Two masses (5.3kg and 7.5kg) are fastened together with a small amount of explosive. They are loaded into a spring gun that is t
1 answer:
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Answer:
It changes at a rate of 4/3 meter per second
Explanation:
In the given figure below we have
Solving for Y given we get
Answer:
The tension in the two ropes are;
T1 = 23.37N T2 = 35.47N
Explanation:
Given mass of the object to be 4.2kg, the weight acting on the bag will be W= mass × acceleration due to gravity
W = 4.2×10 = 42N
The tension acting on the bag plus the weight are three forces acting on the bag. We need to find tension in the two ropes that will keep the object in equilibrium.
Using triangular law of force and sine rule to get the tension we have;
If rope 1 is at 57.6° with respect to the vertical and rope 2 is at 33.8° with respect to the vertical, our sine rule formula will give;
T1/sin33.8° = T2/sin57.6° = 42/sin{180-(33.8°+57.6°)}
T1/sin33.8° = T2/sin57.6° = 42/sin88.6°
From the equality;
T1/sin33.8° = 42/sin88.6°
T1 = sin33.8°×42/sin88.6°
T1 = 23.37N
To get T2,
T2/sin57.6°= 42/sin88.6°
T2 = sin57.6°×42/sin88.6°
T2 = 35.47N
Note: Check attachment for diagram.
