Boxes A and B are being pulled to the right on a frictionless surface. Box A has a larger mass than B. How do the two tension fo
1 answer:
Answer:
Tension T1 is less than tension T2.
T1 < T2
Explanation:
According to given data,
mass of box A ( mA) is grater than mass of box B (mB)
we can write,
m(A) > m(B)
Newton's second law states that:
Tension of object is directly proportional to the mass of the system.
T ∝ m
here Boxes A and B are being pulled to the right on a frictionless surface,
so Tension T1 generates due to the mass of box A m(A)
and Tension T2 arises due to mass of the system m(A) + m(B)
Thus tension T1 will be less than tension T2
T1 < T2
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Answer:
The tensions in is approximately 4,934.2 lb and the tension in
is approximately 6,035.7 lb
Explanation:
The given information are;
The angle formed by the two rope segments are;
The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°
The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°
Therefore, we have;
The tension in rope segment BC =
The tension in rope segment BD =
The tension in rope segment AB = = Pulling force of tugboat = 1200 lb
By resolution of forces acting along the line A_F gives;
× cos(26.0°) +
× cos(21.0°) =
= 1200 lb
× cos(26.0°) +
× cos(21.0°) = 1200 lb............(1)
Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;
× sin(26.0°) +
× sin(21.0°) = 0...........................(2)
Which gives;
× sin(26.0°) = -
× sin(21.0°)
= -
× sin(21.0°)/(sin(26.0°)) ≈ -
× 0.8175
Substituting the value of, , in equation (1), gives;
- × 0.8175 × cos(26.0°) +
× cos(21.0°) = 1200 lb
- × 0.7348 +
×0.9336 = 1200 lb
×0.1988 = 1200 lb
≈ 1200 lb/0.1988 = 6,035.6938 lb
≈ 6,035.6938 lb
≈ -
× 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb
≈ -4934.1733 lb
From which we have;
The tensions in ≈ -4934.2 lb and
≈ 6,035.7 lb.
(a) +2.60 m/s
The motion of the fish dropped by the seagul is a projectile motion, which consists of two independent motions:
- a horizontal uniform motion, at constant speed
- a vertical motion, at constant acceleration (acceleration of gravity, , downward)
In this part we are only interested in the horizontal motion. As we said the horizontal component of the fish's velocity does not change, therefore its value when the fish reaches the ocean is equal to its initial value, which is the speed at which the seagull was flying (because it was flying horizontally):
(b) -17.2 m/s
The vertical component of the fish's velocity instead follows the equation:
where
is the initial vertical velocity, which is zero
is the acceleration of gravity
t is the time
Since the fish reaches the ocean at t = 1.75 s, we can substitute this time into the formula to find the final vertical velocity:
where the negative sign indicates the direction (downward).
(c)
The horizontal component of the fish's velocity would increase
The vertical component of the fish's velocity would stay the same.
As we said from part (a) and (b):
- The horizontal component of the fish's velocity is constant during the motion and it is equal to the initial velocity of the seagull -> so if the seagull's initial speed increases, the horizontal velocity of the fish will increase too
- The vertical component of the fish's velocity does not depend on the original speed of the seagull, therefore it is not affected.