Answer:
Explanation:
V = I R
V = 0.0418 * 30
V = 1.254
this battery seems dead :P
Hi there!
Since the string is light and there is no friction in the pulley, the acceleration of the system is equal to the acceleration of both blocks.
We can begin by summing the forces of each block:
Block on incline:
- Force of gravity (in the negative direction away from the acceleration)
- Force of Tension
∑F = -M₁gsinФ + T
Block hanging:
- Force of gravity (Positive, in direction of acceleration)
- Force of Tension (Negative, opposite from acceleration)
∑F = M₂g - T
Sum both of these net forces for each block:
∑Fт = -M₁gsinФ + T - T + M₂g
∑Fт = -M₁gsinФ + M₂g
Divide by the mass to solve for acceleration:
Plug in the given values:
Answer:
Work done by gravitational force = 0 joule
Explanation:
Work done is given by the relation
W = F x S joule
Where, F - the force applied in the form of push or pull
S - displacement caused by the force
If a force is acting on a body and it doesn't cause any displacement, then work done will be zero.
Gravitational force acts on the body even if the body is at rest.
<em>Work done by the gravitational is applicable only if there is some vertical component of motion involved.</em>
In this case, some amount of work is done to move the body from point A to B.But, the body is displaced in the horizontal direction. No vertical motion is involved.
So, the work done due to gravitational force on a mass is zero joule.
Answer:
1.73 m/s²
3.0 cm
Explanation:
Draw a free body diagram of the yo-yo. There are two forces: weight force mg pulling down, and tension force T pulling up 10° from the vertical.
Sum of forces in the y direction:
∑F = ma
T cos 10° − mg = 0
T cos 10° = mg
T = mg / cos 10°
Sum of forces in the x direction:
∑F = ma
T sin 10° = ma
mg tan 10° = ma
g tan 10° = a
a = 1.73 m/s²
Draw a free body diagram of the sphere. There are two forces: weight force mg pulling down, and air resistance D pushing up. At terminal velocity, the acceleration is 0.
Sum of forces in the y direction:
∑F = ma
D − mg = 0
D = mg
½ ρₐ v² C A = ρᵢ V g
½ ρₐ v² C (πr²) = ρᵢ (4/3 πr³) g
3 ρₐ v² C = 8 ρᵢ r g
r = 3 ρₐ v² C / (8 ρᵢ g)
r = 3 (1.3 kg/m³) (100 m/s)² (0.47) / (8 (7874 kg/m³) (9.8 m/s²))
r = 0.030 m
r = 3.0 cm