Answer:
My biggest reason is to make it a habit. Even if the ball goes into the endzone it is a live ball and the offensive players must down the ball. Don't leave any room for "I thought he downed it" or "I thought I heard the whistle" just run to the ball always.
If the players slow down and the returner takes it out of the end zone it could be a big return. Players are on a full sprint for 40+ yards sometimes and instead of breaking down, they choose to contine through the goal line to slow down at a decreased rate (possibly limiting a muscle pull injury).
A 500 kg wrecking ball is knocking down a wall. When it is pulled back to its highest point, it is at a height of 6.2 m. When it hits the wall, it is moving at 3.1 m/s. How high is the wrecking ball when it hits the wall? (Show your work and follow all of the steps of the GUESS method. Check your answer after you submit the form - it's in the feedback for this question.) |
PLZ FAST!!
Compare and contrast microscopic and macroscopic energy transfer. Give at least three comparisons for each. THX
Answer:
Macroscopic energy is energy at a level of system while microscopic energy is energy at the level of atoms and molecules
Explanation:
1. Macroscopic energy is possessed by a system as whole while microscopic energy is possessed by its constituents’ atoms or molecules.
2. The common form of macroscopic energy is Kinetic and potential energy while the microscopic form of energy are atomic forces due its random, disordered motion and due to intermolecular forces
3. At microscopic level we consider behaviour of every molecule and in macroscopic approach we consider gross or average effects of various molecular infractions
he potential energy between two atoms in a particular molecule has the form U(s) = 2.6/x^8 - 4.3/x^4 where the units of x are length and the numbers 2.G and 4.3 have appropriate units so that U(x) has units of energy. What b the equilibrium separation of the atoms (that is the distance at which the force between the atoms is zero)?
Answer:
x = 1.04866
Explanation:
Force can be defined from power energy by the expressions
F = [tex]- \frac{ dU}{ dx}[/tex]
in this case we are the expression of the potential energy
U = [tex]\frac{2.6}{x^{8} } - \frac{4.3}{ x^{4} }[/tex]
let's find the derivative
dU / dx = 2.6 ( [tex]\frac{-8}{x^{9} }[/tex]) - 4.3 ([tex]\frac{-4}{ x^{5} }[/tex])
dU / dx = [tex]- \frac{20.8}{ x^{9} } + \frac{17.2 }{ x^{5} }[/tex]
we substitute
F = + \frac{20.8}{ x^{9} } - \frac{17.2 }{ x^{5} }
at the equilibrium point the force is zero, so
[tex]\frac{20.8}{ x^{9} } = \frac{17.2}{ x^{5} }[/tex]
20.8 / 17.2 = x⁴
x⁴ = 1.2093
x = [tex]\sqrt[4]{ 1.2093}[/tex]
x = 1.04866
At an airport, two business partners both walk at 1.5 m/sm/s from the gate to the main terminal, one on a moving sidewalk and the other on the floor next to it. The partner on the moving sidewalk gets to the end in 60 ss, and the partner on the floor reaches the end of the sidewalk in 90s.
Required:
What is the speed of the sidewalk in the Earth reference frame?
Answer:
[tex]v=0.8m/s[/tex]
Explanation:
From the question we are told that
Distance [tex]d=1.5m/sm/s[/tex]
Time [tex]t_1=60s[/tex]
Time [tex]t_2=90s[/tex]
Generally the the equation for the distance traveled is mathematically given as
[tex]d=vt[/tex]
[tex]d=1.5*90[/tex]
[tex]d=138m[/tex]
Generally equation for speed of side walk is mathematically given as
[tex]d=(v+u)t[/tex]
[tex]v=\frac{d}{t}-u[/tex]
[tex]v=\frac{138}{60}-1.5[/tex]
[tex]v=0.8m/s[/tex]
2. Why are numbers better than words in a science experiment?
Why words are more important than numbers: ... Words on the other hand are harder to manipulate, they also tell you why someone voted a particular way and to improve your delivery and thus your customer satisfaction you need to understand the why's...
#pglubestiehere
Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 75.0N, the second child exerts a force of 90.0 N, friction is 12.0 N, and the most of the third child plus wagon is 23.0 kga)what is the system of interest if the acceleration of the child in the wagon is to be calculated
Answer:
Explanation:
75 N and 90 N are acting in opposite direction so net force = 90 - 75 = 15 N .
Friction force will act in the direction opposite to the direction of net force .
So friction force will act in the direction in which 75 N is acting .
Total force acting in the direction of 75 = 75 + 12 = 87 N
Net force acing on the third child = 90 - 87 = 3 N
Its direction will be that in the direction of 90 N .
A mysterious crate has shown up at your place of work, Firecracker Company, and you are told to measure its inertia. It is too heavy to lift, but it rolls smoothly on casters. Getting an inspiration, you lightly tape a 0.60-kg iron block to the side of the crate, slide a firecracker between the crate and the block, and light the fuse. When the firecracker explodes, the block goes one way and the crate rolls the other way. You measure the crate's speed to be 0.058 m/s by timing how long it takes to cross floor tiles. You look up the specifications of the firecracker and find that it releases 7 J of energy. That's all you need, and you quickly calculate the inertia of the crate.
What is that inertia?
Answer:
the inertia of the crate is (49.67 kg)r²
Explanation:
Given the data in the question;
First; we will use the law of conservation of momentum to determine the mass of the crate;
m₁v₁ - m₂v₂ = 0
given that; m₁ = 0.60 kg and v₂ = 0.058 m/s
we substitute
0.60 × v₁ = m₂ × 0.058 = 0
m₂ = 0.60v₁ / 0.058 ----------- EQU 1
Next, we use the energy conservation relation to find the velocity
According to conservation of energy;
1/2m₁v₁² + 1/2m₂v₂² = 7 J
we substitute
1/2×0.60×v₁² + 1/2×m₂×(0.058)² = 7 J
0.3v₁² + 0.001682m₂ = 7 J ----- EQU 2
substitute value of m₂ form equ 1 into equ 2
0.3v₁² + 0.001682(0.60v₁ / 0.058) = 7 J
0.3v₁² + 0.0174v₁ = 7 J
0.3v₁² + 0.0174v₁ - 7 J = 0
we solve the quadratic equation;
{ x = [-b±√( b² - 4ac)] / 2a }
v₁ = [-0.0174 ±√( 0.0174² - 4×0.3×-7)] / 2×0.3
= [-0.0174 ±√(8.4003)] / 0.6
= [-0.0174 ± 2.8983 ] / 0.6
= -4.8595 or 4.8015 but{ v₁ ≠ - }
so v₁ = 4.8015 m/s ≈ 4.802 m/s
next we input value of v₁ into equation 1
m₂ = (0.60×4.8015) / 0.058
m₂ = 2.8809 / 0.058
m₂ = 49.67 kg
So, the moment of inertia of the crate will be;
I₂ = m₂r²
we substitute value of m₂
I₂ = (49.67 kg)r²
Therefore, the inertia of the crate is (49.67 kg)r²
what measurement do geologists use to find absolute age
Answer:
see below :)
Explanation:
Radiometric dating.
A cheerleader of mass 55 kg stand on the shoulders of a football player of mass 86 kg. The football player is standing in a soft, thin layer of mud that does not permit air under his shoes. If each of his shoes has an area of 264 cm2, calculate the absolute pressure exerted on the surface underneath one of the shoes. Answer in Pascal, assuming g = 9.80 m/s2 and atmospheric pressure is 101,000 Pa.
A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. A 77.0-kg water-skier has an initial speed of 6.3 m/s. Later, the speed increases to 10.9 m/s. Determine the work done by the net external force acting on the skier.
Answer:
the work done by the net external force acting on the skier is 3046.12 J.
Explanation:
Given;
initial speed of the water skier, u = 6.3 m/s
final speed of the water skier, v = 10.9 m/s
mass of the water skier, m = 77 kg
The work done by the net external force is calculated as;
W = ΔK.E
[tex]W = \frac{1}{2} m(v^2 - u^2)\\\\W = \frac{1}{2} \times \ 77.0(10.9^2 - 6.3^2)\\\\ W= 3046.12 \ J[/tex]
Therefore, the work done by the net external force acting on the skier is 3046.12 J.
20
A person walks 2.0 m east, then turns and goes 4.0 m west, then turns
and goes back 6.0 m east. What is that person's total displacement?
(Remember to include the correct units) *
Your answer
A 50.0-g Super Ball traveling at 29.5 m/s bounces off a brick wall and rebounds at 20.0 m/s. A high-speed camera records this event. If the ball is in contact with the wall for 4.00 ms, what is the magnitude of the average acceleration of the ball during this time interval
Answer:
The magnitude of the average acceleration of the ball during this time interval is 1.238 x 10⁴ m/s².
Explanation:
Given;
mass of the super ball, m = 50 g = 0.05 kg
initial velocity of the ball, u = 29.5 m/s
final velocity of the ball, v = -20.0 m/s (negative because it rebounds)
time of contact of the ball and the wall, t = 4 ms = 4 x 10⁻³ s
The force exerted on the brick wall by the ball is given as;
[tex]F = ma\\\\ma = \frac{m(v-u)}{t} \\\\a = \frac{v-u}{t} \\\\a = \frac{(-20) - 29.5}{4.0 \ \times \ 10^{-3}} \\\\a = \frac{-49.5}{4.0 \ \times \ 10^{-3}} \\\\a = -1.238 \times 10^4 \ m/s^2\\\\|a| = 1.238 \times 10^4 \ m/s^2[/tex]
Therefore, the magnitude of the average acceleration of the ball during this time interval is 1.238 x 10⁴ m/s².
Consider two points in an electric field. The potential at point 1, V1, is 24V. The potential at point 2, V2, is 154V. A proton is moved from point 1 to point 2.
(a) Write an equation for the change of electric potential energy AU of the proton, in terms of the symbols given and the charge of the proton e.
(b) Find the numerical value of the change of the electric potential energy in electron volts (eV).
(c) Express v2, the speed of the electron at point 2, in terms of AU, and the mass of the electron me.
(d) Find the numerical value of v2 in m/s
Answer:
[tex]\triangle U=-e (V_2-V_1)[/tex]
[tex]\triangle U=130eV[/tex]
[tex]V_2=\sqrt{ \frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)}[/tex]
Explanation:
From the question we are told that
The potential at point 1, [tex]V_1 = 24V[/tex]
The potential at point 2, [tex]V_2 = 154V[/tex]
a)Generally work done by proton is given as
[tex]w=-\triangle U[/tex]
[tex]e\triangle V=-\triangle U[/tex]
[tex]\triangle U=-e (V_2-V_1)[/tex]
Generally the Equation for the change of electric potential energy AU of the proton, in terms of the symbols given and the charge of the proton e is mathematically given as
[tex]\triangle U=-e (V_2-V_1)[/tex]
b)Generally the electric potential energy in electron volts (eV). is mathematically given as
[tex]\triangle U=-e (154-24)V[/tex]
[tex]|\triangle U| =|-e (130)V|[/tex]
[tex]\triangle U=130eV[/tex]
c) Generally according to the law of conservation of energy
[tex](K.E+P.E)_1=(K.E+P.E)_2[/tex]
[tex]\frac{1}{2}meV_1^2+eV_1 =\frac{1}{2}mev_2^2+eV_2[/tex]
[tex]V_2^2=\frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)[/tex]
[tex]V_2=\sqrt{ \frac{2}{me}(\frac{1}{2}meV_1^2+e(V_2-V_1)}[/tex]
QUESTION 4.
If
you have 2 randomly selected vectors like R and R;
Show that R. RX 5) = 0
(102)
Answer:
Follows are the solution to this question:
Explanation:
Please find the correct question in the attachment file.
Let:
[tex]\overrightarrow{R}= R_i\hat{i}+R_j\hat{j}+R_k\hat{k}\\\\\overrightarrow{S}= S_i\hat{i}+S_j\hat{j}+S_k\hat{k}\\\\[/tex]
Calculating the value of [tex]\overrightarrow{R} \times \overrightarrow{S}:[/tex]
[tex]\to \left | \begin{array}{ccc}\hat{i}&\hat{j}&\hat{K}\\R_i&R_j&R_k\\S_i&S_j&S_k\end{array}\right | = \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j][/tex]
Calculating the value of [tex]\overrightarrow{R} \cdot (\overrightarrow{R} \times \overrightarrow{S}):[/tex]
[tex]\to (R_i\hat{i}+R_j\hat{j}+R_k\hat{k}) \cdot ( \hat{i}[R_j S_k-S_jR_k]-\hat{j}[R_i S_k-S_iR_k]+\hat{k}[R_i S_j-S_iR_j])[/tex]
by solving this value it is equal to 0.
At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power generation potential of a wind turbine with 80-m-diameter (D) blades at that location. Take the air density to be 1.25 kg/m3. The mechanical energy of air per unit mass is kJ/kg. The power generation potential of the wind turbine is kW.
Answer:
[tex]0.05\ \text{kJ/kg}[/tex]
[tex]3141.6\ \text{kW}[/tex]
Explanation:
v = Velocity of wind = 10 m/s
A = Swept area of blade = [tex]\dfrac{\pi}{4}d^2[/tex]
d = Diameter of turbine = 80 m
[tex]\rho[/tex] = Density of air = [tex]1.25\ \text{kg/m}^3[/tex]
Wind energy per unit mass of air is given by
[tex]E=\dfrac{v^2}{2}\\\Rightarrow E=\dfrac{10^2}{2}\\\Rightarrow E=50\ \text{J/kg}[/tex]
The mechanical energy of air per unit mass is [tex]0.05\ \text{kJ/kg}[/tex]
Power is given by
[tex]P=\rho AvE\\\Rightarrow P=1.25\times \dfrac{\pi}{4}\times 80^2\times 10\times 50\\\Rightarrow P=3141592.65=3141.6\ \text{kW}[/tex]
The power generation potential of the wind turbine is [tex]3141.6\ \text{kW}[/tex].
1. A particle is projected vertically upwards with a velocity of 30 ms from a point 0. Find (a) the maximum height reached(b) the time taken for it to return to 0 (c) the taken for it to be 35m below 0
Assuming the particle is in free fall once it is shot up, its vertical velocity v at time t is
v = 30 m/s - g t
where g = 9.8 m/s² is the magnitude of the acceleration due to gravity, and its height y is given by
y = (30 m/s) t - 1/2 g t ²
(a) At its maximum height, the particle has 0 velocity, which occurs for
0 = 30 m/s - g t
t = (30 m/s) / g ≈ 3.06 s
at which point the particle's maximum height would be
y = (30 m/s) (3.06 s) - 1/2 g (3.06 s)² ≈ 45.9184 m ≈ 46 m
(b) It takes twice the time found in part (a) to return to 0 height, t ≈ 6.1 s.
(c) The particle falls 35 m below its starting point when
-35 m = (30 m/s) t - 1/2 g t ²
Solve for t to get a time of about t ≈ 7.1 s
An engineer claims to have measured the characteristics of a heat engine that takes in 150 J of thermal energy and produces 50 J of useful work. What is the smallest possible ratio of the temperatures (in kelvin) of the hot and cold reservoirs?
Answer:
1.4999
Explanation:
Efficiency can be calculated using below expresion
Efficiency = W/Q.............eqn(1)
Where W= work = 50 J
Q= thermal energy= 150 J
But
W/Q= (Th-Tc)/Th ...........Eqn(2)
Th= temperature of the hot
Tc= temperature of the cold
Where Th/ Tc= ratio of the temperature hot and cold reservoirs?
If we simplify eqn(2) we have
W/Q = 1-Tc/Th.........eqn(3)
If we make the ratio subject of the formula we have
Tc/Th = 1-(W/Q)
Th/Tc = 1/(1-W/Q )
Then substitute the values
= 1/(1-50/150) = 1.4999
Hence, the smallest possible ratio of the temperatures (in kelvin) of the hot and cold reservoirs is 1.4999
To measure work, you must ______ the force by the distance through which it acts.
Answer:
To measure work, you must multiply the force by the distance through which it acts.
All charged objects exert a force that can cause other charges to move. What is the force that
charged objects give off called? What else can it do?
Answer:
exerts force
Explanation:
The accumulation of excess electric charge on an object is called static electricity. ... An electric field surrounds every electric charge and exerts the force that causes other electric charges to attract or repel. Electric fields are represented by arrows showing the electric field would make a positive charge move.
Two particles, an electron and a proton, are initially at rest in a uniform electric field of magnitude 554 N/C. If the particles are free to move, what are their speeds (in m/s) after 51.6 ns
Answer:
the speed of electron is 5.021 x 10⁶ m/s
the speed of proton is 2733.91 m/s
Explanation:
Given;
magnitude of electric field, E = 554 N/C
charge of the particles, Q = 1.6 x 10⁻¹⁹ C
time of motion, t = 51.6 ns = 51.6 x 10⁻⁹ s
The force experienced by each particle is calculated as;
F = EQ
F = (554)(1.6 x 10⁻¹⁹)
F = 8.864 x 10⁻¹⁷ N
The speed of the particles are calculated as;
[tex]F = ma\\\\F = \frac{mv}{t} \\\\v = \frac{Ft}{m} \\\\v_e = \frac{Ft}{m_e}\\\\v_e = \frac{(8.864 \times 10^{-17})(51.6\times 10^{-9})}{9.11 \times \ 10^{-31}}\\\\v_e = 5.021 \times 10^{6} \ m/s[/tex]
[tex]v_p = \frac{Ft}{m_p}\\\\v_p = \frac{(8.864 \times 10^{-17})(51.6\times 10^{-9})}{1.673 \times \ 10^{-27}}\\\\v_p = 2733.91 \ m/s[/tex]
In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a further step to isolate it. Let's say we now have
3C+4D=5
2C+5D=2
None of these terms has a coefficient of 1. Instead, we'll pick the variable with the smallest coefficient and isolate it. Move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient. Which of the following expressions would result from that process?
Now that you have one of the two variables in Part D isolated, use substitution to solve for the two variables. You may want to review the Multiplication and Division of Fractions and Simplifying an Expression Primers.
Answer:
D = -4/7 = - 0.57
C = 17/7 = 2.43
Explanation:
We have the following two equations:
[tex]3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)[/tex]
First, we isolate C from equation (2):
[tex]2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)[/tex]
using this value of C from equation (3) in equation (1):
[tex]3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}[/tex]
D = - 0.57
Put this value in equation (3), we get:
[tex]C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C = \frac{17}{7}\\[/tex]
C = 2.43
What health consequences is most likely to result from alcohol school?
Answer:
difficulty concentrating
A compact car has a mass of 1310 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that the mass is equally distributed over the four springs.
Required:
a. What is the spring constant of each spring if the empty car bounces up and down 2.0 times each second?
b. What will be the car’s oscillation frequency while carrying four 70 kg passengers?
Answer:
a) k= 3232.30 N / m, b) f = 4,410 Hz
Explanation:
In this exercise, the car + spring system is oscillating in the form of a simple harmonic motion, as the four springs are in parallel, the force is the sum of the 4 Hocke forces.
The expression for the angular velocity is
w = √k/m
the angular velocity is related to the period
w = 2π / T
we substitute
T = 2[tex]\pi[/tex] √m/ k
a) empty car
k = 4π² m / T²
k = 4 π² 1310/2 2
k = 12929.18 N / m
This is the equivalent constant of the short springs
F1 + F2 + F3 + F4 = k_eq x
k x + kx + kx + kx = k_eq x
k_eq = 4 k
k = k_eq / 4
k = 12 929.18 / 4
k= 3232.30 N / m
b) the frequency of oscillation when carrying four passengers.
In this case the plus is the mass of the vehicle plus the masses of the passengers
m_total = 1360 + 4 70
m_total = 1640 kg
angular velocity and frequency are related
w = 2pi f
we substitute
2 pi f = Ra K / m
in this case the spring constant changes us
k_eq = 12929.18 N / m
f = 1 / 2π √ 12929.18 / 1640
f = π / 2 2.80778
f = 4,410 Hz
Consider a turnbuckle that has been tightened until the tension in wire AD is 350 N. Draw the FBD that is required to determine the internal forces at point J. (You must provide an answer before moving on to the next part.) The FBD that is required to determine the internal forces at point J is
Answer:
yes
Explanation:
yes
According to question this is a riddle and the doctor was the boy's mother so she could not operate on him.
What is statement?A statement, question, or phrase that is presented as a problem to be solved and has a dual or disguised meaning is called a riddle. Enigmas, which are difficulties typically presented in metaphoric or allegorical language that call for inventiveness and careful thought to solve, and conundra, which are problems that rely on puns either in the question or the answer, are two different forms of riddles.
Across many nations and even entire continents, many riddles take on a similar format. Riddles might be borrowed close to home as well as long distances. A man and his son were rock climbing on a particularly dangerous mountain when they slipped and fell. the man was killed, but the son lived and was rushed to a hospital.
Therefore, According to question this is a riddle and the doctor was the boy's mother so she could not operate on him.
Learn more about riddle here:
brainly.com/question/478260
#SPJ5
The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero. (a) Determine the forces and and the couple (b) Determine the sum of the moments about the right end of the beam. (c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the
This question is incomplete, the complete question is;
The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero.
(a) Determine the forces and and the couple
(b) Determine the sum of the moments about the right end of the beam.
(c) If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the beam and a couple M, what is F and M?
Answer:
a)
the x-component of the force at A is [tex]A_{x}[/tex] = 0
the y-component of the force at A is [tex]A_{y}[/tex] = 400 N
the couple acting at A is; [tex]M_{A}[/tex] = 146 N-m
b)
the sum of the momentum about the right end of the beam is; ∑[tex]M_{R}[/tex] = 0
c)
the equivalent force acting at the left end is; F = -400J ( N)
the couple acting at the left end is; M = - 146 N-m
Explanation:
Given that;
The sum of the forces acting on the beam is zero ∑f = 0
Sum of the moments about the left end of the beam is also zero ∑[tex]M_{L}[/tex] = 0
Vector force acting at A, [tex]F_{A}[/tex] = [tex]A_{x}i[/tex] + [tex]A_{y}j[/tex]
Now, From the image, we have;
a)
∑f = 0
[tex]F_{A}[/tex] - 600j + 200j = 0i + 0j
[tex]A_{x}i[/tex] + [tex]A_{y}j[/tex] - 600j + 200j = 0i + 0j
[tex]A_{x}i[/tex] + ([tex]A_{y}[/tex] - 400)j = 0i + 0j
now by equating i- coefficients'
[tex]A_{x}[/tex] = 0
so, the x-component of the force at A is [tex]A_{x}[/tex] = 0
also by equating j-coefficient
[tex]A_{y}[/tex] - 400 = 0
[tex]A_{y}[/tex] = 400 N
hence, the y-component of the force at A is [tex]A_{y}[/tex] = 400 N
we also have;
∑[tex]M_{L}[/tex] = 0
[tex]M_{A}[/tex] - ( 30 N-m ) - ( 0.380 m )( 600 N ) + ( 0.560 m )( 200 N ) = 0
[tex]M_{A}[/tex] - 30 N-m - 228 N-m + 112 Nm = 0
[tex]M_{A}[/tex] - 146 N-m = 0
[tex]M_{A}[/tex] = 146 N-m
Therefore, the couple acting at A is; [tex]M_{A}[/tex] = 146 N-m
b)
The sum of the moments about right end of the beam is;
∑[tex]M_{R}[/tex] = (0.180 m)(600N) - (30 N-m) - ( 0.56 m)([tex]A_{y}[/tex] ) + [tex]M_{A}[/tex]
∑[tex]M_{R}[/tex] = (108 N-m) - (30 N-m) - ( 0.56 m)(400 N ) + 146 N-m
∑[tex]M_{R}[/tex] = (108 N-m) - (30 N-m) - ( 224 N-m ) + 146 N-m
∑[tex]M_{R}[/tex] = 0
Therefore, the sum of the momentum about the right end of the beam is; ∑[tex]M_{R}[/tex] = 0
c)
The 600-N force, the 200-N force and the 30 N-m couple by a force F which is acting at the left end of the beam and a couple M.
The equivalent force at the left end will be;
F = -600j + 200j (N)
F = -400J ( N)
Therefore, the equivalent force acting at the left end is; F = -400J ( N)
Also couple acting at the left end
M = -(30 N-m) + (0.560 m)( 200N) - ( 0.380 m)( 600 N)
M = -(30 N-m) + (112 N-m) - ( 228 N-m))
M = 112 N-m - 258 N-m
M = - 146 N-m
Therefore, the couple acting at the left end is; M = - 146 N-m
state four law of photoelectric effect
Answer:
LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation.
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LAW 2: For a given metal, there exists a certain frequency below which the photoelectric emission does not take place. This frequency is called threshold frequency.
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LAW 3: For a frequency greater than the threshold frequency, the kinetic energy of photoelectrons is dependent upon frequency or wavelength but not on the intensity of light.
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LAW 4: Photoelectric emission is an instantaneous process. The time lag between incidence of radiations and emission of electron is 10^-9 seconds.
Explanation:
Answer:
LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation. ... LAW 4: Photoelectric emission is an instantaneous process.
during a typical afternoon thunderstorm in the summer, an area of 66.0 km2 receives 9.57 108 gal of rain in 18 min. how many inches of rain fell during this 18 min period
Answer:
2.16 inch
Explanation:
area under water = 66 km²
= 66 x ( 3280.84 x 12 )² inch²
= 1.023 x 10¹¹ sq inch
volume of rain = 9.57 x 10⁸ gallon = 9.57 x 10⁸ x 231 inch³
= 2.21 x 10¹¹ inch³
If depth of rainfall be t
volume of rain = surface area x depth
= 1.023 x 10¹¹ x t
So ,
1.023 x 10¹¹ x t = 2.21 x 10¹¹
t = 2.16 inch
What day of the year is solar time the same as sidereal time?
Answer:
I think the answers March 21
Answer:
Once a year, mean solar time and sidereal time will be the same.
A three-phase line, which has an impedance of (2 + j4) ohm per phase, feeds two balanced three-phase loads that are connected in parallel. One of the loads is Y-connected with an impedance of (30 + j40) ohm per phase, and the other is connected with an impedance of (60 - j45) ohm per phase. The line is energized at the sending end from a 60-Hz, three-phase, balanced voltage source of 120 √3V (rms, line-to-line).
Determine:
a. the current, real power and reactive power delivered by the sending-end source
b. the line-to-line voltage at the load
c. the current per phase in each load
d. the total three-phase real and reactive powers absorbed by each load and by the
Answer:
hello your question has a missing information
The other is Δ-connected with an impedance of (60 - j45) ohm per phase.
answer : A) 5A ∠0° ,
p( real power ) = 1800 and Q ( reactive power ) = 0 VAR
B) 193.64 v
C) current at load 1 = 2.236 A , current at load 2 = 4.472 A
D) Load 1 : 450 watts(real power ) , 600 VAR ( reactive power )
Load 2 : 1200 watts ( real power ), -900 VAR ( reactive power )
Explanation:
First convert the Δ-connection to Y- connection attached below is the conversion and pre-solution
A) determine the current, real power and reactive power delivered by the sending-end source
current power delivered (Is) = 5A ∠0°
complex power delivered ( s ) = 3vs Is
= 3 * 120∠0° * 5∠0° = 1800 + j0 ---- ( 1 )
also s = p + jQ ------ ( 2 )
comparing equation 1 and 2
p( real power ) = 1800 and Q ( reactive power ) = 0 VAR
B) determine Line-to-line voltage at the load
Vload = √3 * 111.8
= 193.64 v
c) Determine current per phase in each load
[tex]I_{l1} = Vl1 / Zl1[/tex]
= [tex]\frac{111.8<-10.3}{50<53.13}[/tex] = 2.236∠ 63.43° A hence current at load 1 = 2.236 A
[tex]I_{l2} = V_{l2}/Z_{l2}[/tex]
= [tex]\frac{111.8<-10.3}{25<-36.87}[/tex] = 4.472 ∠ 26.57° A hence current at load 2 = 4.472 A
D) Determine the Total three-phase real and reactive powers absorbed by each load
For load 1
3-phase real power = [tex]3I_{l1} ^{2} R_{l1}[/tex] = 3 * 2.236^2 * 30 = 450 watts
3-phase reactive power = [tex]3I_{l1} ^{2} X_{l1}[/tex] = 3 * 2.236^2 * 40 = 600 VAR
for load 2
3-phase real power = [tex]3I_{l1} ^{2} R_{l2}[/tex] = 1200 watts
3-phase reactive power = [tex]3I_{l1} ^{2} X_{l2}[/tex] = -900 VAR
The sum of load powers and line losses, 1800 W+ j0 VAR and The line voltage magnitude at the load terminal is 193.64 V.
(a) The impedance per phase of the equivalent Y,
[tex]\bar{Z}_{2}=\frac{60-j 45}{3}=(20-j 15) \Omega[/tex]
The phase voltage,
[tex]\bold { V_{1}=\frac{120 \sqrt{3}}{\sqrt{3}}=120 VV }[/tex]
Total impedance from the input terminals,
[tex]\bold {\begin{aligned}&\bar{Z}=2+j 4+\frac{(30+j 40)(20-j 15)}{(30+j 40)+(20-j 15)}=2+j 4+22-j 4=24 \Omega \\&\bar{I}=\frac{\bar{V}_{1}}{\bar{Z}}=\frac{120 \angle 0^{\circ}}{24}=5 \angle 0^{\circ} A\end{aligned} }[/tex]
The three-phase complex power supplied [tex]\bold {=\bar{S}=3 \bar{V}_{1} \bar{I}^{*}=1800 W}[/tex]
P =1800 W and Q = 0 VAR delivered by the sending-end source.
(b) Phase voltage at load terminals will be,
[tex]\bold {\begin{aligned}\bar{V}_{2} &=120 \angle 0^{\circ}-(2+j 4)\left(5 \angle 0^{\circ}\right) \\&=110-j 20=111.8 \angle-10.3^{\circ} V\end{aligned} }[/tex]
The line voltage magnitude at the load terminal,
[tex]\bold{\left(V_{ LOAD }\right)_{L-L}=\sqrt{3} 111.8=193.64 V(V }[/tex]
(c) The current per phase in the Y-connected load,
[tex]\bold {\begin{aligned}&\bar{I}_{1}=\frac{\bar{V}_{2}}{\bar{Z}_{1}}=1-j 2=2.236 \angle-63.4^{\circ} A \\&\bar{I}_{2}=\frac{\bar{V}_{2}}{\bar{Z}_{2}}=4+j 2=4.472 \angle 26.56^{\circ} A\end{aligned} }[/tex]
The phase current magnitude,
[tex]\bold {\left(I_{p h}\right)_{\Delta}=\frac{I_{2}}{\sqrt{3}}=\frac{4.472}{\sqrt{3}}=2.582 }[/tex]
(d) The three-phase complex power absorbed by each load,
[tex]\bold {\begin{aligned}&\bar{S}_{1}=3 \bar{V}_{2} \bar{I}_{1}^{*}=430 W +j 600 VAR \\&\bar{S}_{2}=3 \bar{V}_{2} \bar{I}_{2}^{*}=1200 W -j 900 VAR\end{aligned}}[/tex]
The three-phase complex power absorbed by the line is
[tex]\bold{\bar{S}_{L}=3\left(R_{L}+j X_{L}\right) I^{2}=3(2+j 4)(5)^{2}=150 W +j 300 VAR }[/tex]
Since, the sum of load powers and line losses,
[tex]\bold {\begin{aligned}\bar{S}_{1}+\bar{S}_{2}+\bar{S}_{L} &=(450+j 600)+(1200-j 900)+(150+j 300) \\&=1800 W +j 0 VAR\end{aligned} }[/tex]
To know more about voltage,
https://brainly.com/question/2364325
What energy store is in the human
BEFORE he/she lifts the hammer?
I believe the answer would be protentional because they have the potential energy in them to lift the hammer.