It refers to a quantity changed by the experimenter.
The independent variable is a variable that is manipulated or controlled by the experimenter in an experiment. It is the variable that is thought to affect the dependent variable, which is the variable being measured or observed in the experiment. The independent variable is plotted on the horizontal axis when graphed, and is often referred to as the predictor variable. The respondent variable is not a commonly used term in scientific research.
An aircraft has to fly between two cities, one of which is 600.0 km north of the other. The pilot starts from the southern city and encounters a steady 100.0 km/h wind that blows from the northeast. The plane has a cruising speed of 236.0 km/h in still air. In what direction (relative to east) must the pilot head her plane
Answer:
72.57° North of east
Explanation:
From the given information:
We can compute the velocity plane that is related to the ground in air in the North direction as;
[tex]v^{\to} _{PG} = v \\ \\ v^{\to} _{PG,x} = 0 \\ \\ v^{\to} _{PG,y} = v[/tex]
However, the velocity of the wind-related to the ground from the NorthEast direction is;
[tex]v^{\to}_{wG}=100 \ km/h \\ \\ \text{from North East} \\ \\ v_{wG,x} = (-100 \ km/h ) cos 45 = -70.7 km/h \\ \\ v_{wG,y} = (-100 \ km/h ) sin 45 = -70.7 km/h[/tex]
Now,
Since the plane is moving with a 236 km/h speed in the Northeast direction;
Then;
[tex]v^{\to} _{pw} = 236 \ km/h \\ \\ v^{\to} _{pw.x} = (236 m/s) cos \theta \\ \\ v^{\to} _{pw,y} = (236\ m/s) sin \theta \\ \\ v_{pG,x} = v_{pw,x} + v_{w G,x} \\ \\ \implies 0 = (236 \ km/h) sin \theta -( 70.7 \ km/h) \\ \\ \implies cos \theta = \dfrac{70.7 \ km/h}{236 \ km/h} \\ \\ \theta = cos^{-1} (0.2996) \\ \\ \mathbf{\theta = 72.57}[/tex]
The work done is a vector quantity and SI base unit is J
Answer:
Is this your question? Also I think work done is a scalar quantity.
Explanation:
PROJECT: VIRTUAL LABS — CIRCUITS
Answer:
Table A
Measuring Current as a Function of Voltage with a 20 Ω Resistor
Voltage
(V)
Current: Calculated
(A)
Current: Experimental
(A)
1 0.05 0.05
5 0.25 0.25
10 0.50 0.50
20 1.00 1.00
50 2.50 2.50
Table B
Measuring Current as a Function of Resistance at 25 V
Resistance
(Ω)
Current: Calculated
(A)
Current: Experimental
(A)
10 2.50 2.50
20 1.25 1.25
100 0.25 0.25
200 0.12 0.12
Table C
Measuring Current in a Parallel Circuit
Resistor Set
(Ω)
Total
Resistance
(Ω)
Calculated
Current
(A)
Observed
Current
(A)
Observed Current
through Each Resistor
(A)
20, 20, 20 6.67 3.75 3.74 1.25, 1.25, 1.25
20, 20, 200 9.52 2.63 2.62 1.25, 1.25, 0.12
Voltage needed to raise current to 3.75 A (20, 20, 200 resistor set):
Calculated: 35.7
Observed: 36
Table D
Calculating Power of Circuit ComponentsTeacher Guide (continued)
Observed Total Current
(A)
Current through Each Bulb
(A)
Power Usage per Bulb
(W)
2.00 0.67 6.7
Explanation:
got this from the teachers guide
Calculate the acceleration of a car if the force on the car is 450 Newtons and the mass is 1300 kilograms.
[tex] \Large {\underline { \sf {Required \; Solution :}}}[/tex]
We have ―
Force, F = 450 NMass of the car, m = 1300 kgWe have been asked to calculate the acceleration of the car.
[tex]\qquad \implies\boxed{\red{\sf{ F = ma }}}\\[/tex]
F denotes Forcem denotes massa denotes acceleration[tex] \quad \twoheadrightarrow\sf { 450 = 1300a} \\ [/tex]
[tex] \quad \twoheadrightarrow\sf {\cancel{ \dfrac{450}{1300}} = 1300a} \\ [/tex]
[tex]\quad\twoheadrightarrow\boxed{\red{\sf{0.346 \; ms^{-2} = a }}}\\[/tex]
Therefore, acceleration of the car is 0.346 m/s².
Determiner l'interfrange i sur le plan d'observation π distant de L de D:
Determine the interfringe i on the observation plane π distant from L from D:
1) pour D=f
if D=f
2)pour D=2f
if D=2f
Answer:
can you explain in Hindi language
because i learn hindi
Which statement is true about how early humans met their needs?
Answer:
they were hunter gatherers
Explanation:
A 0.545-kg ball is hung vertically from a spring. The spring stretches by 3.56 cm from its natural length when the ball is hanging at equilibrium. A child comes along and pulls the ball down an additional 5cm, then lets go. How long (in seconds) will it take the ball to swing up and down exactly 4 times, making 4 complete oscillations before again hitting its lowest position
Answer:
t = 9.52 s
Explanation:
This is an oscillatory motion exercise, in which the angular velocity is
w = [tex]\sqrt{ \frac{k}{m} }[/tex]
Let's use hooke's law to find the spring constant, let's write the equilibrium equation
F_e - W = 0
F_e = W
k x = m g
k = [tex]\frac{m g}{x}[/tex]
k = 0.545 9.8 /0.0356
k = 150 N / m
now the angular velocity is related to the period
W = 2π / T
we substitute
4π² T² = k /m
T = 4pi² [tex]\sqrt{ \frac{m}{k} }[/tex]
we substitute
T = 4 pi² [tex]\sqrt{ \frac{0.545}{150} }[/tex]
T = 2.38 s
therefore for the spring to oscillate 4 complete periods the time is
t = 4 T
t = 4 2.38
t = 9.52 s
PLEASE HELP! I'LL GIVE BRAINLEST
Answer:
In this conversation the Neil astronaut is right
Acellus
A motion sensor emits sound, and
detects an echo 0.0115 s after. A
short time later, it again emits a
sound, and hears an echo after
0.0183 s. How far has the
reflecting object moved?
Help Resources
(Speed of sound = 343 m/s)
(Unit = m)
Answer:
1.17m
Explanation:
The formula to find distance is d=vt/2
This problem is asking for how far the reflecting object has moved so you need to find the distance from the motion sensor at both times.
(343)(0.115) / (2) = 1.97
(343)(0.0183) /(2) =3.14
After that, all you have to do is find the difference so
3.14 - 1.97
= 1.17
What is the mass of a mallard duck whose speed is 9.1 m/s and whose momentum has a magnitude of 12 kg⋅m/s?
Answer:
m = 1.31 kg
Explanation:
Given that,
The speed of duck, v = 9.1 m/s
The magnitude of momentum, p = 12 kg-m/s
We need to find the mass of the duck. We know that the momentum of an object is given by :
p = mv
Where
m is the mass of the duck
[tex]m=\dfrac{p}{v}\\\\m=\dfrac{12\ kg-m/s}{9.1\ m/s}\\\\m=1.31 kg[/tex]
So, the mass of the duck is equal to 1.31 kg.
Protons, neutrons, electrons, and a nucleus are
A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal disk is placed in contact with a wall, and the disk comes to rest after 10s. Which of the following situations associated with linear impulse is analogous to the angular impulse that is described?
a. A 3kg block is initially at rest. An applied force of 3N is applied to the block, but the block does not move.
b. A 3kg block is initially at rest. A net force of 3N is applied to the block until it has a speed of 10m/s.
c. A 3kg block is initially traveling at 10m/s. An applied force of 3N is applied to the block in the direction of its velocity vector for 10s.
d. A 3kg block is initially traveling at 10m/s. The block encounters a 3N frictional force until the block eventually stops.
Answer:
D
Explanation:
From the information given:
The angular speed for the block [tex]\omega = 50 \ rad/s[/tex]
Disk radius (r) = 0.2 m
The block Initial velocity is:
[tex]v = r \omega \\ \\ v = (0.2 \times 50) \\ \\ v= 10 \ m/s[/tex]
Change in the block's angular speed is:
[tex]\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s[/tex]
However, on the disk, moment of inertIa is:
[tex]I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2[/tex]
The time t = 10s
∴
Frictional torques by the wall on the disk is:
[tex]T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10}) \\ \\ =0.6 \ N.m[/tex]
Finally, the frictional force is calculated as:
[tex]F = \dfrac{T}r{}[/tex]
[tex]F= \dfrac{0.6}{0.2} \\ \\ F = 3N[/tex]
Frequency more than 20,000 HZ
Answer:
dvhd
Explanation:
xhxjjdvcbxhjddvifidid
Answer:
The units of frequency are called hertz (Hz). Humans with normal hearing can hear sounds between 20 Hz and 20,000 Hz. Frequencies above 20,000 Hz are known as ultrasound
using the human species as an example, explain what is meant by variation of traits
The indices of refraction for her contact lens, cornea, and the fluid behind her cornea are 1.6, 1.4, and 1.3, respectively. Light is incident from air onto her contact lens at an angle of 30 ∘∘ from the normal of the surface. At what angle is the light traveling in the fluid behind her cornea?
Answer:
[tex]23^{\circ}[/tex]
Explanation:
n = Refractive index of air = 1
[tex]n_1[/tex] = Refractive index of contact lens = 1.6
[tex]n_2[/tex] = Refractive index of cornea = 1.4
[tex]n_3[/tex] = Refractive index of fluid = 1.3
From Snell's law
[tex]n\sin30^{\circ}=n_1\sin\theta\\\Rightarrow \theta=\sin^{-1}\dfrac{1\sin30^{\circ}}{1.6}\\\Rightarrow \theta=18.21^{\circ}[/tex]
[tex]n_1\sin\theta=n_2\sin\theta_1\\\Rightarrow \theta_{1}=\sin^{-1}\dfrac{1.6\times \sin18.21^{\circ}}{1.4}\\\Rightarrow \theta_1=20.92^{\circ}[/tex]
[tex]n_2\sin\theta_1=n_3\sin\theta_3\\\Rightarrow \theta_3=\sin^{-1}\dfrac{1.4\sin20.92^{\circ}}{1.3}\\\Rightarrow \theta_3=22.62^{\circ}\approx 23^{\circ}[/tex]
The angle is the light traveling in the fluid behind her cornea is [tex]23^{\circ}[/tex].
The angle is the light traveling in the fluid will be 23⁰. Light is traveling in a particular direction with an angle.
What is snell law?"The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, for the light of a given color and for a given set of media,
The given data in the problem is;
n is the refractive index of air = 1
n₁ is the refractive index of contact lens = 1.6
n₂ is the refractive index of cornea = 1.4
n₃ is the refractive index of fluid = 1.3
According to Snell's law. The formula for Snell's law is
[tex]\rm n sin30^0 = n_1 sin\theta \\\\ \theta = sin^{- 1}(\frac{1sin30^0}{1.6} )\\\\ \theta = 18.21 ^0[/tex]
For contact lenses;
[tex]\rm n_1sin\theta = n_2 sin\theta_1 \\\\ \theta_1 = sin^{-1}\frac{1.6 \times sin 18.21^0}{1.4} \\\\ \theta_1 =20.92 ^0[/tex]
For fluid;
[tex]n_2 sin\theta_1 = n_2 sin \theta_3\\\\ \theta_3 = sin^{-1}\frac{1.4 sin 20.92^0}{1.3} \\\\ \theta_3 = 22.62 ^ 0 =23^0[/tex]
Hence the angle is the light traveling in the fluid will be 23⁰.
To learn more about snell's law refer to the link;
https://brainly.com/question/10112549
If a virtual image is formed 10.0 cm along the principle axis from a convex mirror of focal length-15.0 cm, how far is the object from the mirror
Answer:
U=30cm
Explanation:
All you have to do is to put
Mirror formula , 1/f=1/u + 1/v
You should be careful in sign convention .
Virtual image is negative
we take focal length of convex lens negative even if its not given and so on...
What do meteorologists call the lines
that join places with the same
temperature?
A. isobars
B. isotherms
C. anisobars
D. anisotherms
In the process of fluorescence, a molecule in its ground state will absorb a photon with a certain energy Eex, called the excitation energy, and then emit a photon with energy Eem, the emission energy. Obviously, the molecule cannot emit more energy than was absorbed.
a) You wrote in your notebook that you excited a fluorescent bead at λ = 640 nm and found that it fluoresced (emitted) at λ = 520 nm. Or was it the other way around? Do a calculation and see if those figures should be switched.
b) You shine a laser on your bead at the excitation wavelength. The laser has a power of 1 mW. How many photons are emitted by the laser in one second?
c) You wish to build a microscope that allows you to excite the bead at its excitation wavelength and detect the fluorescence at the emission wavelength. This requires a pair of filters, one that allows only the excitation wavelength to pass through, and the other that allows only the emission wavelength to pass through. What color will these filters appear (i.e. what color light will they transmit?) You may have to look at an electromagnetic spectrum to figure this out…
Answer:
a) the excitation energy is E₂ λ = 520 nm
the emission energy is E₁, λ= 640 nm
b) #_photons = 2.6 10¹⁸ photons,
c) he excitation wavelength λ = 520 nm is green, therefore the filter is also green
the emission wavelength is lam = 640 nm is orange
Explanation:
a) the energy of a photo is given by the planck relation
E = h f
the speed of light is
c = λ f
f = c /λ
we substitute
E = hc /λ
let's calculate the energy for the two photons
λ = 640 nm = 640 10⁻⁰ m
E₁ = 6.63 10⁻³⁴ 3 10⁸/640 10⁻⁹
E₁ = 3.1 10⁻¹⁹ J
λ = 520 nm = 520 10⁻⁹ m
E₂ = 6.63 10⁻³⁴ 3 10⁸/520 10⁻⁹
E₂ = 3.825 10⁻¹⁹ J
therefore the excitation energy is E₂ λ = 520 nm
the emission energy is E₁, λ= 640 nm
b) For this part let's use a direct proportion rule (rule of three). If a photon (lam = 520 nm) has an energy of 3.825 10⁻¹⁹ J, how many photons have an energy of E = 1 10-3 J. Remember that the power is the energy per unit of time
#_photons = 1 10⁻³ J (1 photon / 3.825 10⁻¹⁹ J)
#_photons = 2.6 10¹⁸ photons
c) the excitation wavelength λ = 520 nm is green, therefore the filter is also green
the emission wavelength is lam = 640 nm is orange
Plz help
What factors determine
how the speed of the marbles changes in a
collision?
Answer:
Force,friction,inertia and momentum
Explanation:
The speed that the marble is moving at can be determined by the amount of force used when pushed or pulled and what kind of surface it's on.Momentum is also a factor because of the mass of the marbles.
what do solar winds and the earths magnetic field create
Answer:
bc earth rotates
3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P3==D---------- :P
Explanation:
Answer:
The interaction between the solar wind and Earth's magnetic field, and the influence of the underlying atmosphere and ionosphere, creates various regions of fields, plasmas, and currents inside the magnetosphere such as the plasmasphere, the ring current, and radiation belts.
Explanation:
A very long straight current-carrying wire produces a magnetic field of 20 mT at a distance d from the wire. To measure a field of 5 mT due to this wire, you would have to go to a distance from the wire of A very long straight current-carrying wire produces a magnetic field of 20 mT at a distance d from the wire. To measure a field of 5 mT due to this wire, you would have to go to a distance from the wire of:_____.
a. 4d.
b. 16d.
c. 2d.
d. 8d.
Answer:
A. 4d
Explanation:
Let's begin with the formula for the magnetic field produced by a long wire.
[tex]B = \frac{\mu_0I}{2\pi d}[/tex]
So [tex]d=\frac{\mu_0 I}{2\pi B }[/tex]
at point d_{1} is
[tex]d_{1}=\frac{\mu_{0} i}{2 \pi B_{1}} \\ \frac{d_{1}}{d}=\frac{\frac{\mu_{0} i}{2 \pi B_{1}}}{\frac{\mu_{0} i}{2 \pi B}} \\ d_{2}=d\left(\frac{B}{B}\right) \\ =d\left(\frac{20 \mathrm{mT}}{5 \mathrm{mT}}\right) \\ =4 d[/tex]
Hence, option (A) is correct answer
A certain type of laser emits light that has a frequency of 4.9 x 1014 Hz. The light, however, occurs as a series of short pulses, each lasting for a time of 2.9 x 10-11 s. The light enters a pool of water. The frequency of the light remains the same, but the speed of light slows down to 2.3 x 108 m/s. In the water, how many wavelengths are in one pulse
Answer:
N = 1.42 × 10⁴ cycles
Explanation:
Given that:
frequency f = 4.9 × 10¹⁴ Hz
Time = 2.9 × 10⁻¹¹ s
Speed = 2.3 × 10⁸ m/s
Recall that:
wavelength [tex]\lambda = \dfrac{c}{f} \\ \\[/tex]
Horizontal distance [tex]\Delta x = ct[/tex]
Number of wavelengths [tex](N) = \dfrac{\Delta x}{\lambda}[/tex]
[tex]N = \dfrac{ct}{c/f} \\ \\ N= ft[/tex]
N = (4.9 × 10¹⁴ cycles/s) (2.9 × 10⁻¹¹ s)
N = 14210
N = 1.42 × 10⁴ cycles
What is Force ?.............
Answer:
Push or pull of an object is considered a force. Push and pull come from the objects interacting with one another. Terms like stretch and squeeze can also be used to denote force.
In Physics, force is defined as:
The push or pull on an object with mass that causes it to change its velocity.
Force is an external agent capable of changing the state of rest or motion of a particular body. It has a magnitude and a direction. The direction towards which the force is applied is known as the direction of the force and the application of force is the point where force is applied.
Please help due today
Answer:
8
Explanation:
(8√2)² = x² + x²
8² × √2² = 2x²
64 × 2 = 2x²
128 = 2x²
64 = x²
x = 8
give me brainliest please
A 25.0kg girl pushes a 50.0kg boy so that he accelerates at 4.00m/s2. What is the force of the boy on the girl? A. 200N B. 100N C. 12.5 D. 400N
Answer:
a
Explanation:
so the answer is 200N
and I hope it is correct
7) If someone behaves against your company's code of ethics, what should you do?
A) Ignore it and mind your own business.
B) Suggest that they talk to the human resources department.
C) Talk to them about the situation.
D) Report the problem to your supervisor
Answer:
D) Report the problem to your supervisor.
Explanation:
This is probably the most efficient way get them to stop or to get them to follow the rules <3
A metal can containing condensed mushroom soup has mass 215 g, height 10.8 cm, and diameter 6.38 cm. It is placed at rest on its side at the top of a 3.00 m long incline that is at 25.0 degrees to the horizontal, and it is then released to roll straight down. Assuming mechanical energy conservation, calculate the moment of inertia of the can if it takes 1.50 s to reach the bottom of the incline. Which pieces of data, if any, are unnecessary for calculating the solution
Answer:
I = 1.093 x 10⁻⁴ kg.m²
Here, all the other data, namely, the height of the can, length of the inclined plane, angle of inclination, time to reach the bottom, are unnecessary.
Explanation:
The can which is filled with the soup can be modelled as a solid cylinder. The moment of inertia of this solid cylinder about its axis of rotation can be given by the following formula:
[tex]I = \frac{1}{2}mr^2[/tex]
where,
I = moment of inertia of can = ?
m = mass of can with soup = 215 g = 0.215 kg
r = radius of can = diameter/2 = 6.38 cm/2 = 3.19 cm = 0.0319 m
Therefore,
[tex]I = \frac{1}{2}(0.215\ kg)(0.0319\ m)^2 \\[/tex]
I = 1.093 x 10⁻⁴ kg.m²
Here, all the other data, namely, the height of the can, length of the inclined plane, angle of inclination, time to reach the bottom, are unnecessary.
Hydrocarbons are by-products of which of the following:
a. Fossil oils, mines and fossil fuels
b. Burning fires, water treatment plants and fossil oils
C. Combustion in fossil, lighting and exhaust fumes
d. Petrol, sea waves and combustion in vehicles
e. Burning fires and combustion in factories and vehicles
Answer:
A
Explanation:
Medical devices implanted inside the body are often powered using transcutaneous energy transfer (TET), a type of wireless charging using a pair of closely spaced coils. And emf is generated around a coil inside the body by varying the current through a nearby coil outside the body, producing a changing magnetic flux. Calculate the average induced emf, of each 10-turn coil has a radius of 1.50 cm and the current in the external coil varies from its maximum value of 10.0 A to zero in 6.25 x10-6s.
Answer:
[tex]0.475\ \text{V}[/tex]
Explanation:
n = Number of turns = 10
r = Radius = 1.5 cm
I = Current = 10 A
t = Time = [tex]6.25\times 10^{-6}\ \text{s}[/tex]
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi\times 10^{-7}\ \text{H/m}[/tex]
Magnetic field is given by
[tex]B=\dfrac{\mu_0I}{2r}\\\Rightarrow B=\dfrac{4\pi 10^{-7}\times 10}{2\times 1.5\times 10^{-2}}\\\Rightarrow B=0.00042\ \text{T}[/tex]
EMF is given by
[tex]\varepsilon=\dfrac{nBA}{t}\\\Rightarrow \varepsilon=\dfrac{10\times 0.00042\times \pi (1.5\times 10^{-2})^2}{6.25\times 10^{-6}}\\\Rightarrow \varepsilon=0.475\ \text{V}[/tex]
The average induced emf is [tex]0.475\ \text{V}[/tex].
A textbook of mass 2.10 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.100 m , to a hanging book with mass 3.10 kg . The system is released from rest, and the books are observed to move a distance 1.29 m over a time interval of 0.850 s . Part A What is the tension in the part of the cord attached to the textbook
Answer:
the tension in the part of the cord attached to the textbook is 7.4989 N
Explanation:
Given the data in the question;
As illustrated in the image below;
first we determine the value of the acceleration,
along vertical direction; we use the second equation of motion;
y = ut + [tex]\frac{1}{2}[/tex]a[tex]_y[/tex]t²
we substitute;
0 m/s for u, 1.29 m for y, 0.850 s for t,
1.29 = 0×0.850 + [tex]\frac{1}{2}[/tex]×a[tex]_y[/tex]×(0.850)²
1.29 = 0.36125a[tex]_y[/tex]
a[tex]_y[/tex] = 1.29 / 0.36125
a[tex]_y[/tex] = 3.5709 m/s²
Now when the text book is moving with acceleration , the dynamic equation will be;
T₁ = m₁a[tex]_y[/tex]
where m₁ is the mass of the text book ( 2.10 kg )
a[tex]_y[/tex] is the vertical acceleration ( 3.5709 m/s² )
so we substitute
T₁ = 2.10 × 3.5709
T₁ = 7.4989 N
Therefore, the tension in the part of the cord attached to the textbook is 7.4989 N