What is renewable and non-renewable energy?
2 answers:
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<h2>Answer:</h2>
(a) 3.2 x 10²s
(b) 0.9 m/s (S 13 E)
(c) 2.9 x 10²m
<h2>Explanation:</h2>The sketch illustrating the scenario has been attached to this response.
As shown;
The fish swims due east with a velocity = 0.2m/s
The river current has a velocity due South = 0.9m/s
The resultant of the velocity is V
The width of the river is x = 64m
(a) To calculate how long it took the fish to get across the river, we know that velocity is the rate of change in distance, therefore we can use the relation;
V = -------------(i)
Where;
V = velocity of the fish = = 0.2m/s
d = distance from the start to the end = width of the river = x = 64m
t = time taken to move for that distance
Make t subject of the formula in equation (i);
t =
Substitute the values of d and V into the equation;
t =
t = 320 s
t = 3.20 x 10²s
Therefore, the time taken for the fish to get across the river is 3.20 x 10²s
(b) The resulting vector of the fish is V whose magnitude is the algebraic sum of vectors and
, and direction is given by θ. i.e
<em>The magnitude of the resulting vector is;</em>
|V| =
|V| =
|V| =
|V| =
|V| = 0.92m/s
|V| ≅ 0.9m/s
<em>The direction of the resulting vector θ and is given by;</em>
tan θ =
tan θ =
tan θ = 4.5
θ = tan⁻¹ ( 4.5)
θ = 77.47° South of East.
θ ≈ 77.5° South of East.
Subtracting θ = 77.5° from 90° gives its value East of South
i.e
90 - 77.5 = 12.5° East of South
<em>This can also be written as S12.5°E</em>
<em>Approximating to the nearest whole number gives </em>S 13 E
Therefore, the resulting velocity of the fish is 0.9m/s in the direction S13°E
(c) When the fish arrives on the opposite bank, its distance from being at the point directly across from where it started is the product of the velocity of the river current and the time taken by the fish to get across the river. This point is equivalent to k as shown in the diagram.
Therefore;
distance = velocity of river current x time taken
distance = 0.9m/s x 3.20 x 10²s
distance = 2.88 x 10²m
distance ≅ 2.9 x 10²m
<em>Notice that the velocity of the river current is used since that's the velocity of the fish on the y-axis.</em>
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Answer:
a) a = 3.06 10¹⁵ m / s , b) F= 1.43 10⁻¹⁰ N, c) F_total = 14.32 10⁻²⁶ N
Explanation:
This exercise will average solve using the moment relationship.
a ) let's use the relationship between momentum and momentum
I = ∫ F dt = Δp
F t = m - m v₀
F = m (v_{f} -v₀o) / t
in the exercise indicates that the speed module is the same, but in the opposite direction
F = m (-2v) / t
if we use Newton's second law
F = m a
we substitute
- 2 mv / t = m a
a = - 2 v / t
let's calculate
a = - 2 4.59 10²/3 10⁻¹³
a = 3.06 10¹⁵ m / s
b) F= m a
F= 4.68 10⁻²⁶ 3.06 10¹⁵
F= 1.43 10⁻¹⁰ N
c) if we hit the wall for 1015 each exerts a force F
F_total = n F
F_total = n m a
F_total = 10¹⁵ 4.68 10⁻²⁶ 3.06 10¹⁵
F_total = 14.32 10⁻²⁶ N
Answer:
Orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.
Explanation:
The gravitational force is responsible for the orbital motion of the planet, satellite, artificial satellite, and other heavenly bodies in outer space.
When an object is applied with a velocity that is equal to the velocity of the orbit at that location, the body continues to move forward. And, this motion is balanced by the gravitational pull of the second object.
The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.
The velocity of the orbit is given by the relation,
Where
V - velocity of the orbit at a height h from the surface
R - Radius of the second object
G - Gravitational constant
h - height from the surface
The body will be in orbital motion when its kinetic motion is balanced by gravitational force.
Hence, the orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.