Answer: 339.148N
Explanation:
Given data:
Time (t) = 47s
Initial speed (U) = 0m/s
Final speed (V) = 9.5m/s
Mass of B = 540kg
Frictional force on B = 230N
Since both boats are linked, movement of A causes B to move as well.
What is the acceleration of boat A?
Applying the motion formula:
V = u + at
9.5 = 0 + a * 47
a = 9.5 / 47
a = 0.2021 m/s²
To determine the force necessary to accelerate boat B, as both boats experience the same force:
F = Mass * acceleration
F = 540 * 0.2021 = 109.14N
Given that there is a frictional force of 230N acting on boat B, the overall force (Tension) becomes:
Tension = frictional force + applied force = (109.14 + 230)N = 339.148N
Answer:
1.5 m/s²
Explanation:
Begin by sketching a free body diagram. Three forces are at play on the sea lion: the force of gravity acting downwards, the normal force that is perpendicular to the ramp, and the frictional force parallel to the ramp.
Considering the forces perpendicular to the incline:
∑F = ma
N − mg cos θ = 0
This gives us N = mg cos θ
Next, examining the forces parallel to the incline:
∑F = ma
mg sin θ − Nμ = ma
Substituting for N yields:
mg sin θ − (mg cos θ) μ = ma
g sin θ − g cos θ μ = a
hence a = g (sin θ − μ cos θ)
If we set θ = 23° and μ = 0.26:
a = 9.8 (sin 23 − 0.26 cos 23)
this results in a = 1.48
When rounded to two significant figures, the acceleration of the sea lion is 1.5 m/s².
This is due to the fact that below 4°c, water behaves differently than other substances and decreases in density as its temperature drops further.
B. The charge on A is -q; B has no charge. Given that a positive charge is situated at the center of an uncharged metallic sphere which is insulated and disconnected from the ground, a negative charge (-q) will appear on the inner surface A of the sphere. Should the exterior surface B be grounded, it will become neutral, resulting in no charge remaining on surface B.
25.82 m/s
Explanation:
Given:
Force applied by the baseball player; F = 100 N
Distance the ball travels; d = 0.5 m
Mass of the ball; m = 0.15 kg
To find the velocity at which the ball is released, we will equate the work done with the kinetic energy involved.
It's important to recognize that work done reflects the energy the baseball player has used. Thus, the relationship can be represented as follows:
F × d = ½mv²
100 × 0.5 = ½ × 0.15 × v²
Solving gives:
v² = (2 × 100 × 0.5) / 0.15
v² = 666.67
v = √666.67
v = 25.82 m/s.
