to cause it to accelerate. The frictional force is dependent on the mass of the object which means that the frictional force acting on the car is also great. Finally, with every push of the car, the frictional force will always be present and acting on the opposite direction. The push that will be supplied must be sustained all throughout.
Answer:
sorry but I can understand the question
Answer:
1. 20.54m/s
2. 1.52s
Explanation:
QUESTION 1:
The speed the stone impact the ground is the final speed/velocity, which can be calculated using the formula:
v² = u² + 2as
Where;
v = final velocity (m/s)
u = initial velocity (m/s)
a = acceleration due to gravity (m/s²)
s = distance (m)
From the provided information, u = 5.65m/s, v = ?, s = 19.9m, a = 9.8m/s²
v² = 5.65² + 2 (9.8 × 19.9)
v² = 31.9225 + 2 (195.02)
v² = 31.9225 + 390.04
v² = 421.9625
v = √421.9625
v = 20.5417
v = 20.54m/s
QUESTION 2:
Using v = u + at
Where v = final velocity (m/s) = 20.54m/s
t = time (s)
u = initial velocity (m/s) = 5.65m/s
a = acceleration due to gravity (m/s²)
v = u + at
20.54 = 5.65 + 9.8t
20.54 - 5.65 = 9.8t
14.89 = 9.8t
t = 14.89/9.8
t = 1.519
t = 1.52s
Do they give answer choices? or is it free write? i’ll help if you tell me!!
