The time component is needed. The acceleration is the change of velocity divided by the time in when this change of velocity happens.
Answer:
D) the second at the doorknob
Explanation:
The torque exerted by a force is given by:
where
F is the magnitude of the force
d is the distance between the point of application of the force and the centre of rotation
is the angle between the direction of the force and d
In this problem, we have:
- Two forces of equal magnitude F
- Both forces are perpendicular to the door, so 
- The first force is exerted at the midpoint of the door, while the 2nd force is applied at the doorknob. This means that d is the larger for the 2nd force
--> therefore, the 2nd force exerts a greater torque
Answer:
v = 54 m/s
Explanation:
Given,
The maximum height of the flight of golf ball, h = 150 m
The velocity at height h, u = 0
The velocity of the golf ball right before it hits the ground, v = ?
Using the III equations of motion
<em> v² = u² + 2gh</em>
Substituting the given values in the above equation,
v² = 0 + 2 x 9.8 x 150 m
= 2940
v = 54 m/s
Hence, the speed of the golf ball right before it hits the ground, v = 54 m/s
Answer:
a) k = 2231.40 N/m
b) v = 0.491 m/s
Explanation:
Let k be the spring force constant , x be the compression displacement of the spring and v be the speed of the box.
when the box encounters the spring, all the energy of the box is kinetic energy:
the energy relationship between the box and the spring is given by:
1/2(m)×(v^2) = 1/2(k)×(x^2)
(m)×(v^2) = (k)×(x^2)
a) (m)×(v^2) = (k)×(x^2)
k = [(m)×(v^2)]/(x^2)
k = [(3)×((1.8)^2)]/((6.6×10^-2)^2)
k = 2231.40 N/m
Therefore, the force spring constant is 2231.40 N/m
b) (m)×(v^2) = (k)×(x^2)
v^2 = [(k)(x^2)]/m
v = \sqrt{ [(k)(x^2)]/m}
v = \sqrt{ [(2231.40)((1.8×10^-2)^2)]/(3)}
= 0.491 m/s
Answer:
nreaker
Explanation:
A switch that automatically interrupts or shuts off an electric current at the first indication of a overload
