Answer:
Answer
Explanation:
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Answer:
The found acceleration in terms of h and t is:
Explanation:
(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)
We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.
<h3>
</h3><h3>
Stage 1</h3>
Constant acceleration, starts from rest.
Distance = 
Velocity = 
<h3>Stage 2</h3>
Constant velocity where
Velocity = 
Distance =
<h3>
</h3><h3 /><h3>Stage 3</h3>
Constant deceleration where
Velocity = 
Distance =
<h3>Total Height</h3>
Total height = y₁ + y₂ + y₃
Total height = 
<h3 /><h3>Acceleration</h3>
Find acceleration by rearranging the found equation of total height.
Total Height = h
h = 5a(t₁)²
Muscles function only by contracting. This makes it necessary for one end of the muscle to be fixed and the other mobile.
Take the bicep for example.
Its origin is at the shoulder and its two heads connect to the bones of the forearm, the radius and ulna.
Now, had the muscle not been fixed at one end, and contracted, it would pull both our shoulder and forearm together resulting in an ineffective movement. The desired motion is to lift the forearm (proximal and distal movement) which can only be achieved if the bicep is fixed at the shoulder and allowed to move at the forearm.
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>
Answer:
The direction of the field is downward, and negatively charged particles will experience an upwards force due to the field.
F = N e E where E is the value of the field and N e the charge Q
M g = N e E and M g is the weight of the drop
N = M g / (e E)
N = 1.1E-4 * 9.8 / (1.6E-19 * 370) = 1.1 * 9.8 / (1.6 * 370) * E15 = 1.82E13
.00011 kg is a very large drop
Q = N e = M g / E = .00011 * 9.8 / 370 = 2.91E-6 Coulombs
Check: N = Q / e = 2.91E-6 / 1.6E-19 = 1.82E13 electrons
