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
=> 
=>
Answer:
A. 1.4 m/s to the left
Explanation:
To solve this problem we must use the principle of conservation of momentum. Let's define the velocity signs according to the direction, if the velocity is to the right, a positive sign will be introduced into the equation, if the velocity is to the left, a negative sign will be introduced into the equation. Two moments will be analyzed in this equation. The moment before the collision and the moment after the collision. The moment before the collision is taken to the left of the equation and the moment after the collision to the right, so we have:
where:
M = momentum [kg*m/s]
M = m*v
where:
m = mass [kg]
v = velocity [m/s]
where:
m1 = mass of the basketball = 0.5 [kg]
v1 = velocity of the basketball before the collision = 5 [m/s]
m2 = mass of the tennis ball = 0.05 [kg]
v2 = velocity of the tennis ball before the collision = - 30 [m/s]
v3 = velocity of the basketball after the collision [m/s]
v4 = velocity of the tennis ball after the collision = 34 [m/s]
Now replacing and solving:
(0.5*5) - (0.05*30) = (0.5*v3) + (0.05*34)
1 - (0.05*34) = 0.5*v3
- 0.7 = 0.5*v
v = - 1.4 [m/s]
The negative sign means that the movement is towards left
The higher you go the more potential energy there is, and the lower it is the more kinetic energy there is, so the more kinetic energy there is the higher the ball will bounce.
Answer:
734.215N
Explanation:
First we calculate the angle that corresponds to a 5% slope using the Tan-1 function
then we use the component that corresponds to the direction parallel to the road, additionally we must multiply by the gravity value to find the weight(g=9.81m/s^2)
Wx=M*g*sen(2.86)=1500kg*9.81*sen(2.86)=734.215N
Answer:
La velocidad de la luz en el vacío es una constante universal con el valor de 299 792 458 m/s (186 282,397 mi/s),aunque suele aproximarse a 3·108 m/s. Se simboliza con la letra c, proveniente del latín celéritās (en español, celeridad o rapidez).
¿Cuál es la consecuencia que a velocidad de la luz sea constante?
Respuesta. En modificaciones del vacío más sutiles, como espacios curvos, efecto Casimir, poblaciones térmicas o presencia de campos externos, la velocidad de la luz depende de la densidad de energía de ese vacío.
