Answer:
v = 1.08 m/s
Explanation:
What is the linear speed of the 0.0500-kg sphere as its passes through its lowest point?
The decrease in PE is
d = 80.0cm * 1 / 1000m = 0.80m
h = 0.80 m /2 = 0.40 m
ΔPE = m*g*h
ΔPE = (0.0500 - 0.0200)kg * 9.8m/s² * 0.400 m
ΔPE = 0.1176 J
The moment of inertia of the assembly is
I = 1/12*m*L² + (m1 + m2)*(L/2)²
I = 1/12*0.390kg*(0.800m)² + 0.0700kg*(0.400m)²
I = 0.032 kg·m²
KE = ½Iω²
0.1176 J = ½ * 0.032kg·m² * ω²
ω = 2.71 rad/s
v = ωr = 2.71 rad/s * 0.400m
The linear velocity
v = 1.08 m/s
Answer:
a) puck is subjected to both the forces of the hockey sticks in a horizontal direction,
b)the puck does not move since the sum of the forces is zero
c) changing the magnitude or direction of its applied force
Explanation:
a) The puck is subjected to both the forces of the hockey sticks in a horizontal direction, these forces are of equal magnitude and opposite direction since the puck is at rest.
In the direction of the y-axis (perpendicular to the ice) you have the weight of the disk and the normal to this weight that are also in equilibrium.
b) the puck does not move since the sum of the forces is zero, which implies that the forces of the hockey sticks are of equal magnitude and opposite direction.
c) the player has several ways to make the puck move
* slightly changing the angle of the club and therefore the direction of the force, in this case the disc comes out in the direction of this component
* inclined the stick slightly so that the force has a vertical component and the puck jumps in this direction
* Increasing the magnitude of the force so that the puck comes out in the opposite direction to the player
* The worst case, decreasing its force to zero and the disk comes out in its direction by the other force that had the same magnitude.
The work done by the man pushing the car over the given distance is 1000J.
Given the data in the question;
- Mass of car;
- Acceleration of the car;
- Distance covered by the car;
Work done; 
<h3>Work done</h3>
Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;
Where f is force applied and d is distance travelled.
To determine the work done by the man, we first solve for the force applied F.
From Newton's Second Law; 
We substitute our given values into the expression
Next we substitute our values into the expression of work done above.
Therefore, the work done by the man pushing the car over the given distance is 1000J.
Learn more about work done: brainly.com/question/26115962
Answer:
<em>specific</em><em> </em><em>heat</em><em> </em><em>capacity</em><em> </em><em>(</em><em>c</em><em>)</em><em> </em><em>=</em>107.2J/kg K
The formula we can use here is:
g = G m / r^2
where g is gravity, G is gravitational constant, m is
mass, and r is radius
Since G is constant, therefore we can equate two
situations:
g1 m1 / r1^2 = g2 m2 / r2^2
(10 m/s^2) r1^2 / m1 = g2 * (1/2 r1)^2 / (1/8 m1)
<span>g2 = 5 m/s^2</span>
