calculate the acceleration of gravity on the surface of kepler-62e Kepler-62e is an exoplanet that orbits within the habitable z
1 answer:
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Answer:
F = 63N
Explanation:
M= 1.5kg , t= 2s, r = (2t + 10)m and
Θ = (1.5t² - 6t).
magnitude of the resultant force acting on 1.5kg = ?
Force acting on the mass =
∑Fr =MAr
Fr = m(∇r² - rθ²) ..........equation (i)
∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)
The horizontal path is defined as
r = (2t + 10)
dr/dt = 2, d²r/dt² = 0
Angle Θ is defined by
θ = (1.5t² - 6t)
dθ/dt = 3t, d²θ/dt² = 3
at t = 2
r = (2t + 10) = (2*(2) +10) = 14
but dr/dt = 2m/s and d²r/dt² = 0m/s
θ = (1.5(2)² - 6(2) ) = -6rads
dθ/dt =3(2) - 6 = 0rads
d²θ/dt = 3rad/s²
substituting equation i into equation ii,
Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)
∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]
∑F = 1.5(14*3+0) = 63N
F = √(Fr² +FΘ²) = √(0² + 63²) = 63N
Answer:
Third displacement = 2.81 m which is 61.70° north of east.
Explanation:
Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.
She sails 2 km east, displacement = 2 i
Then 3.50 km southeast, means 3.5 km 315⁰ to + ve X axis
Displacement = 3.5 cos 315 i + 3.5 sin 315 = 2.47 i - 2.47 j
Let third displacement be x i + y j
We have final displacement = 5.80 km east = 5.80 i
From summation we have total displacement = 2 i + 2.47 i - 2.47 j + x i + y j
= (4.47+x) i + (y - 2.47) j
Comparing both , we have 4.47+x = 5.80
x = 1.33
y-2.47=0
y = 2.47 j
So third displacement = 1.33 i + 2.47 j
Magnitude of third displacement =
θ = tan⁻¹(2.47/1.33) = 61.70°
So third displacement = 2.81 m which is 61.70° north of east.
The question is incomplete. Here is the complete question.
A uniform electric field of 2kN/C points in the +x-direction.
(a) What is the change in potential energy of a +2.00nC test charge, as it is moved from point a at x = -30.0 cm to point b at x = +50.0 cm?
(b) The same test charge is released from rest at point a. What is the kinetic energy when it passes through point b?
(c) If a negative charge instead of a positive charge were used in this problem, qualitatively, how would your answers change?
Answer: (a) ΔU = 3.2× J
(b) KE = 2× J
Explanation: <u>Potential</u> <u>Energy</u> (U) is the amount of work done due to its position or condition and its unit is Joule (J). <u>Kinetic</u> <u>Energy</u> (KE) is the ability to do work by virtue of velocity and the unit is also (J). <u>Mechanical</u> <u>Energy</u> is the sum of Potential and Kinetic Energies of a system.
(a) Related to electricity, Potential Energy can be calculated as:
ΔU = Eqd
where E is the electric field (in N/C);
q is the charge (in C);
d is the distance between plaques (in m);
For a at x = - 30cm and b at x = 50 cm:
E = 2× N/C
q = 2× C
d = 50 - (-30) = 80× = 8×
m
ΔU = = Eqd
= 2×
. 2×
. 8×
ΔU = 3.2× J
(b) Mechanical Energy is constant, so:
Since the initial position is zero and there is no initial kinetic energy:
- (2×
. 2×
. 5×
)
J
(c) If the charge is negative, electric field does positive work, which diminishes the potential energy. The charge flows from the negative side towards the positive side and stays, not doing anything.