Speed=frequency*wavelength, so frequency=speed/wavelength. frequency=80*0.2
Frequency = 16 Hz
Answer: F=0.075N
Explanation:
T*cos(θ) = mg=0.005X9.8
T*cos(θ)=0.049N
T = 0.049/cosθ
F = k*q1*q2 / r2
F = 8.9875 * 10^9 * q1*q2 / r^2 q1
F = 8.9875 * 10^9 * (5 x 10^-6)^2 / r^2
F=0.2246875/r^2
r= 2Lsinθ = 2 x 1 x sinθ =2sinθ
Kindly check the attached for further solution
60.3° from due south and 5.89 m/s For this problem, first calculate a translation that will put John's destination directly on the origin and apply that translation to Mary's destination. Then the vector from the origin to Mary's new destination will be the relative vector of Mary as compared to John. So John is traveling due south at 6.7 m/s. After 1 second, he will be at coordinates (0,-6.7). The translation will be (0,6.7) Mary is traveling 28° West of due south. So her location after 1 second will be (-sin(28)*10.9, -cos(28)*10.9) = (-5.117240034, -9.624128762) After translating that coordinate up by 6.7, you get (-5.117240034, -2.924128762) The tangent of the angle will be 2.924128762/5.117240034 = 0.57142693 The arc tangent is atan(0.57142693) = 29.74481039° Subtract that value from 90 since you want the complement of the angle which is now 60.25518961° So Mary is traveling 60.3° relative to due south as seen from John's point of view. The magnitude of her relative speed is sqrt(-5.117240034^2 + -2.924128762^2) = 5.893783 m/s Rounding the results to 3 significant digits results in 60.3° and 5.89 m/s
Answer:
a diagonal line on the graph
Explanation:
the slope would be constant at an increasing rate
y=x
uniform= increasing at a constant rate
Answer:
See explanation
Explanation:
Since the original count rate is 600 Bq,
i) after 1 half life, the count rate decreases to 1/2 of 600 Bq = 300 Bq
ii) after 2 half lives, the count rate decreases to 1/4 of 600 = 150 Bq
iii) after 3 half lives, the count rate decreases to 1/6 of 600 = 100 Bq
iv) after 4 half lives, the count rate decreases to 1/8 of 600 = 75 Bq
