Answer:
0.7 secs
Explanation:
In this question, the speed does not change as the mass changes. So we can use
Δ<em>t=Δ∨x/χgμ............................equ 1</em>
To stop, the final speed will be 0
Therefore,
<em>Δvx=vf-vt</em>
Δvx=0-4m/s
= -4m/s
Now substitute the various values in equ 1
Δ<em>t=Δ∨x/χgμ</em>
Δ<em>t= -</em>4m/s/(9.8m/s∧2) (0.6)
Δ<em>t=</em>0.7 secs
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I = pressure amplitude given = 0.2 W/m²
dB = decibel reading
decibel reading from the pressure amplitude is given as
dB = 10 log₁₀ (I/10⁻¹²)
inserting the values in the above equation
dB = 10 log₁₀ (0.2/10⁻¹²)
dB = 10 log₁₀ (2 x 10⁻¹/10⁻¹²)
dB = 10 log₁₀ (2 x 10⁻¹.10¹²)
dB = 10 log₁₀ (2 x 10¹²⁻¹)
dB = 10 log₁₀ (2 x 10¹¹)
dB = 113.01 db
hence the decibel reading comes out to be 113.01 db
Answer:
The magnitude of the impulse delivered to the baseball is 7.0 Ns
Explanation:
Given;
mass of the foul ball, m = 0.14 kg
initial velocity, u = 40 m/s
final velocity, v = 30 m/s in perpendicular direction
Impulse is given as change in momentum;
initial momentum in horizontal direction, Pi = mu
Pi = 0.14 x 40 = 5.6 Ns
final momentum in perpendicular direction, Pf = mv
Pf = 0.14 x 30
Pf = 4.2 Ns
The resultant impulse is given by;
J² = 5.6² + 4.2²
J² = 49
J = √49
J = 7.0 Ns
Therefore, the magnitude of the impulse delivered to the baseball is 7.0 Ns
