Answer:
149.05J
Explanation:
Hello,
Data;
Weight = 12N
x = 80miles
Radius of the moon = 1079.4mi
W = mg
But gravity at moon = ⅙ the gravity on earth
12 = ⅙ × m
M = 2kg
Mass of the module = 2kg
K = F.x
F(x) = k / x²
2 = k / (1079.4)²
k = 2.33×10⁶
Work = ∫f.dx
work = ∫₁₀₇₉.₄¹¹⁵⁹ . (2.33×10⁶/x²).dx
Work = (-2.33×10⁶) ×(¹/₁₁₅₉ - ¹/₁₀₇₉)
work = 149.05J
time taken by the object dropped = 2s
this time depends on the height of the plane from ground
it is given by
now the distance covered horizontally is given as 190 m
now the speed of the object is
now when plane is moving at same height but with double speed
so it will take same time to hit the ground again
so the time is given as
so it will take t = 2 s again to hit the ground
If you slow the motion of water molecules down, they are more likely to become ice
Answer:
(a)
(b)
Explanation:
<u>For Earth we have:</u>
- mass of earth,
- radius of earth,
- orbital radius,
- period of rotation,
- period of revolution,
(a)
Angular momentum,
∵...............................(1)
For a particle of mass m moving in a circular path at a distance r from the axis,
&
Putting respecstive values in eq. (1)
To model earth as a particle is reasonable because the distance between the sun and the earth is very large s compared to the radius if the earth.
(b)
For a uniform sphere of mass M and radius R and an axis through its center,
using eq. (1)
Answer:
a) vₓ = 6,457 m / s
, v_{y} = 0.518 m / s
, b) v = 6.478 m / s, θ = 4.9°
Explanation:
a) This is a kinematic problem, let's use trigonometry to find the components of acceleration
sin 31 = / a
cos 31 = aₓ = a
a_{y} = a sin31
aₓ = a cos 31
Now let's use the kinematic equation for each axis
X axis
vₓ = v₀ₓ + aₓ (t-t₀)
vₓ = v₀ₓ + a cos 31 (t-t₀)
vₓ = 2.6 + 0.45 cos 31 (20-10)
vₓ = 6,457 m / s
Y Axis
v_{y} = v_{oy} + a_{y} t
v_{y} = v_{oy} + a_{y} sin31 (t-to)
v_{y} = -1.8 + 0.45 sin31 (20-10)
v_{y} = 0.518 m / s
b) let's use Pythagoras' theorem to find the magnitude of velocity
v = √ (vₓ² + v_{y}²)
v = √ (6,457² + 0.518²)
v = √ (41.96)
v = 6.478 m / s
We use trigonometry for direction
tan θ = v_{y} / vₓ
θ = tan⁻¹ v_{y} / vₓ
θ = tan⁻¹ 0.518 / 6.457
θ = 4.9°
c) let's look for the vector at the initial time
v₁ = √ (2.6² + 1.8²)
v₁ = 3.16 m / s
θ₁ = tan⁻¹ (-1.8 / 2.6)
θ₁ = -34.7
We see that the two vectors differ in module and direction, and that the acceleration vector is responsible for this change.
a = (v₂ -v₁) / (t₂-t₁)