Answer:
199.2J
Explanation:
To calculate the the amount of energy stored in a capacitor, we use the following formula;
W = ½ QV
Where;
W = energy stored (joules)
Q = charge (Coulombs)
V = voltage (volts)
According to the information provided in this question, the capacitor of a camera flash unit has a charge (Q) of 0.83 C and a voltage (V) of 480 V.
Using W = ½ QV
W = ½ × 0.83 × 480
W = ½ × 398.4
W = 398.4/2
W = 199.2
The amount of energy stored in the capacitor to convert electrical energy into light energy is 199.2J.
If the current in each wire is the same, which wire produces the strongest magnetic field?
-a wire that is 1 mm thick and not coiled
-a wire that is 2 mm thick and not coiled
-a 1-mm-thick coiled wire with ten loops
-a 2-mm-thick coiled wire with two loops
Answer:
C
Explanation:
I did this before
Answer:
it is c
Explanation:
cuz i know
a crane lifts a 75 kg mass at a height of 8m. calculate the gravitational potential energy gained by the mass (g= 10N)
Answer:
6000 J
Explanation:
Potential energy = mass*gravity*height (U = m*g*h)
In this case g = 10 so
U = 75*10*8 = 6000 J (joules) [kg * m^2/s^2]
nWhat are four key features of our solar system that any theory of solar system formation must be able to explain
Answer: There are vast numbers of asteroids and comets.
Explanation:
There are clear patterns to the rotation and orbits of large bodies in the solar system.
There are a few notable "exceptions to the rules."
Planets fall into two major categories.
why is the compass needle oriented as it is in the diagram
Answer:
its north pole is attracted to the south pole of the magnet
Explanation:
just did the quiz :)
In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between station stops. The more stops there are, the slower the train’s average speed. To get an idea of this problem, calculate the time it takes a train to make a 15.0-km trip in two situations: (a) the stations at which the trains must stop are 3.0 km apart (a total of 6 stations, including those at the ends); and (b) the stations are 5.0 km apart (4 stations total). Assume that at each station the train accelerates at a rate 1.1 m/s2 of until it reaches 95 km/h, then stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at-2.0 m/s2 Assume it stops at each intermediate station for 22 s.
Answer:
a) t = 746 s
b) t = 666 s
Explanation:
a)
Total time will be the sum of the partial times between stations plus the time stopped at the stations.Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:[tex]v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s (1)[/tex]
Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:[tex]t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)[/tex]
Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:[tex]x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m (3)[/tex]
In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:[tex]t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)[/tex]
We can find the distance traveled while the train was decelerating as follows:[tex]x_{3} = (v_{1} * t_{3}) + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m (5)[/tex]
Finally, we need to know the time traveled at constant speed.So, we need to find first the distance traveled at the constant speed of 26.4m/s.This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:[tex]t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s (7)[/tex]
Total time between two stations is simply the sum of the three times we have just found:t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:tm = t*5 = 131.6 * 5 = 658.2 s (9)Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)b)
Using all the same premises that for a) we know that the only difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.We can find the distance traveled at constant speed, rewriting (6) as follows:x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:[tex]t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s (12)[/tex]
Total time between two stations is simply the sum of the three times we have just found:t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:tm = t*3 = 207.4 * 3 = 622.2 s (14)Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)The skater lowers her arms as shown in the adjacent
figure decreasing her radius to 0.15 m. Find her new speed.
Answer:
is there more?
Explanation:
How does the size of a wind turbine affect its energy output?
A.)Smaller turbines spin slower.
B.) Larger turbines have a greater storage capacity.
C.) Larger turbines generate more electricity.
D.)Smaller turbines are better for capturing strong winds.
Answer:
Larger tubines generate more electricity.
Explanation:
Larger blades allow the turbine to capture more of the kinetic energy of the wind by moving more air through the rotors. However, larger blades require more space and higher wind speeds to operate. This distance is necessary to avoid interference between turbines, which decreases the power output.
Is it safer for a car to decelerate quickly or slowly? And why?
Answer:
It depends on the situations. Stopping at a red light you should start decelerating slowly because if you just stop if a car is behind you they may be unprepared for your sudden stop and slam into you. Decelerating quickly when needed to like if you are about to run into a car or someone pulls out in front of you stopping quickly might be none or less damage.
Explanation:
Answer:
Hello!!! erz here ^^
Explanation:
Have you ever seen someone driving? They usually decelerate slowly for a stop signs/traffic lights... But thats the wrong way to do it, Instead you speed right past that light!
(no but seriously the guy/girl above me is correct xD)
3. Batteries create electricity and generators create electricity. *
True
False
Calculate the acceleration due to gravity at Earth due to the Moon. (b) Calculate the acceleration due to gravity at Earth due to the Sun.
Answer:
a) g’= 3.44 10⁻⁵ m / s²
b) g ‘’ = 5.934 10⁻³ m / s²
Explanation:
For this exercise let's use the law of universal gravitation
F = [tex]G \frac{m M}{r^2}[/tex]
where m is the mass of the body under study, M the mass of the body that creates the force and r the distance between the bodies
F = [tex]m \ ( G \frac{M}{r^2} )[/tex]
the attractive force is called weight W = m g,
Thus
g = [tex]G \frac{M}{r^2}[/tex]
is called the acceleration of gravity
a) the acceleration created by the moon
g' = G \frac{M}{r^2}
the mass of the moon is M = 7.36 10²² kg
the distance from the moon to the Earth's surface is
r = D -R_e
r = 3.84 10⁸ -6.37 10⁶
r = 3.7763 10⁸ m
we calculate
g’= [tex]6.67 \ 10^{-11} \ \frac{7.36 \ 10^{22}}{ (3.7763 \ 10^8)^2}[/tex]
g ’= 3.44 10⁻⁵ m / s²
b) the acceleration created by the sun
mass of the sun M = 1,9991 10³⁰ ka
the distance from the sun wears down the Earth's surface
r = D -R_e
r = 1.496 10¹¹ -6.37 10⁶
r = 1.4959 10¹¹ m
let's calculate
g ’’ = [tex]6.67 \ 10^{-11} \frac{1.991 \ 10^{30}}{ (1.4959 \ 10^{11})^2 }[/tex]
g ‘’ = 5.934 10⁻³ m / s²
What is the net force on the purple ring in the picture above. _________
Answer:
equal
Explanation:
Because of Newton's third law
fasttt plsss!!!!
Energy A car engine drives a generator, which produces and stores electric charge in the car's battery. The headlamps use the electric charge stored in the car battery. List the forms of energy in these three operations
which form of energy is an example of kinetic energy
Answer:
1. realizing of arrow
2. kicking of ball
3. punching the punching bag
uniform acceleration means the velocity of the body is either increasing or decreasing changing at a constant rate TRUE OR FALSE? thank youu
Answer:
False.
Explanation:
In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.
This simply means that, acceleration is given by the subtraction of final speed from the initial speed all over time.
Hence, if we subtract the final speed from the initial speed and divide that by the time, we can calculate an object’s acceleration.
Mathematically, acceleration is given by the equation;
[tex]Acceleration (a) = \frac{initial speed - final speed}{time}[/tex]
[tex]a = \frac{v - u}{t}[/tex]
Where,
a is acceleration measured in [tex]ms^{-2}[/tex]
v and u is initial and final speed respectively, measured in [tex]ms^{-1}[/tex]
t is time measured in seconds.
Additionally, acceleration is a vector quantity because it has both magnitude and direction.
Hence, uniform acceleration means the velocity of the body is increasing at a constant rate. For example, an object or body that is experiencing a free fall in a uniform gravitational field is said to be in uniform acceleration.
The water side of the wall of a 60-m-long dam is a quarter-circle with a radius of 7 m. Determine the hydrostatic force on the dam and its line of action when the dam is filled to the rim. Take the density of water to be 1000 kg/m3.
Answer:
[tex]26852726.19\ \text{N}[/tex]
[tex]57.52^{\circ}[/tex]
Explanation:
r = Radius of circle = 7 m
w = Width of dam = 60 m
h = Height of the dam will be half the radius = [tex]\dfrac{r}{2}[/tex]
A = Area = [tex]rw[/tex]
V = Volume = [tex]w\dfrac{\pi r^2}{4}[/tex]
Horizontal force is given by
[tex]F_x=\rho ghA\\\Rightarrow F_x=1000\times 9.81\times \dfrac{7}{2}\times 7\times 60\\\Rightarrow F_x=14420700\ \text{N}[/tex]
Vertical force is given by
[tex]F_y=\rho gV\\\Rightarrow F_y=1000\times 9.81\times 60\times \dfrac{\pi 7^2}{4}\\\Rightarrow F_y=22651982.59\ \text{N}[/tex]
Resultant force is
[tex]F=\sqrt{F_x^2+F_y^2}\\\Rightarrow F=\sqrt{14420700^2+22651982.59^2}\\\Rightarrow F=26852726.19\ \text{N}[/tex]
The hydrostatic force on the dam is [tex]26852726.19\ \text{N}[/tex].
The direction is given by
[tex]\theta=\tan^{-1}\dfrac{F_y}{F_x}\\\Rightarrow \theta=\tan^{-1}\dfrac{22651982.59}{14420700}\\\Rightarrow \theta=57.52^{\circ}[/tex]
The line of action is [tex]57.52^{\circ}[/tex].
llustration 2: Aman can run a distance of 100 m in 20 seconds. Find the speed of Aman in m/s.
Answer:
[tex]\boxed {\boxed {\sf 5 \ meters/second}}[/tex]
Explanation:
Speed is equal to distance over time.
[tex]s=\frac{d}{t}[/tex]
The distance is 100 meters and the time is 20 seconds.
[tex]d= 100 \ m \\t= 20 \ s[/tex]
Substitute the values into the formula.
[tex]s=\frac{100 \ m }{20 \ s}[/tex]
Divide.
[tex]s= 5 \ m/s[/tex]
Aman's speed is 5 meters per second.
Two coils have the same number of circular turns and carry the same current. Each rotates in a magnetic field acting perpendicularly to its axis of rotation. Coil 1 has a radius of 4.5 cm and rotates in a 0.21-T field. Coil 2 rotates in a 0.39-T field. Each coil experiences the same maximum torque. What is the radius (in cm) of coil 2
Answer:
Explanation:
Torque acting on a coil in a magnetic field = MBsinθ where M is magnetic moment , B is magnetic field and θ is inclination of the normal to coil with direction of field.
For maximum torque sinθ = 1
Maximum torque = MB
M = NIA where N is no of turns , I is current and A is area of the coil
Maximum torque = NIAB
As maximum torque is same
N₁I₁A₁B₁ = N₂I₂A₂B₂
N₁ = N₂ , I₁ = I₂
A₁B₁ = A₂B₂
π R₁² B₁ = π R₂² B₂
4.5² x .21 = R₂² x .39
R₂² = 10.9
R₂ = 3.3 cm .
She left the cubes in the water for three hours which of the following describes a heat flow that took place during those three hours?
Answer:
veffvevfevve
Explanation:
Explanation:
wht are the choices
attach the choices and illbe happy to help you
in fact i most definitely will
advantages of alcohol as a thermometric substance
Answer:
it helps to measure temperature of body
it contracts on cooling and expands on heating which is suitable to use as
thermometric substance
A spherical shell with a net charge of 3Q surrounds a point charge of -q at the center of the shell. The charges on the inner and outer surfaces of the shell are:
Answer:
1) The charge on the outer shell is +4·Q
2) The charge on the inner shell is +Q
Explanation:
1) The given parameters of the spherical shell are;
The net charge on the spherical shell = 3·Q
The point charge surrounded by the spherical shell = -Q
Let 'x' represent the charge on the outer shell, and let 'y', represent the charge on the inner shell, we have;
The net charge, 3·Q = -q + x
∴ x = 3·Q + Q = 4·Q
The charge on the outer shell, x = 4·Q
2) The net charge in the shell is zero, therefore, the charge on the inner shell, 'y', is given as follows;
-Q + y = 0
∴ y = +Q
The charge on the inner shell, y = +Q
Do we know which has more potential energy? object A or B? Best answer with reasoning gets brainliest.
Answer:
Object C has the most potential energy.
Between A and B, we do not know which has more potential energy.
Explanation:
We know the object with the most potential energy and this is the object at C.
Potential energy is the energy due to the position of a body above the ground surface.
The higher a body is above ground, the more its potential energy.
Potential energy = mass x acceleration due to gravity x height
So;
Object C has the most potential energy.
Between A and B, we do not know which has more potential energy.
This is because, the height and mass of the objects are not quantified using numbers.
Potential energy is a function of mass and height and acceleration due to gravity but acceleration due gravity is a constant.
In the form of radioactive decay known as alpha decay, an unstable nucleus emits a helium-atom nucleus, which is called an alpha particle. An alpha particle contains two protons and two neutrons, thus having mass m
Answer:
[tex]\mathbf{v_a = 4.06 \times 10^7 \ m/s}[/tex]
Explanation:
[tex]\text{The missing part of the question is attached below.} \\ \\ \text{ mass m = 4u and charge q = 2e. Suppose a uranium nucleus with 92 protons decays into }[/tex][tex]\text{into thorium, with 90 protons, and an alpha particle. The alpha particle is initally at rest at }[/tex][tex]\text{ the surface of the thorium nucleus, which is 15 fm in diameter. What is the speed of the alpha}[/tex][tex]\text{particle when it is detected in the laboratory? Assume the thorium nucleus remains at rest. }[/tex]
[tex]\text{So, from the above infromation;}[/tex]
[tex]radius (r) = \dfrac{15 \ fm }{2}[/tex]
[tex]radius (r) = 7.5 \times 10^{-15} \ m[/tex]
[tex]\text{Using the formula for potential energy}[/tex]
[tex]U = \dfrac{(9\times 10^9 \ Nm^2/C^2)(2(1.6\times10^{-19} \ C ))(90)(1.6\times 10^{-19} C)}{7.5 \times 10^{-13} \ m}[/tex]
[tex]U = 5.529 \times 10^{-12} \ J[/tex]
[tex]\text{Now, the speed of the alpha particle can be estimated from the conservation of energy principle}[/tex][tex]\dfrac{1}{2}mv_a^2= 5.529 \times 10^{-12} J[/tex]
[tex]v_a = \sqrt{\dfrac{2(5.529 \times 10^{-12} \ J)}{4(1.67 \times 106{-37} \ kg)}}[/tex]
[tex]\mathbf{v_a = 4.06 \times 10^7 \ m/s}[/tex]
Calculate the force between two objects that have a masses of
1000 kg and 4000 kg that are separated by a distance of 25 m.
1.1 x 10-5 N
4.3 x 1015 N
О
4.3 x 10-7N
What is the momentum of a 6.5 kg bowling ball with a velocity of 3.0 m/s?
Answer:
19.5
Explanation:
momentum = mass * velocity
momentum = 6.5 * 3
momentum = 19.5
A 25kg object accelerates at 5 m/s². Find the force.
Answer:
Force = 125 kg-m/s²
Explanation:
Given:
Mass = 25 kg
Acceleration = 5 m/s²
Find:
Force
Computation:
Force = Mass x Acceleration
Force = 25 x 5
Force = 125 kg-m/s²
can you please tell me what this is
Answer:
200000 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) of roller coaster = 1000 Kg
Velocity (v) of roller coaster = 20 m/s
Kinetic energy (KE) =?
Kinetic energy is simply defined as the energy possess by an object in motion. Mathematically, it can be expressed as:
KE = ½mv²
Where
KE => is the kinetic energy.
m =>is the mass of the object
V => it the velocity of the object.
With the above formula, we can obtain the kinetic energy of the roller coaster as follow:
Mass (m) of roller coaster = 1000 Kg
Velocity (v) of roller coaster = 20 m/s
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 1000 × 20²
KE = 500 × 400
KE = 200000 J
Therefore, the kinetic energy of the roller coaster is 200000 J.
Learning Task 5: Read each situation and encircle the letter of the best
answer. Do this activity on your notebook,
1. Our team is the champion because of our captain. No matter how fast his
Opponent shoot the ball he can do the same, he never misses it and hits it
right back. What emotion is shown in the situation?
A. Happiness
B. Anger
C. Sadness
D. Afraid
Answer:
I think the tone is happiness using the clues of "champion". But depending on who is saying it and background information it could be something different but based on the text here and the answer choices personally the best answer in my opinion is happiness.
If there was jealousy or optimism it would be a harder decision.
Explanation:
Hope this helped :)
1. Find the magnitude of the gravitational force a 66.5 kg person would experience
while standing on the surface of each of the following planets:
Gravit Force
Planet
Earth
Mars
Pluto
Mass
5.97 x 10^24 kg
6.42 x 10^23 kg
1.25 x 10^22 kg
Radius
6.38 x 10^6 m
3.40 x 10^6m
1.20 x 10^6m
Answer:
Gravitational forces: Earth: [tex]W = 650.969\,N[/tex], Mars: [tex]W = 246.449\,N[/tex], Pluto: [tex]W = 38.504\,N[/tex]
Explanation:
The weight ([tex]W[/tex]), measured in newtons, experimented by the person on the surface of the planet is defined by Newton's Laws of Motion:
[tex]W = m\cdot g[/tex] (1)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
From Newton's Law of Gravitation we derive this expression for gravitational acceleration:
[tex]g = \frac{G\cdot M}{R^{2}}[/tex] (2)
Where:
[tex]G[/tex] - Gravitational constant, measured in cubic meters per kilogram-square second.
[tex]M[/tex] - Mass of the planet, measured in kilograms.
[tex]R[/tex] - Radius of the planet, measured in meters.
By (2) we calculate the gravitational acceleration for each planet:
Earth ([tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]M = 5.97\times 10^{24}\,kg[/tex], [tex]R = 6.38\times 10^{6}\,m[/tex])
[tex]g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.97\times 10^{24}\,kg)}{(6.38\times 10^{6}\,m)^{2}}[/tex]
[tex]g = 9.789\,\frac{m}{s^{2}}[/tex]
Mars ([tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]M = 6.42\times 10^{23}\,kg[/tex], [tex]R = 3.40\times 10^{6}\,m[/tex])
[tex]g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (6.42\times 10^{23}\,kg)}{(3.40\times 10^{6}\,m)^{2}}[/tex]
[tex]g = 3.706\,\frac{m}{s^{2}}[/tex]
Pluto ([tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]M = 1.25\times 10^{22}\,kg[/tex], [tex]R = 1.20\times 10^{6}\,m[/tex])
[tex]g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (1.25\times 10^{22}\,kg)}{(1.20\times 10^{6}\,m)^{2}}[/tex]
[tex]g = 0.579\,\frac{m}{s^{2}}[/tex]
Lastly, we calculate the gravitational force for each case by (1):
Earth ([tex]m = 66.5\,kg[/tex], [tex]g = 9.789\,\frac{m}{s^{2}}[/tex])
[tex]W =(66.5\,kg)\cdot \left(9.789\,\frac{m}{s^{2}} \right)[/tex]
[tex]W = 650.969\,N[/tex]
Mars ([tex]m = 66.5\,kg[/tex], [tex]g = 3.706\,\frac{m}{s^{2}}[/tex])
[tex]W = (66.5\,kg)\cdot \left(3.706\,\frac{m}{s^{2}} \right)[/tex]
[tex]W = 246.449\,N[/tex]
Pluto ([tex]m = 66.5\,kg[/tex], [tex]g = 0.579\,\frac{m}{s^{2}}[/tex])
[tex]W = (66.5\,kg)\cdot \left(0.579\,\frac{m}{s^{2}} \right)[/tex]
[tex]W = 38.504\,N[/tex]
NEED HELP When a guitar string is plucked, what part of the standing wave is found at the fixed ends of the string?(1 point)
the nodes
the resonance
the antinodes
the harmonics
The average marathon runner can complete the 42.2-km distance of the marathon in 3 h and 30 min. If the runner's mass is 85 kg, what is the runner's average kinetic energy during the run
Answer:
the runner's average kinetic energy during the run is 476.96 J.
Explanation:
Given;
mass of the runner, m = 85 kg
distance covered by the runner, d = 42.2 km = 42,200 m
time to complete the race, t = 3 hours 30 mins = (3 x 3600s) + (30 x 60s)
= 12,600 s
The speed of the runner, v = d/t
v = 42,200 / 12,600
v = 3.35 m/s
The runner's average kinetic energy during the run is calculated as;
K.E = ¹/₂mv²
K.E = ¹/₂ × 85 × (3.35)²
K.E = 476.96 J
Therefore, the runner's average kinetic energy during the run is 476.96 J.