Question 4 4. Two blocks A of mass 15kg and B of unknown mass are connected by a light inextensible string on a rough horizontal surface. A constant force of magnitude 120N is applied onto block A at an angle of 30° to the horizontal as shown in the diagram below. The coefficient of kinetic friction for both blocks is 0.2 and the system of blocks accelerates to the right at 2.08m.s2. B Question 5 A 15 kg 30° 4.1. State Newton's Third law of motion in words 4.2. Draw a labelled free-body diagram for block A 4.3. Show that the frictional force acting on block A as it accelerates is 14.7N 4.4. Calculate the mass of block B [18] 120N (2) (5) (5) [16]
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Answer:
4.1. Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. In other words, when one object exerts a force on another object, the second object exerts an equal force in the opposite direction on the first object.
4.2. Here's a labeled free-body diagram for Block A:
```
T (tension in the string)
↑
│
│
│
│
│
F (applied force)
──→ (direction of motion)
```
In this diagram, "T" represents the tension in the string, and "F" represents the applied force at an angle of 30° to the horizontal. The arrow indicates the direction of motion.
4.3. To find the frictional force acting on block A as it accelerates, we can use Newton's Second Law:
\[F_{\text{net, A}} = m_A \cdot a\]
Where:
- \(F_{\text{net, A}}\) is the net force acting on block A.
- \(m_A\) is the mass of block A (given as 15 kg).
- \(a\) is the acceleration (given as 2.08 m/s²).
Rearranging the equation to solve for \(F_{\text{net, A}}\):
\[F_{\text{net, A}} = 15 kg \cdot 2.08 m/s² = 31.2 N\]
Now, we need to consider the frictional force, which opposes the motion and acts in the direction opposite to the applied force. So, the frictional force is 31.2 N in the opposite direction of motion, making it:
Frictional force on block A = -31.2 N
However, since you want it in magnitude, it's 31.2 N.
4.4. To calculate the mass of block B, we can use the fact that block A and block B are connected by a string, so they experience the same acceleration. Therefore, we can use the following equation:
\[F_{\text{net, B}} = m_B \cdot a\]
Where:
- \(F_{\text{net, B}}\) is the net force acting on block B, which is the tension in the string.
- \(m_B\) is the mass of block B (unknown).
- \(a\) is the acceleration (given as 2.08 m/s²).
We already calculated that the tension in the string is 31.2 N. Plugging in the values:
\[31.2 N = m_B \cdot 2.08 m/s²\]
Now, solving for \(m_B\):
\[m_B = \frac{31.2 N}{2.08 m/s²} \approx 15 kg\]
So, the mass of block B is approximately 15 kg.
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Final answer:
A deep draught vessel sailing in a narrow channel can experience a sudden sheer due to squat, which is the downward displacement caused by hydrodynamic forces in shallow water. This can lead to the ship sinking lower in water and potentially losing control. Ship operators should be aware of this phenomenon and take precautions when navigating in narrow channels.
Explanation:
A deep draught vessel sailing in a very narrow channel can develop a sudden sheer as it slows down due to the phenomenon called squat. Squat refers to the downward displacement of a ship caused by hydrodynamic forces when it is sailing in shallow water. As the vessel slows down, the decrease in speed leads to a decrease in the hydrodynamic forces supporting the ship, causing it to sink lower in the water and potentially lose control.
This sudden sheer can be dangerous as it can result in the ship deviating from its intended course and potentially colliding with the channel walls or disturbing other vessels in the vicinity.
It is important for ship operators to be aware of this phenomenon and take necessary precautions when navigating in narrow channels, such as maintaining a safe speed and keeping a proper distance from the channel walls.
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Final answer:
A deep draught vessel sailing in a very narrow channel can develop a sudden sheer as it slows down due to the change in pressure caused by the narrowing of the channel.
Explanation:
A deep draught vessel sailing in a very narrow channel can develop a sudden sheer as it slows down due to the change in pressure caused by the narrowing of the channel. When the channel narrows, there is a pressure difference which results in a net force on the fluid. This net force causes the vessel to experience a sudden sheer as it slows down.
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F = force
m = mass
a = acceleration
1.5 m/s^2 = a
5 kg = m
F = ?
F = 5 * 1.5
F = 7.5 Newtons
Answer would then be : Force = 7.5 N
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Explanation:
Flour typically sticks to surfaces due to adhesive forces, and it's more likely to stick to areas with moisture or oils on the skin. If the scar on the man's chest lacks moisture or oils (which is common for scars), the flour may have difficulty adhering to that specific area. However, it could stick to the surrounding skin where there might be more moisture or natural oils.
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Answer:
Buoyancy or Upthrust
Explanation:
This is an upward force exerted by a fluid that opposes the weight of a partially or fully immersed object.
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When a moving car hits a parked car, causing the parked car to move, the type of collision is elastic collision. An elastic collision is when two bodies collide and separates after collision conserving the total kinetic energy before and after collision.