Answer number four because it is one of the factors evaluating science
what is phase voltage of star connected three phase system in balanced loa is 240 the line voltage
Answer:
Explanation: the Star Connection, the similar ends (either start or finish) of the three windings are connected to a common point called star or neutral point. The three-line conductors run from the remaining three free terminals called line conductor
A small mass is released from rest at a very great distance from a larger stationary mass. Draw a graphs best represents the gravitational potential energy U of the system of the two masses as a function of time T.
Answer:
attached below
Explanation:
Gravitational potential
energy = [tex]- \frac{GmM}{r}[/tex] ,
V ∝ [tex]- \frac{1}{r}[/tex]
attached below is a graph that represents the gravitational potential energy U
Gravitational potential energy is the energy held in an entity as a result of its vertical position or length. This gravitational pull of a planet on an object causes the information to be stored.
This is connected with gravitational pull since it takes work to lift anything against the gravity of the Earth.
Freshwater in a raised lake or kept behind a dam demonstrates gravitational force.
According to the formula:
Gravitational potential energy [tex]\bold{=-\frac{GMm}{r^2}}[/tex]
Therefore
[tex]\to \bold{V \propto -\frac{1}{r}}[/tex]
Therefore, the answer is "Option d"
Learn more about the Gravitational potential :
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Suppose the maximum safe intensity of microwaves for human exposure is taken to be 1.00 W/m2 . (a) If a radar unit leaks 10.0 W of microwaves (other than those sent by its antenna) uniformly in all directions, how far away must you be to be exposed to an intensity considered to be safe
Answer:
We must be approximately at least 1.337 meters away to be exposed to an intensity considered to be safe.
Explanation:
Let suppose that intensity is distributed uniformly in a spherical configuration. By dimensional analysis, we get that intensity is defined by:
[tex]I = \frac{\dot W}{\frac{4\pi}{3}\cdot r^{3}}[/tex] (1)
Where:
[tex]I[/tex] - Intensity, measured in watts per square meter.
[tex]r[/tex] - Radius, measured in meters.
If we know that [tex]\dot W = 10\,W[/tex] and [tex]I = 1\,\frac{W}{m^{2}}[/tex], then the radius is:
[tex]r^{3} = \frac{\dot W}{\frac{4\pi}{3}\cdot I }[/tex]
[tex]r = \sqrt[3]{\frac{3\cdot \dot W}{4\pi\cdot I} }[/tex]
[tex]r = \sqrt[3]{\frac{3\cdot (10\,W)}{4\pi\cdot \left(1\,\frac{W}{m^{2}} \right)} }[/tex]
[tex]r \approx 1.337\,m[/tex]
We must be approximately at least 1.337 meters away to be exposed to an intensity considered to be safe.
A 20 kg sled is pulled up a 10m tall hill. What work is done against gravity?
Answer:
1962
Explanation:
w = f × d
= 20×9.81 × 10
= 1962
54:29
Mason notices that his boat sinks lower into the water in a freshwater lake than in the ocean. What could explain
this?
The boat is more dense in salt water.
The boat is less dense in salt water.
Salt water is less dense than freshwater
O Salt water is more dense than freshwater
Save and Exit
Sube
Mark this and retum
Answer:
Salt water is more dense than freshwater
Explanation:
The only and best explanation for this phenomenon for this occurrence is that salt water is more dense than freshwater.
Density is the mass per unit volume of a substance. When a substance is A more dense than the other substance B, B will float on A. So, the density of the salt water is higher than that of the fresh water because it has more particles per unit volume compared to fresh water. Therefore, the boat will go farther into the fresh water compared to the salt water.A finch rides on the back of a Galapagos tortoise, which walks at the stately pace of 0.060 m>s. After 1.5 minutes the finch tires of
Complete Question:
A finch rides on the back of a Galapagos tortoise, which walks
at the stately pace of 0.060 m/s. After 1.5 minutes the finch tires of
the tortoise’s slow pace, and takes flight in the same direction for
another 1.5 minutes at 11 m/s.
What was the average speed of the finch for this 3.0-minute interval?
Answer:
[tex]Speed = 5.53 m/s[/tex]
Explanation:
Distance is calculated as:
[tex]Distance = Speed * Time[/tex]
First, we calculate the distance for the first 1.5 minutes
For the first 1.5 minutes, we have:
[tex]Speed = 0.060m/s[/tex]
[tex]Time = 1.5\ mins[/tex]
[tex]D_2= 0.060m/s * 1.5\ mins[/tex]
Convert 1.5 mins to seconds
[tex]D_2= 0.060m/s * 1.5 * 60s[/tex]
[tex]D_2= 5.4m[/tex]
Next, we calculate the distance for the next 1.5 minutes
[tex]Speed = 11m/s[/tex]
[tex]Time = 1.5\ mins[/tex]
[tex]D_2= 11m/s * 1.5\ mins[/tex]
Convert 1.5 mins to seconds
[tex]D_2 = 11m/s * 1.5 * 60s[/tex]
[tex]D_2= 990m[/tex]
Total distance is:
[tex]Distance = 990m + 5.4m[/tex]
[tex]Distance = 995.4m[/tex]
The average speed for the 3.0 minute interval is:
[tex]Speed = \frac{Distance}{Time}[/tex]
[tex]Speed = \frac{995.4\ m}{3.0\ mins}[/tex]
Convert 3.0 minutes to seconds
[tex]Speed = \frac{995.4\ m}{3.0 * 60 secs}[/tex]
[tex]Speed = \frac{995.4\ m}{180 secs}[/tex]
[tex]Speed = 5.53 m/s[/tex]
PLEASE PLEASE HELP!!!!
Two stones are thrown from the top edge of a building with a speed of 20 m/s, one straight down and the other straight up. The first one hits the street in a time t1. How much later is it before the second stone hits
Answer:
Second stone hist at 4.08 s later.
Explanation:
The second stone straight up, so we need to find the rise and descent time.
[tex]v_{f}=v_{i}-gt[/tex]
At the top, the final speed is 0 and t is the rise time.
[tex]0=20-gt_{r}[/tex]
[tex]t_{r}=\frac{20}{9.81}[/tex]
[tex]t_{r}=2.04 s[/tex]
Let's recall that the rise time is equal to the descent time.
So the total time of the stone will be
[tex]t_{t}=4.08 s[/tex]
Now, the second stone hits the street in t2 = 4.08 s + t1 s
But we just need the time after the first stone hist the street. So we just subtract the time t1 to the second time t2, which means t2 = 4.08 s.
Therefore stone hist at 4.08 s later.
The second stone hits the wall 4.08 seconds later.
To solve this question, we will use the equation of motion. In particular, the first equation of motion which states that;
v = u + at, where
v = the final velocity
u = the initial velocity
a = acceleration of gravity
t = time taken by the stone.
From the question, we're told that the stone is thrown with a speed of 20 m/s, which is the initial velocity. And a final velocity, v of 0 m/s. Using a constant of 9.81 for acceleration of the stone. We have,
0 = 20 - 9.81 × t
20 = 9.81t
t = [tex]\frac{20}{9.81}[/tex]
t = 2.039or 2.04 seconds.
Assuming that the time taken to ascent = time taken to descent, this means that the total time taken will be 2.039 + 3.039 = 4.078 seconds or 4,08 seconds.
visit https://brainly.com/question/24040430 to learn more about the first equation
A light ray in glass (n=1.5) hits the air-glass interface at an angle of 10 degrees from the normal. What angle from the normal is the light ray in the air (n=1.0)? (You can use the small angle approximation.)
Answer:
The angle from the normal is 15.1°.
Explanation:
We can find the angle by using Snell's law:
[tex] n_{1}sin(\theta_{1}) = n_{2}sin(\theta_{2}) [/tex]
Where:
n₁: is the first medium (glass) = 1.5
n₂: is the second medium (air) = 1.0
θ₁: is the first angle (in the glass) = 10°
θ₂: is the second angle (in the air) =?
[tex] \theta_{2} = arcsin(\frac{n_{1}sin(\theta_{1})}{n_{2}}) = arcsin(\frac{1.5*sin(10)}{1.0}) = 15.1 ^{\circ} [/tex]
Therefore, the angle from the normal is 15.1°.
I hope it helps you!
The resistance of a wire depends on its length i and on its cross sectional area A the resistance is
Answer:
The resistance of a wire is directly proportional to its length and inversely proportional to its cross-sectional area
Explanation:
Given an area of 8 ft^2. Express this area in the following terms. (a)m^2 5.9458 Incorrect: Your answer is incorrect. m^2 (b) in^2 WebAssign will check your answer for the correct number of significant figures. 96 Incorrect: Your answer is incorrect. in^2 (express to 4 significant figures)
Answer:
0.7432m²
1152in²
Explanation:
a) Usng the conversion
1ft² = 0.0929m²
8ft² = y
y = 8 × 0.0929
y = 0.7432m²
Hence 8ft² in m² is 0.7432m²
For ft² to in²
1ft² = 144in²
8ft² = x
X =8×144
x = 1152in²
Hence 8ft² expressed in in² is 1152in²
What are the three longest wavelengths for standing waves on a270-cm-long string that is fixed at both ends
Answer:
The answer is below
Explanation:
a) What are the three longest wavelengths for standing waves on a 270-cm-long string that is fixed at both ends? b. If the frequency of the second-largest wavelength is 50.0 Hz, what is the frequency of the third-longest wave length?
Solution:
a) The wavelengths (λ) for standing waves is given by the formula:
[tex]\lambda_m=\frac{2*length\ of\ string}{m}\\\\Where\ m=1,2,3,.\ .\ .\\\\Given\ that\ length\ of\ string = 270\ cm=2.7\ m,\ m=1,2,3(three\ longest\ wavelengths)\\\\Hence:\\\\\lambda_1=\frac{2(2.7)}{1}=5.4\ m\\\\\lambda_2=\frac{2(2.7)}{2}=2.7\ m \\\\\lambda_3=\frac{2(2.7)}{3}=1.8\ m[/tex]
b) The frequency (f) and wavelength (λ) is given by:
fλ = constant
Hence:
[tex]f_2\lambda_2=f_3\lambda_3\\\\f_2=50\ Hz\\\\2.7*50=f_3(1.8)\\\\f_3=\frac{2.7*50}{1.8} \\\\f_3=75\ Hz[/tex]
The three longest wavelengths for the standing waves on a 270-cm long string that is fixed at both ends are:
1. 5.4 meters.
2. 2.7 meters.
3. 1.8 meters.
Given the following data:
Length of string = 270 cm to m = [tex]\frac{270}{100} =2.7\;m[/tex]To determine the three (3) longest wavelengths for these standing waves:
Mathematically, the wavelength for standing waves is given by the formula:
[tex]\lambda_n = \frac{2L}{n}[/tex]
Where:
[tex]\lambda_n[/tex] is the wavelength for standing waves.L is the length of string.Note: n = 1, 2, and 3.
When n = 1:
[tex]\lambda_1 = \frac{2\times 2.7}{1} \\\\\lambda_1 = 5.4 \;meters[/tex]
When n = 2:
[tex]\lambda_2 = \frac{2\times 2.7}{2} \\\\\lambda_2 = 2.7 \;meters[/tex]
When n = 3:
[tex]\lambda_3 = \frac{2\times 2.7}{3} \\\\\lambda_3 =\frac{5.4}{3} \\\\\lambda_3 = 1.8 \;meters[/tex]
Read more: https://brainly.com/question/14708169
Two floors in a building are separated by 4.1 m. People move between the two floors on a set of stairs. (a) Determine the change in potential energy of a 3.0 kg backpack carried up the stairs. (b) Determine the change in potential energy of a person with weight 650 N that descends the stairs.
Answer:
a) The change in potential energy of a 3.0 kilograms backpack carried up the stairs.
b) The change in potential energy of a persona with weight 650 newtons that descends the stairs is -2665 joules.
Explanation:
Let consider the bottom of the first floor in a building as the zero reference ([tex]z = 0\,m[/tex]). The change in potential energy experimented by a particle ([tex]\Delta U_{g}[/tex]), measured in joules, is:
[tex]\Delta U_{g} = m\cdot g\cdot (z_{f}-z_{o})[/tex] (1)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]z_{o}[/tex], [tex]z_{f}[/tex] - Initial and final height with respect to zero reference, measured in meters.
Please notice that [tex]m\cdot g[/tex] is the weight of the particle, measured in newtons.
a) If we know that [tex]m = 3\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]z_{o} = 0\,m[/tex] and [tex]z_{f} = 4.1\,m[/tex], then the change in potential energy is:
[tex]\Delta U_{g} = (3\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.1\,m-0\,m)[/tex]
[tex]\Delta U_{g} = 120.626\,J[/tex]
The change in potential energy of a 3.0 kilograms backpack carried up the stairs.
b) If we know that [tex]m\cdot g = 650\,N[/tex], [tex]z_{o} = 4.1\,m[/tex] and [tex]z_{f} = 0\,m[/tex], then the change in potential energy is:
[tex]\Delta U_{g} = (650\,N)\cdot (0\,m-4.1\,m)[/tex]
[tex]\Delta U_{g} = -2665\,J[/tex]
The change in potential energy of a persona with weight 650 newtons that descends the stairs is -2665 joules.
Water runs out of a horizontal drainpipe at the rate of 135 kg/min. It falls 3.1 m to the ground. Assuming the water doesn't splash up, what average force does the water exert on the ground
Answer:
The average force exerted by the water on the ground is 17.53 N.
Explanation:
Given;
mass flow rate of the water, m' = 135 kg/min
height of fall of the water, h = 3.1 m
the time taken for the water to fall to the ground;
[tex]h = ut + \frac{1}{2}gt^2\\\\h = 0 + \frac{1}{2}gt^2\\\\t = \sqrt{\frac{2\times 3.1}{9.8} } \\\\t = 0.795 \ s[/tex]
mass of the water;
[tex]m = m't\\\\m = 135 \ \frac{kg}{min} \ \times \ 0.795 \ s \ \times \ \frac{1 \ \min}{60 \ s} \ = 1.789 \ kg[/tex]
the average force exerted by the water on the ground;
F = mg
F = 1.789 x 9.8
F = 17.53 N
Therefore, the average force exerted by the water on the ground is 17.53 N.
A cart's initial velocity is +3.0 meters per second. What is its final velocity after accelerating at a rate of 1.5 m/s2 for 8.0 seconds?
A. 36 m/s
B. 9 m/s
C. 72 m/s
D. 15 m/s
Answer:
V = 15m/s
Explanation:
Given the following data;
Initial velocity = 3m/s
Time = 8secs
Acceleration = 1.5m/s²
To find the final velocity, we would use the first equation of motion;
V = U + at
Substituting into the equation, we have
V = 3 + 1.5*8
V = 3 + 12
V = 15m/s
When you graph displacement vs. time, velocity is represented by
A. the x-intercept of the graph
B. the y-intercept of the graph
C. the slope of the graph
D. the curve of the graph
Help please..................
why is there a difference between potential and kinetic energy?
Potential energy is stored energy because it has the potential to do something which laters turns into kinetic energy which is the moving energy.
A small car of mass 833 kg is parked behind a small truck of mass 1767 kg on a level road. The brakes of both the car and the truck are off so that they are free to roll with negligible friction. A 41 kg woman sitting on the tailgate of the truck shoves the car away by exerting a constant force on the car with her feet. The car accelerates at 1.5 m/s 2 . What is the acceleration of the truck
Answer: the acceleration of the truck a_t is 0.6911 m/s²
Explanation:
Given the data in the question;
There is no external force on the system; net force on the system is 0
Mass of the truck with the woman M = 1767 kg + 41 kg = 1808 kg
mass of the car m = 833 kg
car acceleration a_c = 1.5 m/s²
now let a_t be the acceleration of the truck in opposite direction
Action force on the car = Reaction force on the car
ma_c = Ma_t
a_t = ma_c / M
we substitute
a_t = (833 × 1.5) / 1808
a_t = 1249.5 / 1808
a_t = 0.6911 m/s²
Therefore, the acceleration of the truck a_t is 0.6911 m/s²
A block weighing 9.3 N requires a force of 3.7 N to push it along at constant velocity. What is the coefficient of friction for the surface
Answer:
0.398
Explanation:
According to friction, the frictional force is directly proportional to the normal reaction
Ff = nR
Ff is the frictional force
n is the coefficient of friction
R is the reaction
Reaction is equal to the weight
R= W = 9.3N
Fm = Ff = 3.7N
Fm is the moving force
Get the coefficient of friction
n = Ff/R
n = 3.7/9.3
n = 0.398
Hence the coefficient of friction for the surface is 0.398
Which example best matches the term refraction? (17 Points)
O A. Light spreads out after it travels through a keyhole
O B. A straw in a glass of water looks bent
O C. Seeing your image in a lake
O D. An orchestra of different sounds coming together to make a larger sound
Answer:
B
Explanation:
Just answerd that question!!!
Hope it helped!!!
.
A 9.0 kg test rocket is fired vertically from Cape Canaveral. Its fuel gives it a kinetic energy of 1905 J by the time the rocket engine burns all of the fuel. What additional height will the rocket rise
Answer:
21.6m
Explanation:
Since the rocket engine burns all the fuel hence the kinetic energy will be converted to potential energy
Potential Energy = mass × acceleration due to gravity × height
Given
PE = 1905J
Mass = 9.0kg
Acceleration due to gravity =9.8m/s²
Required
Height h
Substitute into the formula
1905 = 9(9.8)h
1905 = 88.2h
h =1905/88.3
h = 21.6m
Hence the required height is 21.6m
A motorcycle moving 18.8 m/s has
57800 J of KE. What is its mass?
Answer:
m = 327.07 kg
Explanation:
Given that,
Kinetic energy of a motorcycle, E = 57800 J
Velocity of the motorcycle, v = 18.8 m/s
We need to find the mass of the motorcycle. The kinetic energy of an object is given by :
[tex]E=\dfrac{1}{2}mv^2[/tex]
m is mass
[tex]m=\dfrac{2E}{v^2}\\\\m=\dfrac{2\times 57800 }{(18.8)^2}\\\\m=327.07\ kg[/tex]
So, the mass of the motorcycle is 327.07 kg.
A person is standing on and facing the front of a stationary skateboard while holding a construction brick. The mass of the person is 67.0 kg, the mass of the skateboard is 4.10 kg, and the mass of the brick is 2.50 kg. If the person throws the brick forward (in the direction they are facing) with a speed of 23.0 m/s relative to the skateboard and we ignore friction, determine the recoil speed of the person and the skateboard, relative to the ground.
Answer:
v₁ = -0.8087 m / s
Explanation:
To solve this problem we can use the conservation of momentum, for this we define a system formed by the man, the skateboard and the brick, therefore the force during the separation is internal and the momentum is conserved
Initial instant. When they are united
p₀ = 0
Final moment. After throwing the brick
[tex]p_{f}[/tex] = (m_man + m_skate) v1 + m_brick v2
the moment is preserved
p₀ = p_{f}
0 = (m_man + m_skate) v₁ + m_brick v₂
v₁ = - [tex]\frac{ m_{brick} }{m_{man} + m_{skate} } v_{2}[/tex]
the negative sign indicates that the two speeds are in the opposite direction
let's calculate
v₁ = - [tex]\frac{2.5}{67 + 4.10} 23.0[/tex]
v₁ = -0.8087 m / s
help mmmemememeeee I'M BEING TIMED
Answer:
it's c. or A. not b.
Explanation:
.................
How much work would be needed to lift the ball from the 2-m shelf to the 5-m shelf, and how much potential energy would it have on the 5-m shelf?
Answer:
a) 2.94 N
b) 4.90 N
Explanation:
Let us assume that the weight of the ball is 0.98 N
Solution:
a) An object’s gravitational potential energy depends on two factors which are height and its weight (or mass). The equation for gravitational potential energy (PE) is given as:
Potential energy = weight (w) * height (h)
PE = wh
Potential energy at 2 m shelf = weight * height = 0.98 N * 2 m = 1.96 N
Potential energy at 5 m shelf = weight * height = 0.98 N * 5 m = 4.90 N
The work needed to lift the ball from the 2-m shelf to the 5-m shelf = Potential energy at 5 m shelf - Potential energy at 2 m shelf
The work needed to lift the ball from the 2-m shelf to the 5-m shelf = 4.90 N - 1.96 N = 2.94 N
b) Potential energy at 5 m shelf = weight * height = 0.98 N * 5 m = 4.90 N
what makes 97 of the water on earth unfit for drinking
Answer:
presence of salt in oceanic water makes 97 of the water on the earth unfit for drinking
All of the following are ways in which sports psychologists help athletes except __________.
A.
staying motivated
B.
managing fear of failure
C.
improving performance
D.
enhancing memory
Please select the best answer from the choices provided
A
B
C
D
Answer:
D-Enhancing memory
Explanation:
An object with velocity 141 ft/s has a kinetic energy of 1558.71 ft∙lbf, on a planet whose gravity is 31.5 ft/s2. What is its mass in pounds
Answer:
The mass of the object is 5.045 lbm.
Explanation:
Given;
kinetic energy of the object, K.E = 1558.71 ft.lbf
velocity of the object, V = 141 ft/s
The kinetic energy of the object is calculated as;
[tex]K.E = \frac{1}{2} mV^2\\\\mV^2 = 2K.E\\\\m = \frac{2K.E}{V^2} \\\\1 \ lbf = 32.174 \ lbm.ft/s^2\\\\m = \frac{2 \ \times \ 1558.71 \ ft.lbf \ \times \ 32.174 \ lbm.ft/s^2 }{(141 \ ft/s)2 \ \ \times \ \ \ \ 1 \ lbf\ }[/tex]
[tex]m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm.ft^2/s^2 }{(141 )^2\ ft^2/s^2 }\\\\m = \frac{(2 \ \times \ 1558.71 \ \times \ 32.174) \ lbm }{(141 )^2 }\\\\m = 5.045 \ lbm[/tex]
Therefore, the mass of the object is 5.045 lbm.
please help me this is timed .
Find x if a = 3.0 m/s^2
Answer:
x = 50 N
Explanation:
Given that we have a net force, a mass, and acceleration, we can use the fundamental formula for force found in newton's second law which is F = m × a.
Given a mass of 150 kg, and an acceleration 3.0m/s². We can substitute these two values in our formula to calculate the magnitude of these forces or it's net force to identify the unknown force acting on our known force for this situation to work.
_______
F (Net force) = F2 (Second force which we are given) - F1 (First force) = m × a
m (mass which we are given) = 150 kg
a (acceleration which we are given) = 3.0m/s
________
So F = m × a → F2 - F1 = m × a →
500 - F1 = 150 × 3.0 → 500 - F1 = 450 →
-F1 = -50 → F1 = 50