How do I solve for the maximum speed and height given those accelerations? (please give the formula so I can solve these types o
f questions later on)
1 answer:
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Answer:
Vfx = 0.42 √x and Vfy = - 0.729 √x
Explanation:
To solve this problem let's use the coulomb law applied to each load
F = k q1 q2 / r²
Let's look for the distance from a corner to the center of the square
r² = (a / 2) ² + (a / 2)²
r = √ (a / 2)² 2
r = a / 2 √2
Let's write the forces, in the attached there is a diagram of the charges
Charge q in the corner
F1 = k q q / r²
F1 = k q² / r²
F1 = k q² 2 / a²
F1 = 8.99 10⁹ (2.8 10⁻⁶)² 2 / 1.5²
F1 = 62.65 10⁻³ N = 6.265 10⁻² N
charge 2q in another corner
F2 = k 2q q / r²
F2 = 2 k q² / r²
Notice that this force is twice the force F1, since the distances are equal
F2 = 2 F1
3q charge
F3 = k 3q q / r²
F3 = 3 F1
6q charge
F4 = k 6q q / r²
F4 = 6 F1
Let's calculate the total force, for this it is useful to break down each force into its components and then add them, let's use trigonometry
cos θ = Fx / F
Fx = F cos θ
sin θ = Fy / F
Fy = F sin θ
Let's add each force
X axis
Fx = F1x - F2x - F3x + F4x
Fx = F1 cos 45 - 2F1 cos 45 - 3F1 cos 45 + 6 F1 cos 45
Fx = 2 F1 cos 45
Fx = 2 (6,265 10-2) cos 45
Fx = 8.86 10-2 N
Axis y
Fy = F1y + F2y -F3y -F4y
Fy = F1 sin 45 + 2 F1 sin 45 - 3 F1 sin 45 - 6 F1 sin 45
Fy = - 6Fi sin 45
Fy = -6 (6,265 10⁻²) sin45
Fy = - 26.54 10⁻² N
As we have the accelerations, we can find the speed on each axis
X axis
Vf² = Vo² + 2aₓ x
Vfx² = 2aₓ x
Vfx² = 2 8.86 10⁻² x
Vfx = 0.42 √x
Axis y
Vfy² = vfy² + 2ay Y
Vfy² = 2 ay Y
Vfy² = -2 26.54 10-2 x
Vfy = - 0.729 √x
Answer:
F = 402.18 N
Explanation:
Given that,
A particular satellite with a mass of m is put into orbit around Ganymede (the largest moon of Jupiter) at a distance 300 km from the surface. Let the mass of the satellite is 350 kg.
We need to find the gravitational force of attraction between the satellite and the moon.
The formula for the gravitational force is given by :
M is mass of Ganymede
m is mass of satellite
R is Radius of Ganymede
h is distance = 300 km
Putting all the values,
So, the required force of attraction between the satellite and the moon is 402.18 N.
Answer:
Time, t = 0.104 seconds
Explanation:
Frequency of the click of the Dolphin, f = 55.3 kHz
A dolphin sends out a series of clicks that are reflected back from the bottom of the ocean 80 m below, d = 80 m
The speed of sound in seawater is, v = 1530 m/s
Once the sound is send and reflects, the total distance covered by it is 2d such that,
So, the time elapses before the dolphin hears the echoes of the clicks is 0.104 seconds.