A steel cable lying flat on the floor drags a 20 kg block across a horizontal, frictionless floor. A 100 N force applied to the
1 answer:
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If we have the angle and magnitude of a vector A we can find its Cartesian components using the following formula
Where | A | is the magnitude of the vector and is the angle that it forms with the x axis in the opposite direction to the hands of the clock.
In this problem we know the value of Ax and Ay and we need the angle .
Vector A is in the 4th quadrant
So:
So:
So:
= -47.28 ° +360° = 313 °
= 313 °
Option 4.
Answer:
required distance is 233.35 m
Explanation:
Given the data in the question;
Sound intensity = 1.62 × 10⁻⁶ W/m²
distance r = 165 m
at what distance from the explosion is the sound intensity half this value?
we know that;
Sound intensity is proportional to 1/(distance)²
i.e
∝ 1/r²
Now, let r² be the distance where sound intensity is half, i.e ₂ =
₁/2
Hence,
₂/
₁ = r₁²/r₂²
1/2 = (165)²/ r₂²
r₂² = 2 × (165)²
r₂² = 2 × 27225
r₂² = 54450
r₂ = √54450
r₂ = 233.35 m
Therefore, required distance is 233.35 m