The point obviously is in the 3rs quadrant
So
စ= tan^-1( y/x)-180
စ= -89.7°
What is he minumum coating of thickness needed to ensure that lifght of waveelntght 5660 mbnd si
<h3><u>Question</u><u>:</u></h3>
A racing car is travelling at 70 m/s and accelerates at -14 m/s^2. What would the car’s speed be after 3 s?
<h3><u>Statement:</u></h3>
A racing car is travelling at 70 m/s and accelerates at -14 m/s^2.
<h3><u>Solution</u><u>:</u></h3>
- Initial velocity (u) = 70 m/s
- Acceleration (a) = -14 m/s^2
- Time (t) = 3 s
- Let the velocity of the car after 3 s be v m/s
- By using the formula,
v = u + at, we have

- So, the velocity of the car after 3 s is 28 m/s.
<h3><u>Answer:</u></h3>
The car's speed after 3 s is 28 m/s.
Hope it helps
Complete Question
The Question diagram is attached below
Answer:
a) 
b) 
c) 
Explanation:
From the question we are told that:
Mass 
Distance 
Generally the equation for Work done is mathematically given by

For 


For 


For 

Answer:
= 8
Explanation:
noise from power mower = 106 dB
noise from rock concert = 115 dB
we would be applying the formula for the decibel level of sound to get the intensities.
(B) = 10 log \frac{I}{I₀}
I = intensity of sound
I₀= reference intensity = 10^{-12} W/m^{2}
- intensity of the power mower
106 = 10 log \frac{I}{10^{-12}}
10.6 = log \frac{I}{10^{-12}}
10^{10.6} = \frac{I}{10^{-12}}
I = 10^{10.6} x 10^{-12}
I of power mower = 0.0398 W/m^{2}
- intensity of the rock music
115 = 10 log \frac{I}{10^{-12}}
11.5 = log \frac{I}{10^{-12}}
10^{11.5} = \frac{I}{10^{-12}}
I = 10^{11.5} x 10^{-12}
I of rock music = 0.316 W/m^{2}
now the ratio of the intensity of the rock music to the power mower
= \frac{0.316}{0.0398}
= 8