Answer:
4500W
Explanation:
[tex]P = {I}^{2} R[/tex]
where P = Power consumed , I = Current & R = Resistance.
In the question it's given that
I = 15A ; R = 20Ω
So ,
[tex]P = {15}^{2}\times 20 = 225 \times 20 = 4500W[/tex]
What is the water cycle ?
Answer:
The water cycle shows the continuous movement of water within the Earth and atmosphere. ... Liquid water evaporates into water vapor, condenses to form clouds, and precipitates back to earth in the form of rain and snow. Water in different phases moves through the atmosphere (transportation).
Explanation:
It's the water cycle.
A 1.13 kg skateboard is coasting along the pavement at a speed of 4.28 m/s when a 0.93
kg cat drops from a tree vertically downward onto the skateboard. What is the speed of the
skateboard-cat combination?
By conservation of momentum,
Pinitial = Pfinal
m1v1 + m2v2 = (m1 + m2)*vf
m1 = mass of skateboard = 1.13 kg
m2 = mass of cat = 0.93 kg
v1 = initial velocity of skateboard = 4.28 m/s
v2 = initial velocity of cat = 0 m/s
vf = final velocity of skateboard-cat combo
So plug in the values and solve for vf,
1.13(4.28) + 0.93(0) = (1.13 + 0.93)vf
vf = 2.35 m/s
Based on the law of conservation of energy, which statement is correct?
A.
Energy is always being added to all parts of the Universe.
B.
Energy is often destroyed in some parts of the Universe.
C.
Energy in a closed system cannot change forms.
D.
Energy in an isolated system remains constan
Answer:
D
Explanation:
Nothing can enter or leave so it remains constant
A box of weighing 60N is placed on the ground. The bottom of the box measures 2m by 1m. What is the pressure on the ground? *
Answer:
pressure=force/area
p=60N/(2m×1m)
=35N/M
What are the effects of hot rolling on the metal grain structure?
Answer:Mark me as the brainliset
Explanation:The rollers impart compression stresses, while squeezing the metal. These results in changes of grains, grain fragmentation, and lattice distortion. In hot rolling the coarse grains are converted into smaller grains. Care has been taken that the correct range of temperature is maintained during the process.
A circular wire loop lies inside a region of space containing a magnetic field. The direction of the magnetic field is out of the screen and parallel to the central axis of the loop. The magnitude of the magnetic field increases as a function of time. A circular loop oriented parallel to the plane of the screen lies inside a region containing magnetic field B. The field is directed out of the screen and is increasing. What is the direction of the induced current in the loop
Answer:
clockwise
Explanation:
According to the law given by Lenz, known as the Lenz law, it is said that a current induced in the circuit which is due to the change in the magnetic field and is so directed so as to oppose the change in the flux and to apply a force in the opposite direction if the force.
Here, as the magnetic field is directed out of the screen, the current flows in the direction which is clockwise in the loop and it opposes the increasing magnetic field.
The clockwise induced current will produce magnetic field in to the screen.
A bicyclist is riding a constant speed 62.0 seconds for 200. meters what is his average speed
Answer:
Explanation::
Some metals have a molecular structure that makes them good conductors. Explain how understanding this relationship can help engineers make more powerful batteries.
Answer:
Explained below.
Explanation:
Conductors can be defined as materials that permit electricity to flow through them easily.
Now, metals have a molecular structure that makes them good conductors because electrons in the atoms of these conductors tend to move freely from one atom to the other. So a majority of metals make good conductors because these metals tend to hold their electrons loosely. In short, it can help engineers make powerful batteries because then it means that they are capable of giving much more electrical energy since nowadays, advanced batteries make use of ion charges for the batteries.
A 2800 kg speedboat starting from rest attains a speed of 16 m/s in 8.0 s as a combination of 1200 N of air resistance and water drag act on the boat. How much power is being expended by speedboat at the maximum speed?
Answer:
For me
Explanation:
first of all
find the distance=speed×time
16×8=128
power= workdone /time
power=force×distance/time
power=1200×128/8
power=153600/8
powere=19200
ASAP PLS HELP
What is chemical potential energy?
Answer:
chemical potential of a species is energy that can be absorbed or released due to a change of the particle number of the given species, e.g. in a chemical reaction or phase transition.
Explanation:
explain and derive the equation for capillary action in the phenomenon of surface tension
Answer:
Explanation:Capillary action is the ability of a liquid to flow in narrow spaces without the assistance of, ... This article is about the physical phenomenon. ... If the diameter of the tube is sufficiently small, then the combination of surface tension (which is caused by cohesion ... They derived the Young–Laplace equation of capillary action.
The phrase "ten meters per second squared" describes the
Answer:
Acceleration
Explanation;
change in velocity over timeTherefore metre per second over second, which is meter per second squaredA ball weighing 60 N is swinging like a pendulum at the end of a 6.0 m rope. As the ball swings through its lowest point, its speed is measured at 5.0 m/s. What is the tension in the rope at this point?
A. 58.6 N
B. 85.4 N
C. 90.2 N
D. 97.3 N
Answer:
85.4 N
Explanation:
Weight of the ball, W = 60 N
W = mg, m is mass
m = W/g
m = 60/9.8 = 6.12 kg
Length of a rope, r = 6 m
Speed, v = 5 m/s
We need to find the tension in the rope at this point. Tension is equal to the centripetal force It is given by :
[tex]F=\dfrac{mv^2}{r}+mg\\\\F=\dfrac{6.12\times (5)^2}{6}+60\\\\=85.5\ N[/tex]
So, the correct option is (b) " 85.4 N".
what are the importance of informal education?
Answer:
Informal education is important because it can help individual to learn how to react and control situations.
It help individual to improve on its existing knowledge, new skills or ideas. This kind of education can happen any where and it can add values to the learner.
Explanation:
Informal education is a type of education that is learned from different life experiences, happenings outside a structured curriculum.
Informal education is important because it can help individual to learn how to react and control situations.
It help individual to improve on its existing knowledge, new skills or ideas. This kind of education can happen any where and it can add values to the learner.
Two locomotives approach each other on parallel track. Each has speed of 95 km/h with respect to the ground.If they are initially 8.5 km apart. How long will it take before they reach each other?
Answer:
It will take 2.68 minutes for them to reach each other.
Explanation:
We use the two following kinematic equations, making the final position the same (for the moment they meet each other):
locomotive 1 --> [tex]x_f=v*t[/tex]
locomotive 2 --> [tex]x_f=8.5\,-\,v*t[/tex]
we make the two xf equal, and solve for the time (t) using v = 95 km/h:
[tex]95*t=8.5-95*t\\2*95*t=8.5\\t=8.5/(190)\\t = 0.0447\,\,hours[/tex]
converting the hours into minutes by multiplying this value times 60;
t = 2.68 minutes
Parent: My son spends three hours on homework every night. He has no time to see his friends. He has no time to relax. We are watching him turn into a homework machine instead of a human being.
Answer:
this is a homework helping website I don't know If I can help you with this problwm
Which example provides a complete scientific description of an object in motion?
(A) The hiker followed the north trail a distance of two kilometers in thirty minutes
(B) The tiger had to run at 50 kilometers per hour across the field to catch the zebra
(C) The hockey player sent the puck flying toward the north goal to score the winning point
(D) The vibration of the jackhammer broke through the south-facing rock in under 10 seconds
Answer:
C
Explanation:
it explains how an object that was in motion continued motion after the force
As an astronaunt travels from the surface of the earth to a postion that is four times
as far away from the center of the earth, the astronaut's
a) mass decreases
b)
mass remains the same
c) weight increases
d) weight remains the same
Answer: mass remains the same
Explanation:
My teacher gave me the answer
As an astronaut travels from the surface of the earth to a position that is four times as far away from the center of the earth, the astronaut's:
B. Mass remains the sameAccording to the given question, we can see that an astronaut is on a journey to a new location which is four times further from the center of the earth. We need to find out if there is a change to the mass of the astronaut.
As a result of this, we can see that the mass of the astronaut remains the same as he travels to a position which is four times further from the center of the earth, because the gravitational forces are the same.
Therefore, the correct answer is option B
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In the Fresnel circular aperture setup, the distances from the aperture to the light source and the reception screen are 1.5 m and 0.6 m, respectively. The wavelength is 630 nm. Suppose that the radius of the aperture can be increased from 0.5 mm, determine: (a) The first two radii when the center intensity at the reception screen is maximum. (b) The first two radii when the center intensity is minimum.
Explanation:
The width of the central maximum is given by
W = 2 λ L / a
where W is the width of the central maximum
λ is the wavelength of the light used.
L is the distance between the aperture and screen
a is the width of the slit or aperture
So we can see that if any one quantity is varied by keeping others constant in the above formula , there would be a change in width of central maximum.
A mass m is located at the origin; a second mass m is at x = d. A third mass m is above the first two so the three masses form an equilateral triangle. What is the net gravitational force on the third mass?
Answer:
√3 * Gm²/d²
Explanation:
m1 = m, m2= m, distance = d. hence:
F = Gm²/d²
Let the origin be A, the point x = d be B and the point above the first two is C.
The net force acting on the third mass (point C) [tex]F_{net}[/tex] = [tex]F_A+F_B[/tex]
Let j represent the vertical component and i the horizontal component. Hence:
[tex]F_B=-F_j\\\\F_A=-F(icos\frac{\pi}{6}+jsin\frac{\pi}{6} )=-F(i\frac{\sqrt{3} }{2}+j\frac{1}{2} )\\\\F_{net} =F_A+F_B\\\\F_{net} =-F_j+{-F(i\frac{\sqrt{3} }{2}+j\frac{1}{2} )}\\\\F_{net} =-\frac{F}{2} \sqrt{3}(i+j\sqrt{3} )\\\\The\ magnitude\ of\ the\ net\ force\ is:\\\\|F_{net}|=\frac{F}{2}\sqrt{3}(\sqrt{1^2+\sqrt{3}^2 })=\frac{F}{2} \sqrt{3}(\sqrt{4})\\\\|F_{net}|=\frac{F}{2} \sqrt{3}*2=F*\sqrt{3}\\\\|F_{net}|=\sqrt{3}*\frac{Gm^2}{d^2}[/tex]
Answer:
The net gravitational force on the third mass = [tex]3^{0.5} * \frac{Gm^2}{d^2}[/tex]Explanation:
For equilateral triangle,
[tex]\theta = 30^o[/tex]
Force between masses,
[tex]F_1 = \frac{G*m*m}{d^2}\\\\F_! = \frac{Gm^2}{d^2}[/tex]
Therefore,
[tex]F_net = 2F_1cos\theta\\\\F_net = 2 * \frac{Gm^2}{d^2} * cos30\\\\F_net = 3^{0.5} * \frac{Gm^2}{d^2}[/tex]
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It is now 9:11 a.m. but when the bell rings at 9:12 a.m. Susie will be late for Mrs. Garner's U.S. History class for the 3rd time this week. She must get from one side of the school to the other by hurrying down three different hallways. She runs down the first hallway (D-Hall), a distance of 47.0 meters. The second hallway (C-Hall) is filled with students, and she covers its 63.0 m length quickly. The final hallway (B- Hall) is empty, and Susie sprints its 76.0 m length. How fast does Susie need to go to make it to class on time (Hint: Calculate the total distance. Then calculate her total average speed rounded to the nearest tenths in meters/seconds.)?
Answer:
3.1 m/s
Explanation:
The total distance she has to run is the addition of the three lengths:
47 + 63 + 76 = 186 meters.
She needs to cover it one minute (60 seconds). Therefore her speed must be:
186 m / 60 s = 3.1 m/s
A race car starts from rest and accelerates down a track at a rate of 3.0 m/s2
How fast is the car moving after 10 seconds?
30 m/s
Explanation:We are given:
Initial Velocity (u) = 0 m/s
Acceleration of the Car (a) = 3 m/s²
Time Interval (t) = 10 seconds
Speed of the Car After 10 seconds:
From the First equation of motion:
v = u + at
replacing the given values
v = 0 + (3)(10)
v = 30 m/s
Hence, the car is moving at a velocity of 30 m/s after 10 seconds
ANSWER ALL QUESTIONS and show your work ON THE ATTACHMENT AWARD 50 pts if you don’t know the answer to all of them I have posted them individually so I can still mark you brainlessly (however you spell the darn thing)
Answer:
the velocity is 210 - hq squared which equals to 14
Which best describes the law of conservation of mass?
O The coefficients in front of the chemicals in the reactants should be based on the physical state of the products,
O Products in the form of gases are not considered a part of the total mass change from reactants to products
O When reactants contain both a solid and a liquid, the solid counts toward the overall mass and the liquid does not
O The mass of the reactants and products is equal and is not dependent on the physical state of the substances.
Explanation:
The law of conservation of mass states that mass in an isolated system is neither created nor destroyed by chemical reactions or physical transformations. According to the law of conservation of mass, the mass of the products in a chemical reaction must equal the mass of the reactants.
A 200 g air-track glider is attached to a spring. The glider is pushed 10.0 cm against the spring, then released. A student with a stopwatch finds that 10 oscillating take 12.0 s. What is the spring constant?
Answer:
5.5N/m
Explanation:
Calculation for What is the spring constant
First step is to calculate the time period
T = 12 second/10
T = 1.2 second
Now let calculate the spring constant using this formula
k=4π²m/T²
Where,
m=0.2kg
T=1.2second
k represent spring constant=?
Let plug in the formula
k=4π²×0.2kg/(1.2)²
k=39.48×0.2kg/1.44
k=7.90/1.44
k=5.48N/m
k=5.5N/m ( Approximately)
Therefore the spring constant will be 5.5N/m
A hammer strikes a nail with a 10N force for .01 seconds. Calculate the impulse of the hammer
Answer:
Impulse = 0.1 Kgm/s
Explanation:
Given the following data;
Force, F = 10N
Time, t = 0.01 seconds
To find impulse
An impulse can be defined as the net force acting an object for a very short period of time.
Mathematically, impulse is given by the formula;
Impulse = force * time
Substituting into the equation, we have;
Impulse = 10 * 0.01
Impulse = 0.1 kgm/s
Therefore, the impulse of the hammer is 0.1 kilogram meter per seconds.
The impulse of the hammer is 0.1 Ns.
To solve the given problem we need to use the formula for calculating impulse.
Impulse: This can be defined as the product of force and time on a body.
The formula of impulse is given below.
I = Ft....................... Equation 1
Where :
I = Impulse of the harmerF = Force on the nailt = timeFrom the question,
Given:
F = 10 Nt = 0.01 sSubstitute these values into equation 1
I = 10(0.01)I = 0.1 NsHence, The impulse of the hammer is 0.1 Ns.
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A pail in a water well is hoisted by means of a frictionless winch, which consists of a spool and a hand crank. When Jill turns the winch at her fastest water-fetching rate, she can lift the pail the 28.0 m to the top in 11.0 s. Calculate the average power supplied by Jill's muscles during the upward ascent. Assume the pail of water when full has a mass of 7.30 kg.
Answer:
182.28 W
Explanation:
Here ,
m = 7.30 Kg
distance , d= 28.0 m
time , t = 11.0 s
average power supplied = change in potential energy/time
average power supplied = m×g×d/time
average power supplied = 7.30×9.81×28/11
average power supplied = 182.28 W
the average power supplied is 182.28 W
A chimpanzee climbs a vine up and to the right for a total displacement of 50 m 50m50, start text, m, end text that makes a 60 ° 60°60, degree angle from the ground. What was the horizontal displacement of the chimpanzee in m mstart text, m, end text?
Answer:
25m
Explanation
Let the horizontal displacement formula be expressed as;
Dx = Dcos theta
theta is the angles subtended by the displacement
Given
D = 50m
theta = 60°
Required
horizontal displacement of the chimpanzee
Substitute into the formula
Recall that: Dx = Dcos theta
Dx = 50cos60°
Dx = 50(0.5)
Dx = 25m
Hence the horizontal displacement of the chimpanzee is 25m
You hit a hockey puck and it slides across the ice at nearly a constant speed.Is a force keeping it in motion?Explain.
Answer:
Explanation:
When the puck is sliding on the ice, there is no force being exerted on the puck to keep it moving forward. Instead, inertia keeps the puck moving forward. Friction between the puck and the ice gradually slows the puck down. You hit a hockey puck and it slides across the ice at nearly a constant speed
At constant speed and varying position of the hockey puck, implies a change in the velocity of the hockey puck and net force is acting on it to keep it in motion.
According to Newton's second law of motion, the force applied to a an object is directly proportional to the product of mass and acceleration of the object.
F = ma
Acceleration is the change in the velocity of an object per change in time of motion.
At constant velocity, the acceleration of an object is zero.When acceleration of an object is zero, the force on the object is zero.A constant speed (magnitude only) and change in the direction of the object, implies a change in velocity of the object.at changing velocity, the acceleration on an object is positive, and hence net force acts on the object.Thus, we can conclude that at constant speed and varying position of the hockey puck, implies a change in the velocity of the hockey puck and net force is acting on it to keep it in motion.
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Consider a star that is a sphere with a radius of 6.32 108 m and an average surface temperature of 5350 K. Determine the amount by which the star's thermal radiation increases the entropy of the entire universe each second. Assume that the star is a perfect blackbody, and that the average temperature of the rest of the universe is 2.73 K. Do not consider the thermal radiation absorbed by the star from the rest of the universe. J/K
Answer:
The value is [tex]\Delta s = 8.537 *10^{25 } \ J/K[/tex]
Explanation:
From the we are told that
The radius of the sphere is [tex]r = 6.32 *10^{8} \ m[/tex]
The temperature is [tex]T_x = 5350 \ K[/tex]
The average temperature of the rest of the universe is [tex]T_r = 2.73 \ K[/tex]
Generally the change in entropy of the entire universe per second is mathematically represented as
[tex]\Delta s = s_r - s_x[/tex]
Here [tex]s_r[/tex] is the entropy of the rest of the universe which is mathematically represented as
[tex]s_r = \frac{Q}{T_r}[/tex]
Here Q is the quantity of heat radiated by the star which is mathematically represented as
[tex]Q = 4 \pi * r^2 * \sigma * T^4_x[/tex]
Here [tex]\sigma[/tex] is the Stefan-Boltzmann constant with value
[tex]\sigma = 5.67 * 10^{-8 }W\cdot m^{-2} \cdot K^{-4}.[/tex]
=> [tex]Q = 4 \pi * (6.32*10^{8})^2 * 5.67 * 10^{-8 } * 5350 ^4[/tex]
=> [tex]Q = 2.332 *10^{26} \ J[/tex]
So
[tex]s_r = \frac{2.332 *10^{26}}{2.73}[/tex]
=> [tex]s_r = 8.5415 *10^{25}\ J/K[/tex]
Here [tex]s_x[/tex] is the entropy of the rest of the universe which is mathematically represented as
[tex]s_x = \frac{Q}{T_x}[/tex]
=> [tex]s_x = \frac{2.332 *10^{26} }{5350}[/tex]
=> [tex]s_x = 4.359 *10^{22} \ J/K[/tex]
So
[tex]\Delta s = 8.5415 *10^{25} - 4.359 *10^{22}[/tex]
=> [tex]\Delta s = 8.537 *10^{25 } \ J/K[/tex]
This question involves the concepts of entropy and the thermal radiation
The entropy of the entire universe is increased by "8.41 x 10²⁵ J/k
".
The increase in entropy is given as follows:
[tex]\Delta s = s-s_T[/tex]
where,
Δs = increase in entropy = ?
σ = Stefan-Boltzman's constant = 5.67 x 10⁻⁸ W/m².k⁴
A = surface area = 4πr² = 4π(6.32 x 10⁸ m)² = 5.01 x 10¹⁸ m²
Tr = Absolute temperature of the star = 5350 K
T = absolute temperature of the rest of the universe = 2.73 k
Q = thermal radiation energy
Q = [tex]\sigma A T_r^4=(5.67\ x\ ^{-8}\ W/m^2.k^4)(5.01\ x\ ^{18}\ m^2)(5350\ k)^4=2.3\ x\ 10^{26}\ J[/tex]
s = entropy of the universe = [tex]\frac{Q}{T}=\frac{2.3\ x\ 10^{26}\ J}{2.73 k}=8.42\ x\ 10^{25}\ J/k[/tex]
[tex]s_T[/tex] = entropy of the star = [tex]\frac{Q}{T_r}=\frac{2.3\ x\ 10^{26}\ J}{5350\ k}=4.3\ x\ 10^{22}\ J/k[/tex]
Therefore,
Δs = 8.42 x 10²⁵ J/k - 4.3 x 10²² J/k
Δs = 8.41 x 10²⁵ J/k
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