Answer:
We can use 2 g H = v2^2 - v1^2 or
v2^2 = 2 g H + v1^2
Since 88 ft/sec = 60mph we have 30 mph = 44 ft/sec
The object will return with the same speed that it had initially so the object
starts out with a downward speed of 44 ft/sec
Then v2^2 = 2 * 32 ft/sec^2 * 160 ft + 44 (ft/sec)^2
v2^2 = (2 * 32 * 160 + 44^2) ft^2 / sec^2 = 12180 ft^2/sec^2
v2 = 110 ft/sec
The answer true I’m guessing. It’s a 50/50 chance
Answer:
Image distance of apple=-6.7 cm
Magnification of apple=0.33
Explanation:
We are given that an apple is placed 20.cm in front of a diverging lens.
Object distance=u=-20 cm
Focal length=f=-10 cm
Because focal length of diverging lens is negative.
We have to find the image distance and magnification of the apple.
Lens formula
Substitute the values then we get
Image distance of apple=-6.7 cm
Magnification=m=
Magnification of apple=
Hence, the magnification of apple=0.33
Answer:
X₃₁ = 0.58 m and X₃₂ = -1.38 m
Explanation:
For this exercise we use Newton's second law where the force is the Coulomb force
F₁₃ - F₂₃ = 0
F₁₃ = F₂₃
Since all charges are of the same sign, forces are repulsive
F₁₃ = k q₁ q₃ / r₁₃²
F₂₃ = k q₂ q₃ / r₂₃²
Let's find the distances
r₁₃ = x₃- 0
r₂₃ = 2 –x₃
We substitute
k q q / x₃² = k 4q q / (2-x₃)²
q² (2 - x₃)² = 4 q² x₃²
4- 4x₃ + x₃² = 4 x₃²
5x₃² + 4 x₃ - 4 = 0
We solve the quadratic equation
x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2 5
x₃ = [-4 ± 9.80] 10
X₃₁ = 0.58 m
X₃₂ = -1.38 m
For this two distance it is given that the two forces are equal
As the shock waves travel in concentric outward circles from the epicenter, and the diameter is measured 120 miles,
area of a circle =<span>π</span><span>r*r</span>
d=120
<span>r=<span>120/2</span></span><span>r=60</span><span><span>60*60</span>=3600</span><span>3600*π=11309.734</span>
<span>11309.734 square miles</span>
