Answer:
The correct option is B
Explanation:
From the question we are told that
The mass of the pile is M
The height is H = 4 y
The vertical distance achieve during the first lift is [tex]h_1 = 3 y[/tex]
The time taken is [tex]t_1 = 4T [/tex]
The vertical distance achieve during the second lift is [tex]h_2 = y[/tex]
The time taken is [tex] t_2 = T [/tex]
Generally the velocity of the crane during the first lift is
[tex]v _1 = \frac{h_1}{t_1 }[/tex]
=> [tex]v _1 = \frac{3 y}{4T }[/tex]
Generally the velocity of the crane during the second lift is
[tex]v _2 = \frac{h_2}{t_2 }[/tex]
=> [tex]v _2 = \frac{ y}{T}[/tex]
Generally the power generated by the crane during the first lift is
[tex]P_1 = F_1 * v_1[/tex]
Here [tex]F_1[/tex] is the weight of the brick which is mathematically represented as
[tex]F_1 = M * g [/tex] , g is the acceleration due to gravity
So
[tex]P_1 = Mg * \frac{3y}{4T}[/tex]
Generally the power generated by the crane during the first lift is
[tex]P_1 = F_2 * v_2[/tex]
Here [tex]F_2[/tex] is the weight of the brick which is mathematically represented as
[tex]F_2 = M * g [/tex] , g is the acceleration due to gravity
So
[tex]P_1 = Mg * \frac{y}{T}[/tex]
The ratio of the first power generated to the second power is
[tex]\frac{P_1}{P_2} = \frac{Mg * \frac{3y}{4T} }{ Mg * \frac{y}{T} }[/tex]
=> [tex]\frac{P_1}{P_2} = \frac{3}{4}[/tex]
=> [tex]P_2 = \frac{4}{3} P_1[/tex]
1
2
3
4
5
6
8
9
10
Dressing appropriately for exercise includes
A. wearing the same clothes for all exercises
B. choosing dark colored clothing when exercising at night
C. wearing sunscreen when exercising outside
D. making sure you wear the best brand-name clothes
Please select the best answer from the choices provided.
A
B.
C.
D.
The answer is C.
Answer:
the answer is C
Explanation:
you said its c
A car with a mass of 2,000 kg travels a distance of 400m as it moves from one stoplight to the next. At its fastest , the car travels 18m/s. What is the kinetic energy at this point ?
Answer: KE= 324,000
Explanation: I hope that this helps! -_-
Answer:
324,000
Explanation:
One student runs with a velocity of +10 m/s while a second student runs with a velocity of –10 m/s. Which student has the faster velocity? Why?
Answer:
The one with the faster velocity is the one with a velocity of -10m/s
the radius of the earth social
3. A wye-connected load has a voltage of 480 V applied to it. What is the voltage dropped across each phase?
Answer:
[tex]E_s = 277.13V[/tex]
Explanation:
Given
[tex]Load\ Voltage = 480V[/tex]
Required
Determine the voltage dropped in each stage.
The relation between the load voltage and the voltage dropped in each stage is
[tex]E_l = E_s * \sqrt3[/tex]
Where
[tex]E_l = 480[/tex]
So, we have:
[tex]480 = E_s * \sqrt3[/tex]
Solve for [tex]E_s[/tex]
[tex]E_s = \frac{480}{\sqrt3}[/tex]
[tex]E_s = \frac{480}{1.73205080757}[/tex]
[tex]E_s = 277.128129211[/tex]
[tex]E_s = 277.13V[/tex]
Hence;
The voltage dropped at each phase is approximately 277.13V
Which factor affects the amount of runoff that occurs in an area?
land use
the water table
the saturation zone
amount of nutrients in soil
Answer:
A. land use
Explanation:
Answer:
a
Explanation:
What is the change in internal energy (in J) of a system that absorbs 0.523 kJ of heat from its surroundings and has 0.366 kcal of work done on it
Answer:
The change in internal energy of the system is 2,054 J
Explanation:
The first law of thermodynamics relates the work and the transferred heat exchanged in a system through internal energy. This energy is neither created nor destroyed, it is only transformed.
Taking into account that the internal energy is the sum of all the energies of the particles that the system has, you have:
ΔU= Q + W
where U is the internal energy of the system (isolated), Q is the amount of heat contributed to the system and W is the work done by the system.
By convention, Q is positive if it goes from the environment to the system, or negative otherwise, and W is positive if it is carried out on the system and negative if it is carried out by the system.
In this case:
Q= 0.523 kJ (because the energy is absorbed, this is,it goes from the environment to the system)W= 0.366 kcal= 1.531 kJ (because the work is done on the system, and being 1 kcal= 4.184 kJ)Replacing:
ΔU= 0.523 kJ + 1.531 kJ
Solving:
ΔU= 2.054 kJ = 2,054 J (being 1 kJ=1,000 J)
The change in internal energy of the system is 2,054 J
how much power is used if it takes frank (a 450 N boy ) 3 seconds to run 2 meters ?
Answer:
300
Explanation:
450Newton × 2Meter ÷ 3sec
Matching type. Send help please. ASAP!
Answer:
46-D
47-C
48-F
49-A
50-B
I am not very sure I am right about those answers though.
If Mary runs 5 miles in 50 minutes, what is her speed with the correct
label?
Victor has 100 trading cards, each with a distinct power level between 101 and 200 inclusive. Whenever he heads out, he always randomly selects 21 cards to bring with him just in case he meets a fellow collector. Prove that no matter which 21 cards he brings, Victor will always be able to select 4 of those cards that exhibit the following property:
Let the average power level of all 4 cards bep. The cards can be split into two pairs, each of which also has an average power level of p.
Answer:
Victor will always be able to select 4 of those cards with the following property
Explanation:
Number of trading cards = 100
victor selects 21 cards
let the 4 cards be labelled : A,B,C and D
The average power level of : A,B,C,D = ( A + B + C + D )/ 4 = P
let the two pairs be : ( A + B ) and ( C + D )
note: average power of each pair = P and this shows that
( A + B ) = ( C + D ) for Victor to select 4 cards out of the 21 cards that exhibit the same property
we have to check out the possible choices of two cards out of 21 cards yield distinct sums.
= C(21,2)=(21x20)/2 = 210.
from the question the number of distinct sums that can be created using 101 through 200 is < 210 .
hence it is impossible to get 210 distinct sums therefore Victor will always be able to select 4 of those cards
What must be true if a wave's wavelength is short?
A
Its frequency is low.
B
It does not carry energy.
с
Its frequency is high.
D
The waves are visible.
Answer:
im sure its A
Explanation:
A pair of glasses are dropped from the top of a 32.0 m high stadium. A pen is dropped 2.00 s later. How high above the ground is the pen when the glasses hit the ground? (Disregard air resistance. a = -g = -9.81 m/s2.)
Answer:
Explanation:
the distance have the following relation:
d = (1/2)gt2
D=32.0 m
t =√ (2D/g) = √(2*32.0m/9.8m/s2) = 2.56s
it take 2.56s from the glasses to hit the ground
when the glasses hit the ground, the pen only travel Δt =2.56s - 2.00s = 0.56s
x = (1/2)g(Δt)2 = 0.5*9.8m/s2*(0.56s)2 = 1.54 m
the pen only travel 1.54m
so the pen is above the ground 32.0m - 1.54m = 30.46m
The pen is 30.46m above the ground. when the glasses hit the ground. It is represented by x.
What is the height?Height is a numerical representation of the distance between two objects or locations on the vertical axis.
The height can refer to a physical length or an estimate based on other factors in physics or common use. |
The given data in the problem is;
h is the height from the top of a stadium = 32.0 m
t is the time period when the pen is dropped later = 2.00 s
x is the height above the ground
a is the air resistance. a = -g = -9.81 m/s²
From the second equation of motion;
[tex]\rm H =ut+\frac{1}{2} gt^2 \\\\ \rm H =\frac{1}{2} gt^2 \\\\ \rm t = \sqrt{\frac{2H}{g} } \\\\ \rm t = \sqrt{\frac{2 \times32.0 }{9.81} } \\\\ \rm t =2.56\ sec[/tex]
When the glasses fall to the ground, the pen only travels a short distance;
[tex]\rm \triangle t = 2.56 -2.00 \\\\ \rm \triangle t = 0.56 \ sec[/tex]
So the pen travel the distance;
[tex]\rm h= \frac{1}{2} g \triangle t^2 \\\\ \rm h= \frac{1}{2} \times 9.81 (0.56)^2 \\\\ h=1.54 \ m[/tex]
The pen above the ground is found as;
[tex]\rm x = H-h \\\\ \rm x = 32.0-1.54 \\\\ \rm x =30.46 \ m[/tex]
Hence the pen is 30.46m above the ground. when the glasses hit the ground.
To learn more about the height refer to the link;
https://brainly.com/question/10726356
calculating light in physics
1. According to its computer, a rocket launched
traveled 1200 m, had an average speed of 100.0
m/s. How-long did the trip take?
Answer:
I think it's = 12 seconds
Explanation:
the formula for speed is:
speed=[tex]\frac{distance}{time}[/tex] SO, time is equal to:
time=[tex]\frac{distance}{speed}[/tex]
(sub the numbers in the formula)
distance=1200m, speed=100m/s
time=[tex]\frac{1200}{100}[/tex]
=12 seconds
Grass and plants get energy from
А
the sun.
B
eating food.
с
windmills.
D
electrons.
Answer:
From the Sun
Explanation:
Plants can't eat any food. They don't ue or need windmills to get energy. They are plants so they don't have any electrons. The only way that they can recive energy from is the sun. Sometimes plants die when they don't get enough sun because they don't have any energy to live.
An object is dropped from a 45 m high building. At the same time, another object is thrown
upward with a velocity of 8.5 ms 1. How high above the ground will the two objects meet?
(With work please)
Answer:
-92.33 (meaning the objects will not meet above the ground).
Explanation:
We can use the kinematic equation displacement = initial velocity*time + 1/2*acceleration*time^2.
We can plug in the known values of the 2 objects into the equation, where t is the time and x is the displacement:
x = 0*t + 1/2*(-9.8)*t^2+45
x = 8.5*t + 1/2*(-9.8)*t^2
We need to first solve for t to solve for x. Since both equations are equal to x, we can set them equal to each other and solve for t:
0*t + 1/2*(-9.8)*t^2+45 = 8.5*t + 1/2*(-9.8)*t^2
-4.9*t^2 +45 = 8.5*t + -4.9*t^2
45 = 8.5*t
t = 45/8.5 ≈5.294
Now, we can plug t as 5.294 into any of the equations above to solve for x:
x = 0*5.294 + 1/2*-9.8*(5.294)^2+45 ≈ -92.33
That means, the objects will not meet above the ground.
Determine the electrical force of attraction between two balloons
that are charged with the opposite type of charge but the same
quantity of charge. The charge on the balloons is 6.0 x 10-7 C and they
are separated by a distance of 0.50 m.
Answer:
F=1.3x10^-2N
Explanation:
Fe= k(6x10^-7C)^2/(0.5)^2
Electrical force of attraction between the balloons is F=1.3x10^-2N
The electric force of attraction between two balloons should be F=1.3x10^-2N.
Calculation of the electric force;Since The charge on the balloons is 6.0 x 10-7 C and they are separated by a distance of 0.50 m.
So, here the electric force is
Fe= k(6x10^-7C)^2/(0.5)^2
F=1.3x10^-2N
hence, The electric force of attraction between two balloons should be F=1.3x10^-2N.
Learn more about force here: https://brainly.com/question/19848845
An object moving 20 m/s
experiences an acceleration of 4 m/s' for 8
seconds. How far did it move in that time?
Variables:
Equation and Solve:
Answer:
We are given:
initial velocity (u) = 20m/s
acceleration (a) = 4 m/s²
time (t) = 8 seconds
displacement (s) = s m
Solving for Displacement:
From the seconds equation of motion:
s = ut + 1/2 * at²
replacing the variables
s = 20(8) + 1/2 * (4)*(8)*(8)
s = 160 + 128
s = 288 m
In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop. (a) What is the kinetic energy of the ball just bef
Answer:
(a) The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.
(b) The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.
(c) Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.
(d) The work done by the mattress on the bowling ball is 113.272 joules.
Explanation:
The statement is incomplete. The complete question is:
In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop.
(a) What is the kinetic energy of the ball just before it hits the mattress?
(b) How much work does the gravitational force of the earth do on the ball as it falls, for the first part of the fall (from the moment you drop it to just before it hits the mattress)?
(c) How much work does the gravitational force do on the ball while it is compressing the mattress?
(d) How much work does the mattress do on the ball? (You’ll need to use the results of parts (a) and (c)）
(a) Based on the Principle of Energy Conservation, we know that ball-earth system is conservative, so that kinetic energy is increased at the expense of gravitational potential energy as ball falls:
[tex]K_{1}+U_{g,1} = K_{2}+U_{g,2}[/tex] (Eq. 1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Kinetic energies at top and bottom, measured in joules.
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Gravitational potential energies at top and bottom, measured in joules.
Now we expand the expression by definition of gravitational potential energy:
[tex]U_{g,1}-U_{g,2} = K_{2}-K_{1}[/tex]
[tex]K_{2}= m\cdot g \cdot (z_{1}-z_{2})+K_{1}[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the bowling ball, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Initial and final heights of the bowling ball, measured in meters.
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex], [tex]z_{2} = 0\,m[/tex] and [tex]K_{1} = 0\,J[/tex], the kinetic energy of the ball just before it hits the matress:
[tex]K_{2} = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (1.5\,m-0\,m)+0\,m[/tex]
[tex]K_{2} = 102.974\,J[/tex]
The kinetic energy of the bowling ball just before it hits the matress is 102.974 joules.
(b) The gravitational work done by the gravitational force of Earth ([tex]\Delta W[/tex]), measured in joules, is obtained by Work-Energy Theorem and definition of gravitational potential energy:
[tex]\Delta W = U_{g,1}-U_{g,2}[/tex]
[tex]\Delta W = m\cdot g\cdot (z_{1}-z_{2})[/tex] (Eq. 2)
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{1}= 1.5\,m[/tex] and [tex]z_{2} = 0\,m[/tex], then the gravitational work done is:
[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (1.5\,m-0\,m)[/tex]
[tex]\Delta W = 102.974\,J[/tex]
The work done by the gravitational force of Earth on bowling ball during the first part of the fall is 102.974 joules.
(c) The work done by the gravitational force of Earth while the bowling when mattress is compressed is determined by Work-Energy Theorem and definition of gravitational potential energy:
[tex]\Delta W = U_{g,2}-U_{g,3}[/tex]
Where [tex]U_{g,3}[/tex] is the gravitational potential energy of the bowling ball when mattress in compressed, measured in joules.
[tex]\Delta W = m\cdot g \cdot (z_{2}-z_{3})[/tex]
Where [tex]z_{3}[/tex] is the height of the ball when mattress is compressed, measured in meters.
If we know that [tex]m = 7\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{2}= 0\,m[/tex] and [tex]z_{3} = -0.15\,m[/tex], the work done is:
[tex]\Delta W = (7\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot [0\,m-(-0.15\,m)][/tex]
[tex]\Delta W = 10.298\,J[/tex]
Work done by gravitational force on bowling ball when mattress is compressed is 10.298 joules.
(d) The work done by the mattress on the ball equals the sum of kinetic energy just before mattress compression and the work done by the gravitational force when mattress is compressed:
[tex]\Delta W' = K_{2}+\Delta W[/tex]
([tex]K_{2} = 102.974\,J[/tex], [tex]\Delta W = 10.298\,W[/tex])
[tex]\Delta W' = 113.272\,J[/tex]
The work done by the mattress on the bowling ball is 113.272 joules.
A car traveling at 32.4 m/s skids to a stop in 4.55 s. Determine the skidding distance of the car (assume uniform acceleration).
Answer:
x = 73.71 [m]
Explanation:
In order to solve this problem we must use two formulas of kinematics. It is important to make it clear that these formulate are for uniformly accelerated motion, i.e. with constant acceleration.
[tex]v_{f }= v_{i}-(a*t)[/tex]
where:
Vf = fnal velocity = 0
Vi = initial velocity = 32.4 [m/s]
t = time = 4,55 [s]
a = acceleration or desacceleration [m/s^2]
0 = 32.4 - (a*4.55)
a = 7.12 [m/s^2]
Note: it is important to clarify that the negative sign in the above equation is because the car stops and decreases its speed to zero, thus its final velocity is equal to zero.
Now using the following equation:
[tex]x = x_{o} + (v_{i}*t)-(\frac{1}{2} )*a*t^{2}[/tex]
where:
xo = initial distance = 0
x = final distance [m]
Therefore we have:
x = 0 + (32.4*4.55) - (0.5*7.12*4.55^2)
x = 73.71 [m]
A diffusion couple, made by welding a thin onecentimeter square slab of pure metal A to a similar slab of pure metal B, was given a diffusion anneal at an elevated temperature and then cooled to room temperature. On chemically analyzing successive layers of the specimen, cut parallel to the weld interface, it was observed that, at one position, over a distance of 5000 nm, the atom fraction of metal A, NA, changed from 0.30 to 0.35. Assume that the number of atoms per m3 of both pure metals is 9 x 10^28. First determine the concentration gradient dnA/dx. Then if the diffusion coefficient, at the point in question and annealing temperature, was 2 10^-14 m^2/s.
Required:
Determine the number of A atoms per second that would pass through this cross-section at the annealing temperature.
Answer:
The value is [tex]H = 18*10^{2} \ Atom / sec [/tex]
Explanation:
From the question we are told that
The atom fraction of metal A at point G is [tex] A = 0.30 \ m[/tex]
The atom fraction of metal A at a distance 5000nm from G is [tex]A_2 = 0.35[/tex]
The number of atoms per [tex]m^3[/tex] is [tex]N_h = 9 * 10^{28}[/tex]
The diffusion coefficient is [tex]D = 2* 10^{-14 } m^2/s[/tex]
Generally of the concentration of atoms of metal A at G is
[tex] N_A = A * N_h [/tex]
=> [tex] N_A = 0.3 * 9 * 10^{28}[/tex]
=> [tex] N_A = 2.7 * 10^{28} 2.7 atoms/m^3[/tex]
Generally of the concentration of atoms of metal A at a distance 5000nm from G is
[tex]D = 0.35 *9 * 10^{28}[/tex]
=> [tex]D = 3.15 * 10^{28} \ atoms / m^3[/tex]
The concentration gradient is mathematically represented as
[tex]\frac{dN_A}{dx} = \frac{(3.15 - 2.7) * 10^{28} }{5000nm - 0 }[/tex]
=> [tex]\frac{dN_A}{dx} = \frac{(3.15 - 2.7) * 10^{28} }{[5000 *10^{-9}] - 0 }[/tex]
=> [tex]\frac{dN_A}{dx} = 9 * 10^{20} / m^4[/tex]
Generally the flux of the atoms per unit area according to Fick's Law is mathematically represented as
[tex]J = -D* \frac{d N_A}{dx}[/tex]
=> [tex]J = -2* 10^{-14 * 9 * 10^{20} [/tex]
=> [tex] J = 18*10^{6}\ atoms\ crossing\ /m^2 s [/tex]
Generally if the cross-section area is [tex] a = 1 cm^2 = 10^{-4} \ m^2[/tex]
Generally the number of atom crossing the above area per second is mathematically is
[tex]H = 18*10^{6} * 10^{-4} [/tex]
=> [tex]H = 18*10^{2} \ Atom / sec [/tex]
if a ramp is 12 meters long has a mechanical advantage of 6 whats its height brainly HELPP!
Answer:
h = 2 m
Explanation:
Mechanical advantage of a ramp is given by :
MA = length of incline/height of incline
Length of the ramp, l = 12 m
MA = 6
We need to find the height of the ramp.
So,
height of ramp = length of incline/MA
h = 12/6
h = 2 m
So, the height of the ramp is 2 m.
A person has a mass of 1000g and an acceleration of 20 m/s/s. What is the force on the person
Answer:
20000
Explanation:
Newtons Second law states that the force acting on an object is equal to its mass times its acceleration, f=ma. To solve for force, plug in your values for m and a, and then solve. f = (1000)*(20) = 20000
a motor boat is traveling at 25 knots towards 340 degree on a river flowing at 20 knots towards 175 degrees. What is the actual speed of the boat as seen by a helicopter piolet hovering above?
Answer:
Vbx = 25 * cos 340 = 23.5 knot x-component of boats speed
Vrx = 20 cos 175 = -19.9 knot x-component of rivers speed
Vx = 3.58 knot x-component of boat and river speed
Vby = 25 sin 340 = -8.55 knot y-component of boat speed
Vry = 20 sin 175 = 1.74 knot y-component of river speed
Vy = -6.81 knot y-component of boat and river speed
V = (Vx^2 + Vb^2)^1/2 = (3.58^2 + 6.81^2)^1/2 = 7.69 knots
why do some athletes get injuries before and after the game?
Answer:
they don't strech so they tear a muscle when they perform
Explanation:
What is the force of a 12 kg object that is accelerating 6 m/s
We are given:
Mass of object (m) = 12 kg
acceleration (a) = 6 m/s²
Solving for the Force:
From newton's second law of motion:
F = ma
replacing the variables
F = 12*6
F = 72N
What does the principle of superposition help scientists determine?
A) The super powers of a rock layer
B) The exact and absolute age of a rock layer
C) The relative age of a rock layer
D) The position of a fossil
Answer:
B
Explanation:
the exact and absolute age of a rock layer
Answer:
The relative age of a rock layer.
Explanation:
The answer is C.
How far will a 600 kg boat travel in 12 s if there is a constant 900 N force on it and it starts from rest?
Answer:
108 metres
Explanation:
Given
[tex]Force = 900N[/tex]
[tex]Mass = 600kg[/tex]
[tex]Time = 12s[/tex]
Required
Determine the distance moved
First, we need to determine the acceleration.
[tex]Force = Mass * Acceleration[/tex]
[tex]900N = 600kg * a[/tex]
Solve for a
[tex]a = 900/600[/tex]
[tex]a = 1.5m/s^2[/tex]
Next, we determine the distance using:
[tex]S = ut + \frac{1}{2}at^2[/tex]
Since it starts from rest,
[tex]u = 0[/tex]
[tex]t = 12[/tex]
[tex]a = 1.5[/tex]
So:
[tex]S = 0 * 12 + \frac{1}{2} * 1.5 * 12^2[/tex]
[tex]S = \frac{1}{2} * 1.5 * 144[/tex]
[tex]S = \frac{1}{2} * 216[/tex]
[tex]S = 108m[/tex]
An object is dropped from rest and falls freely 20 m to Earth. When is the speed of the object 9.8 m/s?
At the end of the first second of its fall
At the end of the first second of its fall
During the entire time of its fall
During the entire time of its fall
At no time is the speed 9.8 m/s
At no time is the speed 9.8 m/s
During the entire first second of its fall
During the entire first second of its fall
After it has fallen 9.8 meters