A charge −1.3 × 10−5 C is fixed on the x-axis at 7 m, and a charge 1 × 10−5 C is fixed on the y-axis at 4 m. Calculate the magni
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Answer:
(a)2.7 m/s
(b) 5.52 m/s
Explanation:
The total of the system would be conserved as no external force is acting on it.
Initial momentum = final momentum
⇒(4.30 g × 943 m/s) + (730 g × 0) = (4.30 g × 484 m/s) + (730 g × v)
⇒ 730 ×v = (4054.9 - 2081.2) =1973.7
⇒v=2.7 m/s
Thus, the resulting speed of the block is 2.7 m/s.
(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.
Thus, the speed of the bullet-block center of mass is 5.52 m/s.
Answer:
1. T₁ is approximately 100.33 N
T₂ is approximately -51.674 N
2. 230°F is 383.15 K
3. Part A
The total torque on the bolt is -4.2 N·m
Part B
Negative anticlockwise
Explanation:
1. The given horizontal force = 86 N
The direction of the given 86 N force = To the left (negative) and along the x-axis
(The magnitude and direction of the 86 N force = -86·i)
The state of the system of forces = In equilibrium
The angle of elevation of the direction of the force T₁ = 31° above the x-axis
The direction of the force T₂ = Downwards, along the y-axis (Perpendicular to the x-axis)
Given that the system is in equilibrium, we have;
At equilibrium, the sum of the horizontal forces = 0
Therefore;
T₁ × cos(31°) - 86 = 0
T₁ = 86/(cos(31°)) ≈ 100.33
T₁ ≈ 100.33 N
Similarly, at equilibrium, the sum of the vertical forces = 0
∴ T₁×sin(31°) + T₂ = 0
Which gives;
100.33 × sin(31°) + T₂ = 0
T₂ = -100.33 × sin(31°) ≈ -51.674
T₂ ≈-51.674 N
2. 230° F to Kelvin
To convert degrees Fahrenheit (°F) to K, we use;
Pluggining in the given temperature value gives;
230°F = 383.15 K
3. Part A
Torque = Force × perpendicular distance from the line of action of the force
Therefore, the clockwise torque = 9 N × 0.4 m = 3.6 N·m (clocwise)
The anticlockeisre torque = 13 N × 0.6 m = 7.8 N·m (anticlockwise)
The total torque o the bolt = 3.6 N·m - 7.8 N·m = -4.2 N·m (clockwise) = 4.2 N·m anticlockwise
Part B
The torque is negative anticlockwise.