Answer:
<h3>130N</h3>
Explanation:
according to Newtons second law;
F = ma
F = m(v-u)/t
m is the mass = 60+5 = 65kg
v is the final velocity = 0m/s
initial velocity = 8m/s
time t = 4.0s
F = 65(8)/4
F = 65*2
F = 130N
Hence he applied a force of 130N
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:
-142m
Explanation:
To find the initial position of the car, we can add the displacement.
-245 + 103 = -142
Best of Luck!
Answer:
3.70242 nm
Explanation:
Using Compton effect formula
Δλ = ( h / mec) ( 1 - cosθ)
where h is planck constant = 6.62607 × 10 ⁻³⁴ m²kg/s
me, mass of an electron = 9.11 × 10⁻³¹ kg
c is the speed of light = 3 × 10⁸ m/s
Δλ = 6.62607 × 10 ⁻³⁴ m²kg/s / (9.11 × 10⁻³¹ kg × 3 × 10⁸ m/s ) ( 1 - cos 90°) = 0.242 × 10 ⁻¹¹ m = 2.42 × 10⁻¹² m = 0.00242 nm
modified wavelength = 3.7 nm + 0.00242 nm = 3.70242 nm
