Which statement describes the direction of the current and the magnetic field when the left hand rule is being used?
Answer:
They are perpendicular.
Explanation:
Why is a control group generally very important in an experiment?
Answer: The control group is the part where you see what happens when you change a variable you want to study/examine. Basically, you need the control group because you need something to see what happens when you change something.
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A car of mass 500kg travelling at 60m/s has it speed reduced to 40m/s by a constant breaking force over a distance of 200m. find car initial kinetic energy. the final kinetic energy
Answer:
Ek1 = 900000 [J]
Ek1 = 400000 [J]
Explanation:
In order to solve this problem we must remember that kinetic energy is defined as the product of mass by velocity squared by a medium. Therefore using the following equation we have:
[tex]E_{k1}=\frac{1}{2}*m*v1^{2}[/tex]
where:
m = mass = 500 [kg]
v1 = 60 [m/s]
So we have:
Ek1 = 0.5*500*(60^2)
Ek1 = 900000 [J]
and:
Ek2 = 0.5*500*(40^2)
Ek2 = 400000 [J]
Which action will cause the induced current to decrease or remain constant?
A. Moving the magnet faster B. Increasing the strength of the magnet
C. Adding more turns to coil. D. Reversing polarity of the magnet.
Answer:
Its Reversing the polarity of the magnet :)
Explanation:
The action that cause the induced current for reduce or be same is when the polarity is reverse with respect to the magnet.
The following actions should not induced the current:
In the case when the magnet moved in the fastest way. When the magnet strength is increased. And, when the more turns to coil is added.Therefore we can conclude that the action that cause the induced current for reduce or be same is when the polarity is reverse with respect to the magnet.
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which of the following units is part of the international system of units, or SI?
Pounds meters
ounces inches
it is meters and ounces
PLEASE HELP!! Salmon often jump waterfalls to reach their
breeding grounds.
Starting downstream, 3.41 m away from a
waterfall 0.397 m in height, at what minimum
speed must a salmon jumping at an angle of
28.4° leave the water to continue upstream?
The acceleration due to gravity is 9.81 m/s2
Answer in units of m/s.
Answer:
7.13781 m/s
Explanation:
X-direction | Y-direction
[tex]x=v_{xo}t+\frac{1}{2}a_{x}t^2[/tex] | [tex]y=v_{yo}t+\frac{1}{2}a_{y}t^2[/tex]
[tex]3.41=v_{o}cos(28.4)t[/tex] | [tex]0.397=v_{yo}sin(28.4)t+\frac{1}{2} (-9.81)t^2[/tex]
[tex]3.41=v_{o}(0.87964)t[/tex] | [tex]0.397=v_{yo}sin(28.4)(\frac{3.87658}{v_{o} })+\frac{1}{2} (-9.81)(\frac{3.87658}{v_{o} })^2[/tex]
[tex]\frac{3.41}{0.87964}=v_{o}t[/tex] | [tex]0.397=1.84379-\frac{73.71171}{v^2}[/tex]
[tex]\frac{3.87658}{v_{o} } =t[/tex] | [tex]7.13781=v[/tex]
Which platform has touch controls?
A.
consoles
B.
arcades
C.
personal computers
D.
mobiles
Answer:
I think it's D
Explanation:
Most mobiles have touch screens
NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA's idea is basicallyan electric slingshot that consists of 4 electrodes arranged in a horizontal square with sides of length d at a height h above the ground. The satellite is then placed on the ground aligned with the center of the square. A power supply will provide each of the four electrodes with a charge of Q/4 and the satellite with a charge -Q. When the satellite is released from rest, it moves up and passes through the center of the square. At the instant it reaches the square's center, the power supply is turned offand the electrodes are grounded, giving them a zero electric charge. To test this idea, you decide to use energy considerations to calculate how big Q will have to be to get a 100 kg satellite to a sufficient orbit height. Assume that the satellite startsfrom 15 meters below the square of electrodes and that the sides of the square are each 5 meters. In your physics text you find the mass of the Earth to be 6.0 x 1024kg.
Answer:
The answer is "[tex]q=0.0945\,C[/tex]".
Explanation:
Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).
[tex]U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}[/tex]
It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:
[tex]U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}[/tex]
Potential energy shifts:
[tex]= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\[/tex]
[tex]=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J[/tex]
Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.
[tex]=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\[/tex]
[tex]=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\[/tex]
[tex]\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C[/tex]
This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.
You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfortunately, you have the density of mercury in units of kilogrammeter3kilogrammeter3 and the density of silicon in other units: 2.332.33 gramcentimeter3gramcentimeter3. You decide to convert the density of silicon into units of kilogrammeter3kilogrammeter3 to perform the comparison. By which combination of conversion factors will you multiply 2.332.33 gramcentimeter3gramcentimeter3 to perform the unit conversion?
Answer:
The conversion factors you will use to multiply is 1/0.001 or simply 1000 to perform the unit conversion.
Explanation:
To convert from gram/centimeter³ to kilogram/meter³
First,
NOTE:
1 kilogram = 1000 gram, that is 1 gram = 0.001 kilogram; and
1 meter = 100 centimeter, that is 1 centimeter = 0.01 meter
then,
1 meter³ = 1000000 centimeter³, that is
1 centimeter³ = 0.000001 meter³
Now, we can write that
1 kilogram/meter³ = 1000 gram / 1000000 centimeter³
= 0.001 gram/centimeter³
If 1 kilogram/meter³ = 0.001 gram/centimeter³
Then, 1 gram/centimeter³ = 1/0.001 kilogram/meter³ = 1000 kilogram/meter³
Hence, 1 gram/centimeter³ = 1000 kilogram/meter³
∴The conversion factors you will use to multiply is 1/0.001 or simply 1000 to perform the unit conversion.
That is,
2.33 gram/centimeter³ will be 2.33 × 1000 kilogram/meter³
= 2330 kilogram/meter³
A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 3.54 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2230 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 5.25 V/m, (b) in the negative z direction and has a magnitude of 5.25 V/m, and (c) in the positive x direction and has a magnitude of 5.25 V/m
Answer:
(a). The magnitude of the net force is [tex](2.1\times10^{-18}\ N)k[/tex]
(b). The magnitude of the net force is [tex](4.23\times10^{-19}\ N)k[/tex]
(c). The magnitude of the net force is [tex](8.4\times10^{-19}\ N)i+(12.6\times10^{-19}\ N)k[/tex]
Explanation:
Given that,
Magnetic field [tex]B=-3.54\times10^{-3}i\ T[/tex]
Velocity = 2230j m/s
We know that,
The net force acting on the proton is equal to the sum of electric and magnetic force.
[tex]F=F_{e}+F_{B}[/tex]
(a). If the electric field is in the positive z direction and has a magnitude of 5.25 V/m,
We need to calculate the magnitude of the net force acting on the proton
Using formula of net force
[tex]F_{net}=e(E+v\times B)[/tex]
Put the value into the formula
[tex]F_{net}=1.6\times10^{-19}(5.25k+2230\times-3.54\times10^{-3}(j\times i))[/tex]
[tex]F_{net}=1.6\times10^{-19}(5.25k+2230\times-3.54\times10^{-3}(-k))[/tex]
[tex]F_{net}=(2.1\times10^{-18}\ N)k[/tex]
(b). If the electric field is in the negative z direction and has a magnitude of 5.25 V/m,
We need to calculate the magnitude of the net force acting on the proton
Using formula of net force
[tex]F_{net}=e(E+v\times B)[/tex]
Put the value into the formula
[tex]F_{net}=1.6\times10^{-19}(-5.25k+2230\times-3.54\times10^{-3}(j\times i))[/tex]
[tex]F_{net}=1.6\times10^{-19}(-5.25k+2230\times-3.54\times10^{-3}(-k))[/tex]
[tex]F_{net}=(4.23\times10^{-19}\ N)k[/tex]
(c). If the electric field is in the positive x direction and has a magnitude of 5.25 V/m
We need to calculate the magnitude of the net force acting on the proton
Using formula of net force
[tex]F_{net}=e(E+v\times B)[/tex]
Put the value into the formula
[tex]F_{net}=1.6\times10^{-19}(5.25i+2230\times-3.54\times10^{-3}(j\times i))[/tex]
[tex]F_{net}=(8.4\times10^{-19}\ N)i+(12.6\times10^{-19}\ N)k[/tex]
Hence, (a). The magnitude of the net force is [tex](2.1\times10^{-18}\ N)k[/tex]
(b). The magnitude of the net force is [tex](4.23\times10^{-19}\ N)k[/tex]
(c). The magnitude of the net force is [tex](8.4\times10^{-19}\ N)i+(12.6\times10^{-19}\ N)k[/tex]
You are piloting a General Aviation aircraft (e.g. Cessna 172) over the Georgia Tech campus in level flight at 3000 ft altitude. Your airspeed indicator shows that the true airspeed is 100 knots. You have trimmed the aircraft to have no sideslip and pitch angle at a 3 degree angle of attack. Your magnetic compass (which is uncorrected for magnetic variation) indicates that you are flying in the direction of local magnetic East. A local weather forecast predicted the winds to be steady at 10 knots coming exactly from the northeast direction at the time you took off.
a. Using the information provided above and any additional information you may require (state what it is), determine your aircraft's north and east velocity components, Vn and ve, in knots. Starting from Tech Tower at t = 0, you continue in this direction and you are planning to fly directly over a local VOR beacon exactly 10 minutes later. However, you note that you don't pass directly over the beacon and after 10 minutes and 15 seconds have elapsed you are exactly 300 meters away from the VOR beacon in the Easterly direction.
b. Assuming the discrepancy is attributable only to wind speed, estimate the actual average wind speed components, Vwind-n and Vwind-e in knots. Use a flat Earth approximation.
c. Based on the instruments you are using, describe 3 common causes of measurement error that could affect the output of either sensor.
Answer:
a) vₓ = 100 + 7.07 = 107.07 knot (East) , v_{y} = 7.07 knot (North)
b) v_wind_ North = 0 , v_wind_West = 61.12 knot (West)
Explanation:
a) This is a vector velocity addition problem, they tell us that the plane goes East at 100 Knots and the wind goes North East at 10 Knots
the components ask the total speed of the plane.
For this we decompose the wind speed
cos 45 = v₂ₓ / V₂
sin 45 = [tex]v_{2y}[/tex] / v₂
v₂ₓ = v₂ cos 45
v_{2y} = v₂ sin 45
v₂ₓ = 10 cos 45 = 7.07 m / s
v_{2y} = 10 sin45 = 7.07 m / s
the speed of the plane is
vₓ = v_plane + v₂ₓ
v_{y} = v_{2y}
vₓ = 100 + 7.07 = 107.07 knot (East)
v_{y} = 7.07 knot (North)
speed is
v = (107.07 i ^ + 7.07 j ^) knot
b) Estimated time for Target VOR t = 10 min
for a time of t = 10 min and 15 s, it is at a distance of d = 300 m from the VOR in an easterly direction
find the average speed of time on the ride
in the North direction there is no deviation so the average wind speed is zero
v_wind_ North = 0
In the east direction
as the estimated time was 10 min and the real time for the distance is 10 min 15 s
the time difference is t = 15 s to travel d = 300 m
with these data we can calculate the speed of the plane
v = x / t
v = 300/15
v = 20 m / s
let's reduce this speed to knot
v_real = 20 m / s (1knot / 0.5144 m / s) = 38.88 knots
therefore the wind speed is
v_tral = v_plane + v_wind_Este
v_wind_Este = v_real - v_avión
v_wind_Este = 38.88 - 100
v_wind_Este = -61.12 knot
the negative sign indicates that the wind is going west
v_wind_West = 61.12 knot (West)
3)
* The compass is not calibrated, to correct the magnetic deviation, therefore it gives an appreciable error in the direction
* The wind speed is taken when leaving, but there is no constant monitoring, so a change in direction or wind speed in the trajectory can significantly affect the results.
* The aircraft may noir throughout the trajectory at level, due to pitch errors
Three vectors, one of 74 km at 30°, the second of 54 km at 135°, and the final of 62 km due South. What is the sum of the three vectors?
- will give brainliest -
Answer:
The sum of the three vectors are 63.18 km.
Explanation:
Given that,
One vector = 74 km at 30°
Second vector = 54 km at 135°
Third vector = 62 km
We need to calculate the sum of the three vectors
Using given data
[tex]R=\vec{A}\sin\theta+\vec{B}\sin\theta+\vec{C}[/tex]
Put the value into the formula
[tex]R=-74\sin30+54\sin135+62[/tex]
[tex]R=63.18\ km[/tex]
Hence, The sum of the three vectors are 63.18 km.
The mass of proton is 1.67x10 kg. How many protons will make a mass of
1,00kg?
1011
Answer: 938
Explanation:
1.67x10^-27kg (938V/c2)
what will happen if a low massive main sequence star runs out of hydrogen fuel?
Answer:
Low mass stars are: hydrogen burning in the core while on the Main Sequence. As the hydrogen fuel runs out, extreme pressure raises the temperature to 100 million degrees, where helium burning becomes possible.
Marie curie investigation method
Answer: To isolate the unknown substances, of which only tiny amounts were present, the Curies were the first to use a new method of chemical analysis. They employed various standard (but sometimes demanding) chemical procedures to separate the different substances in pitchblende.
Explanation:
In 2006, NASA’s Mars Odyssey orbiter detected violent gas eruptions on Mars, where the acceleration due to gravity is 3.7 m/s2. The jets throw sand and dust about 62.0 m above the surface. Scientists estimate that the jets originate as high-pressure gas speeds through vents just underground at about 130 km/h. How much energy per kilogram of material is lost due to nonconservative forces as the high-speed matter forces its way to the surface and into the air? (Express your answer to two significant figures.)
Answer:
The unit energy losses due to nonconservative forces is 881.40 joules per kilogram.
Explanation:
We can estimate the unit energy losses of gas eruption by Principle of Energy Conservation and Work-Energy Theorem:
[tex]U_{g,1} + K_{1} = U_{g,2}+K_{2}+W_{loss}[/tex] (Eq. 1)
Where:
[tex]U_{g,1}[/tex] - Gravitational potential energy of gas eruptions at surface, measured in joules.
[tex]U_{g,2}[/tex] - Gravitational potential energy of gas eruptions at highest height, measured in joules.
[tex]K_{1}[/tex] - Translational kinetic energy of gas eruptions at surface, measured in joules.
[tex]K_{2}[/tex] - Translational kinetic energy of gas eruptions at highest height, measured in joules.
[tex]W_{loss}[/tex] - Energy losses due to nonconservative forces, measured in joules.
We clear the component associated with energy losses in (Eq. 1):
[tex]W_{loss} = U_{g,1}-U_{g,2}+ K_{1}-K_{2}[/tex]
And we expand it afterwards:
[tex]W_{loss} = m\cdot g\cdot (z_{1}-z_{2}) + \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})[/tex] (Eq. 2a)
[tex]w_{loss} = g\cdot (z_{1}-z_{2})+\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})[/tex] (Eq. 2b)
Where:
[tex]W_{loss}[/tex] - Energy losses due to nonconservative forces, measured in joules.
[tex]w_{loss}[/tex] - Unit energy losses due to nonconservative forces, measured in joules per kilogram.
[tex]g[/tex] - Gravitational acceleration, measured in meters per second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Bottom and top height, measured in meters.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Gas eruption speeds at surface and highest heights, measured in meters per second.
If we know that [tex]g = 3.7\,\frac{m}{s^{2}}[/tex], [tex]z_{1} = 0\,m[/tex]. [tex]z_{2} =62\,m[/tex]. [tex]v_{1} = 36.111\,\frac{m}{s}[/tex] and [tex]v_{2} = 0\,\frac{m}{s}[/tex], the unit energy losses are:
[tex]w_{loss} = \left(3.7\,\frac{m}{s^{2}} \right)\cdot (62\,m-0\,m)+\frac{1}{2} \cdot \left[\left(36.11\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right][/tex]
[tex]w_{loss} = 881.40\,\frac{J}{kg}[/tex]
The unit energy losses due to nonconservative forces is 881.40 joules per kilogram.
according to newton’s first law a moving object acted on by a net force of zero....
Answer:
Newton's first law says that if the net force on an object is zero ( Σ F = 0 \Sigma F=0 ΣF=0\Sigma, F, equals, 0), then that object will have zero acceleration. That doesn't necessarily mean the object is at rest, but it means that the velocity is constant.
Explanation:
According to Newton’s first law of motion, a moving object acted on by a net force of zero will continue in motion.
Newton's first law of motion states that an object in state of rest or uniform motion in a straight line will continue in that state unless it is acted upon by an external force.
This first is also known as the law of inertia because it is the reluctance of an object in motion to stop moving or reluctance of an object at rest to start moving which depends on the mass of the object.
Thus, we can conclude that according to Newton’s first law of motion, a moving object acted on by a net force of zero will continue in motion.
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The speed of light is 300,000 kilometres per second. This number isn't easy to get a feel for but it's good it's so fast because it allows us to see things instantly on Earth without waiting for the light to arrive.
The Sun is far enough away that light takes an appreciable time to travel to us. Assuming the sun is 150,000,000 kilometres away, how long would it take us to realize if the sun suddenly stopped shining?
Answer:
499 seconds or a little over 8 minutes
Explanation:
~150 million km/ 300,000 km per second = 500 seconds
Protons have ___charge; they have equal amounts of positive and negative ____charges?
A train travels 600 km in 4 hours. What is the speed of the train?
0.007 km/h
O 604 km/h
O 150 km/h
O 2,400 km/h
The speed of the train is 150 Km/h.
Speed can be defined as the distance travelled over a period of time. Mathematically, it can be expressed as:
Speed = Distance / Time
With the above formula, we can obtain the speed of the train as shown below:
Distance = 600 KmTime = 4 hours Speed =?Speed = Distance / Time
Speed = 600 / 4
Speed = 150 Km/h
Therefore, the speed of the train is 150 Km/h.
Learn more about speed:
https://brainly.com/question/680492
Can someone please help me out?
Explanation:
If you want to get speed, u have to divided distance over time
The lowest speed will lose
What is a point of view with regard to motion called in physics?
O reference grid
O displacement
O distance
O frame of reference
Answer:
Frame of reference
Experiments need to be repeated by the scientist and also replicated by other scientists.
true
false
Answer:
false
Explanation: because the other scientists could've missed something in there experiment
A cell contains 20% solute. If it is placed in a solution with 35% solute, what will happen to the cell?
An object is moving at 2.50 m/s [E]. At a time 3.00 seconds later the object is traveling at 1.50 m/s
[E]. What was the displacement during this 3.00 second time interval?
Given parameters:
First velocity = 2.50m/s
Time of travel = 3s
Second velocity = 1.50m/s
Unknown:
The displacement during the first interval = ?
Velocity is the displacement of a body with time. Displacement is a distance move in a specific direction by a body.
Velocity = [tex]\frac{Displacement}{Time taken}[/tex]
So;
Displacement = Velocity x Time taken
Now input the parameter for the first velocity and time of travel;
Displacement = 2.5 x 3 = 7.5m
The displacement id 7.5m
Which of the following does not affect gas pressure
Answer:
I NEED OPTIONS
Explanation:
In the equation, y = A sin( wt - kx ), obtain the dimensional formula ' w ' and ' k.
Answer:
In the given equation y=Asin(ωt−kx)
Since, trigonometric function is a dimensionless quantity. So, (ωt−kx)is also dimensionless quantity.
So,
(ωt−kx)=1
ωt=1
[ω]= 1=[T] ^−1
[T]
kx=1
[k]= 1=[L]^ −1
[L]
I WILL GIVE YOU BRANILEST!
If a car hits a tree, the tree pushes back on the car. Damage to the car and/or the tree depends on what factors?
Answer:
Acceleration and mass.
Explanation:
Depends on how heavy the car is and how fast it is going.
F=MA
A monarchy is the type of government that the colonist do not want true or false
Answer:
Monarchy is rule from kings and queens
Explanation:
what is 100+10000000000000000000
Answer:
200000000000000000000000
Explanation:
Answer:
10000000000000000100
Explanation: