The satellite exerts a force of 1900 N on the Earth
Explanation:
To answer this question, let's remind Newton's third law:
"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"
In this problem, we can identify:
The Earth as object A
The satellite as object B
This means that if we apply the law, the force that the Earth exerts on the satellite must be equal (in magnitude) and opposite (in direction) to the force the satellite exerts on the Earth.
Since the Earth exerts a force of 1900 N on the satellite, this means that the satellite also exerts a force of 1900 N on the Earth.
Learn more about Newton's third law:
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Answer:
48cm by the minus of difference of 32 CM length 20 CM because of the length of the building
Answer:
437 J
Explanation:
Parameters given:
Weight of child, W = 230 N
Height of swing, h = 1.9 m
Gravitational Potential Energy is given as:
P. E. = m*g*h = W*h
m = mass
h = height above the ground
W = weight
P. E. = 230 * 1.9
P. E. = 437 J
Answer:Chemical
Explanation:Photosynthesis is the biochemical reaction through which all green plants, including trees, make their food to live and grow.
First
let us imagine the projectile launched at initial velocity V and at angle
θ relative to the horizontal. (ignore wind resistance)
Vertical component y:
The
initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum
height of h, the vertical velocity
will be 0, therefore the time t taken to attain this maximum height is:
h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g
where
g is acceleration due to gravity
Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike
the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the
horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g]
D = (2V²sinθ cosθ)/g
D = (V²sin2θ)/g
In order for D (horizontal distance) to be
maximum, dD/dθ = 0
That is,
2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2 or θ = π/4
Therefore it is now<span> shown that the maximum horizontal travelled is attained when
the launch angle is π/4 radians, or 45°.</span>
