Answer:
I₁ / I₂ = 1.43
Explanation:
To find the relationship of the two inertial memits, let's calculate each one, let's start at the moment of inertia with the arms extended
Before starting let's reduce all units to the SI system
d₁ = 42 in (2.54 10⁻² m / 1 in) = 106.68 10⁻² m
d₂ = 38 in = 96.52 10⁻² m
The moment of inertia is a scalar quantity for which it can be added, the moment of total inertia would be the moment of inertia of the man (cylinder) plus the moment of inertia of each arm
I₁ = I_man + 2 I_ arm
Man indicates that we can approximate them to a cylinder where the average diameter is
d = (d₁ + d₂) / 2
d = (106.68 + 96.52) 10-2 = 101.6 10⁻² m
The average radius is
r = d / 2 = 50.8 10⁻² m = 0.508 m
The mass of the trunk is the mass of man minus the masses of each arm.
M = M_man - 0.2 M_man = 80 (1-0.2)
M = 64 kg
The moments of inertia are:
A cylinder with respect to a vertical axis: Ic = ½ M r²
A rod that rotates at the end: I_arm = 1/3 m L²
Let us note that the arm rotates with respect to man, but this is at a distance from the axis of rotation of the body, so we must use the parallel axes theorem for the moment of inertia of the arm with respect to e = of the body axis.
I1 = I_arm + m D²
Where D is the distance from the axis of rotation of the arm to the axis of the body
D = d / 2 = 101.6 10⁻² /2 = 0.508 m
Let's replace
I₁ = ½ M r² + 2 [(1/3 m L²) + m D²]
Let's calculate
I₁ = ½ 64 (0.508)² + 2 [1/3 8 1² + 8 0.508²]
I₁ = 8.258 + 5.33 + 4.129
I₁ = 17,717 Kg m² / s²
Now let's calculate the moment of inertia with our arms at our sides, in this case the distance L = 0,
I₂ = ½ M r² + 2 m D²
I₂ = ½ 64 0.508² + 2 8 0.508²
I₂ = 8,258 + 4,129
I₂ = 12,387 kg m² / s²
The relationship between these two magnitudes is
I₁ / I₂ = 17,717 /12,387
I₁ / I₂ = 1.43
Answer:
Explanation:
Since the field lines are parallel and the electric field is uniform between two parallel plates, a test charge would experience the same force of attraction or repulsion no matter where it is located in the field,
I attached an image that could help to understand the representation of the field. The formula used to calculate it is given by,
(1)
If we want to consider the change in Voltage with respect to the position then it would be,
According to the information provided, the potential is and there is a distance d, therefore
(2)
Taking equation (1) we can clear V, to what we have,
Integrating,
Substituting (2)
Where x is the height from the grounded plate.
2. What is the angle a of the force F in the figure above?
(a) The magnitude of the force F acting on the knot is 5.54 N.
(b) The angle α of the force F is 54.4⁰.
The given parameters:
The net vertical force on the knot is calculated as follows;
The net horizontal force on the knot is calculated as follows;
From the trig identity;
The angle α of the force F is calculated as follows;
Find the image uploaded for the complete question.
Learn more about net force here:brainly.com/question/12582625
The knot is in equilbrium, so there is no net force acting on it. Starting with the unknown force and going clockwise, denote each force by F₁, F₂, F₃, and F₄, respectively. We have
F₁ + F₂ + F₃ + F₄ = 0
Decomposing each force into horizontal and vertical components, we have
F cos(180º - α) + (5.7 N) cos(50º) + (6.2 N) cos(-44º) + (6.7 N) cos(-137º) = 0
F sin(180º - α) + (5.7 N) sin(50º) + (6.2 N) sin(-44º) + (6.7 N) sin(-137º) = 0
Recall that cos(180º - x) = - cos(x) and sin(180º - x) = sin(x), so these equations reduce to
F cos(α) ≈ - 3.22 N
F sin(α) ≈ 4.51 N
(1) Recall that for all x, sin²(x) + cos²(x) = 1. Use this identity to solve for F :
(F cos(α))² + (F sin(α))² = F ² ≈ 30.73 N² → F ≈ 5.5 N
(2) Use the definition of tangent to solve for α :
tan(α) = sin(α) / cos(α) ≈ 1.399 → α ≈ 126º
or about 54º from the horizontal from above on the left of the knot.
0.63 volts
B.
158volts
C.
633 volts
D.
158,000 volts
E.
5.8 volts
The voltage of the electricity will be 632.9 V. Electric power is found as the multiplication of the voltage and current. Option B is correct.
Electric power is the product of the voltage and current. Its unit is the watt. It is the rate of the electric work done.
The given data in the problem is;
V is the voltage = ? Volt (V)
Electric current (I)= 15.8 amps (A)
P is the power =10.0 kilowatts =10⁴ watt
The formula for the power is given as;
The voltage of the electricity will be 63.29 V.
Hence, option B is correct.
To learn more about the electric power, refer to the link;
#SPJ2
Hmmm. Kilowatts should be converted to watts. Simply just move the decimal place to the right three times.
10,000 W / 15.8 A = V
632.9, or 633.
Answer:
4.93 m
Explanation:
According to the question, the computation of the height is shown below:
But before that first we need to find out the speed which is shown below:
As we know that
= 9.92 m/s
Now
98.4064 = 19.96 × height
So, the height is 4.93 m
We simply applied the above formulas so that the height i.e H could arrive
The height of the water slide is 5.04 meters.
The problem described in this question involves a water slide, where swimmers start from rest at the top and leave the slide traveling horizontally. To determine the height of the slide, we can use the equations of motion in the horizontal direction. The horizontal displacement (x) is given as 5.00 m and the time (t) is given as 0.504 s. Assuming no friction or air resistance, we can use the equation x = v*t, where v is the horizontal velocity. Rearranging the equation, we can solve for v, which is equal to x/t. Substituting the given values, we have v = 5.00 m / 0.504 s = 9.92 m/s. The horizontal velocity (v) is constant throughout the motion, so we can use the equation v = sqrt(2*g*H), where g is the acceleration due to gravity (9.8 m/s^2) and H is the height of the slide. Rearranging the equation, we can solve for H, which is equal to v^2 / (2*g). Substituting the known values, we have H = (9.92 m/s)^2 / (2*9.8 m/s^2) = 5.04 m.
Where,
Qh: the absolute value (magnitude) of the heat absorbed from the hot reservoir during one cycle or during some time specified in the problem
Qc: the absolute value (magnitude) of the heat delivered to the cold reservoir during one cycle or during some time specified in the problem
W: the amount of work done by the engine during one cycle or during some time specified in the problem
A) e=QhW
B) e=QcQh
C) e=QcW
D) e=WQh
E) e=WQc
Answer:
Efficiency e = W/Qh
Explanation:
As written above efficiency of a system is calculated as the output per unit input. For heat Engine, Efficiency is calculated by dividing the Work done by Engine by Heat absorbed from hot reservoir.
In theoretical terms The maximum efficiency of a heat engine (which no engine ever attains) is equal to the temperature difference between the hot and cold ends divided by the temperature at the hot end, each expressed in absolute temperature (Kelvin).
But in practical calculations, it is calculated as e = W/Qh , and we define the thermal efficiency, of any heat engine as the ratio of the work it does, W, to the heat input at the high temperature, Qh.