Answer:
average speed = 66.67 km/h
Explanation:
In order to find the average speed of the train, you need to calculate the total distance traveled, divided by the time it took to cover that distance. So for the total distance:
Distance= 120 km + 180 km = 300 km
For the total time we need to add three different quantities, two of which we need to derived based on the information provided:
time for first part of the trip:
[tex]time_1=\frac{D_1}{v_1} =\frac{120}{60} \,h= 2\,h[/tex]
for the time of the stop:
[tex]time_2=0.5\,\,h[/tex]
for the last part of the trip:
[tex]time_3=\frac{180}{90} \,h= 2 \,\,h[/tex]
Which gives a total of 4.5 hours
Then, the average speed is: 300/4.5 km/h = 66.67 km/h
What happens when white light shines through a translucent, red, glass window? a) All colors of light except red are transmitted through the glass. b) Red light is transmitted through and reflected by the glass c) Red light is absorbed by the glass d) all colors except red are reflected by the glass
Answer:
b. Red light is transmitted through and reflected by the glass
Explanation:
Give me brainliest plz!
All of the wavelengths of the red light are absorbed when it passes through a translucent red glass window, but the red light is transmitted and reflected.
What happens when white light shines through red glass?Red light emerges from the other side of a white light source that has been passed through a red filter. This is so that only red light can pass through the red filter. The spectrum's other colors (wavelengths) are absorbed. Similar to this, a green filter only lets through green light.White light is colorless light that contains all the wavelengths of the visible spectrum. Only specific wavelengths of white light are filtered through transparent or translucent things.Through a colorful glass, white light shines through. Except for the color it is shining through, all light wavelengths are absorbed by the glass.Therefore, if the red light passes through a translucent, red, glass, window, all the wavelengths are absorbed but the red light is transmitted and reflected.
Therefore, the correct answer is option b) Red light is transmitted through and reflected by the glass.
To learn more about light refer to:
https://brainly.com/question/12002703
#SPJ2
On a horizontal frictionless surface a mass M is attached to two light elastic strings both having length l and both made of the same material. The mass is displaced by a small displacement Δy such that equal tensions T exist in the two strings, as shown in the figure. The mass is released and begins to oscillate back and forth. Assume that the displacement is small enough so that the tensions do not change appreciably. (a) Show that the restoring force on the mass can be given by F = -(2T∆y)/l (for small angles) (b) Derive an expression for the frequency of oscillation.
Answer:
(a) By small angle approximation, we have;
F = -2×T×Δy/l
(b) [tex]The \ frequency \ of \ oscillation, \ f = \dfrac{1}{2\cdot \pi }\cdot\sqrt{\dfrac{2 \cdot T}{l \cdot M} }[/tex]
Explanation:
(a) The diagram shows the mass, M, being restored by two equal tension, T acting on the elastic strings l, such the restoring force, F acts along the path of motion of the mass, with distance Δy
Therefore, the component of the tension T that form part of the restoring force is given as follows;
Let the angle between the line representing the extension of the elastic strings T and the initial position of the string = ∅
Then we have;
String force, [tex]F_{string}[/tex] = T×sin∅ + T×cos∅ + T×sin∅ - T×cos∅ = 2×T×sin∅
Whereby the angle is small, we have;
sin∅ ≈ tan∅ = Δy/l
Which gives;
[tex]F_{string}[/tex] = 2×T×sin∅ = 2×T×Δy/l (for small angles)
Restoring force F = [tex]-F_{string}[/tex] = -2×T×Δy/l
F = -2×T×Δy/l
(b) Given that the the tensions do not change appreciably as the mass, M, oscillates from Δy we have;
By Hooke's law, F = -k×x
Whereby Δy corresponds to the maximum displacement of the mass, M from the rest position, which gives;
Which gives;
F = M×a = -k×Δy
a = -k×Δy/M
d²(Δy)/dt² = -k×Δy/M
When we put angular frequency as follows;
ω² = k/M
We get;
d²(Δy)/dt² = -ω²×Δy
Which gives;
Δy(t) = A×cos(ωt + Ф)
The angular frequency is thus, ω = √(k/M)
Period of oscillation = 2·π/ω = 2·π/√(k/M)
The frequency of oscillation, f = 1/T = √(k/M)/(2·π)
Where:
k = 2·T/l, we have;
f = √(k/M)/(2·π) = √(2·T/l)/m)/(2·π)
The frequency of oscillation is given as follows;
[tex]f = \dfrac{1}{2\cdot \pi }\cdot\sqrt{\dfrac{2 \cdot T}{l \cdot M} }[/tex]
A 1000 kg car travels on a highway with a speed of 30 m/s. The driver sees a roadblock and applies the brakes, which provide a
constant braking force of 4 kN. What is the acceleration of the car?
A. 4 m/s
Answer:
-4m/s
Explanation:
use the formula
[tex]f = ma[/tex]
where f-force
m-mass
a-accleration
so
1kN=1000N
so apply
4000=1000×a
a=4m/s
(the negative is because the car was braking)
Answer:
Your answer is -4 m/s^2
Explanation:
Set Up: Let +x be the direction the car is traveling.
List the known & unknown quantities:
m = mass of the car = 1000 kg
υ = 30 m/s
Fx = –4 kN = –4000 N (negative since it is a braking force)
ax = acceleration =?
Solve: Use Newton’s second law of motion.
Fx=max
ax=Fx/m = −4000 N /1000 kg = −4000 kg·m/s^2 / 1000 kg =−4m/s^2
Can someone please help me?
An alien spaceship is 650 m above the ground and moving at a constant velocity of 175 m/s upwards.
How high above the ground is the ship after 5 seconds?
Answer:
1525 meters above ground
Explanation:
So to do this you will need to write this in slope intercept form or [tex]y=mx+b[/tex]. So 650 would be the b, 175 would be the m, and the x would be 5 so the equation would be [tex]y=175(5)+650[/tex] so if you solve or simplify the equation you will get 1525 meters above the ground and that would be our final answer.
(a) In electron-volts, how much work does an ideal battery with a 17.0 V emf do on an electron that passes through the battery from the positive to the negative terminal? (b) If 3.88 × 1018 electrons pass through each second, what is the power of the battery?
Answer:
(a) 17.0eV
(b) 10.55W
Explanation:
(a) The amount of work done (W) on an electron by an ideal battery of emf value of V as it moves from the positive to the negative terminal is given by;
W = q x V --------(i)
Where;
q = charge on the electron = 1e
From the question;
V = 17.0 V
Substitute the values of q and V into equation (i) as follows;
W = 1e x 17.0
W = 17.0eV
Therefore, the work done in electron volts is 17.0
(b) The power (P) of the battery as some electrons (n) pass through it at time t, is given as;
P = (n q V) / t --------------(ii)
Where;
n = number of electrons = 3.88 x 10¹⁸
t = 1s
q = 1.6 x 10⁻¹⁹C
V = 17.0V
Substitute these values into equation (ii) as follows;
P = (3.88 x 10¹⁸ x 1.6 x 10⁻¹⁹ x 17.0) / 1
P = 10.55W
Therefore the power of the battery is 10.55W
Show that energy dissipated due to motion of a conductor in the magnetic field is due to mechanical energy.
Explanation:
let us use the explanation below to get the intuition so desired;
According to Faraday's law of electro magnetic induction, when ever a coil/conductor is made to rotate in a magnetic field, voltage or emf is created and current is produced, in the long run energy has be produced or converted.
The conversion of this energy is made possible by the motion of the coil/conductor is the magnetic field, just by the motion of the conductor cutting through the magnetic field, thus creating electro motive force(E.M.F) hence producing current, and ultimately energy is created
formula of minimmum pressure
Answer:
pressure=force/area
The fan on a personal computer draws 0.3 ft3/s ofair at 14.7 psia and 708F through the box containing the CPU and other components. Air leaves at 14.7 psia and 838F.Calculate the electrical power, in kW, dissipatedby the PCcomponents
Answer:
0.12 kW
Explanation:
Given that
The flow rate of air (V)=0.3 ft³/s
V=0.008 m³/s
Pressure, P=14.7 psia
P=1.013529 atm=101.325 kPa
Inlet temperature = 70° F=294.261 K
Exit temperature = 83° F=301.483 K
We know that , specific heat capacity of the air
Cp=1.005 kJ/kg.K
The mass flow rate of air is given as
[tex]\dot{m}=\dfrac{P\times V}{R\times T}\\\dot{m}=\dfrac{101.325\times 0.008}{0.287\times 294.261}\\\dot{m}= 0.0095\ kg/s[/tex]
By using energy conservation
[tex]Electric\ power =\dot{m}\times C_p\times (T_2-T_1)\\Electric\ power =0.0095\times 1.005\times (83-70)=0.12\ kW[/tex]
Therefore electric power dissipate by components will be 0.12 kW.
how much heat is required to raise the temperature of 5kg of iron from 50°c to250°c
Answer:
462000J
Explanation:
Quantity of heat= mass x specific heat capacity of iron x change in temp
specific heat capacity of iron is 462J/Kg/K
change in temp = 250-50= 200°C
200°C is equivalent to 200K since 1°C is 1K
Q= mct
= 5x462x200
= 462000J
an attempt to estimate the height of a tree the Shadow of an upright metre rule was found to be 25 cm and the length of the Shadow of the tree was 7 m what is the height of the tree
Answer:
The actual height of the tree is 28 m
Explanation:
The given information are;
The length of the shadow of an upright meter rule = 25 cm
The actual height of the meter rule = 100 cm
The length of the shadow of the tree = 7 m
The actual height of the tree = h
We have
[tex]\dfrac{The \ length \ of \ the \ shadow \ of \ an \ upright \ metre \ rule}{The \ actual \ height \ of \ the \ metre \ rule} = \dfrac{The \ length \ of \ the \ shadow \ of \ the \ tree}{The \ actual \ height \ of \ the \ tree}[/tex]Which gives;
[tex]\dfrac{25 \ cm}{100 \ cm} = \dfrac{7 \ m}{The \ actual \ height \ of \ the \ tree}[/tex]
Therefore;
[tex]The \ actual \ height \ of \ the \ tree = 7 \ m \times \dfrac{100 \ cm}{25\ cm} = 7 \ m \times 4 = 28 \ m[/tex]
That is the actual height of the tree = 28 m.
Ozone molecules in the stratosphere absorb much of the harmful radiation from the sun. How many ozone molecules are present in 2.00 L of air under the stratospheric ozone conditions of 275 K temperature and 1.89 × 10−3 atm pressure?
Answer:
1.01×10^20 molecules of ozone.
Explanation:
Data obtained from the question include:
Volume (V) = 2 L
Temperature (T) = 275 K
Pressure (P) = 1.89×10¯³ atm
Gas constant (R) = 0.0821 atm.L/Kmol
Number of mole (n) of ozone =.?
Using the ideal gas equation, we can obtain the number of mole of ozone as follow:
PV = nRT
1.89×10¯³ x 2 = n x 0.0821 x 275
Divide both side by 0.0821 x 275
n = (1.89×10¯³ x 2) /(0.0821 x 275)
n = 1.67×10¯⁴ mole.
Therefore the number of mole of ozone in 2 L of air is 1.67×10¯⁴ mole.
Finally, we shall determine the number of molecules present in 1.67×10¯⁴ mole of ozone.
This can be obtained as follow:
From Avogadro's hypothesis, 1 mole of any substance contains 6.02×10²³ molecules. This implies that 1 mole of ozone contains 6.02×10²³ molecules.
If 1 mole of ozone contains 6.02×10²³ molecules,
therefore, 1.67×10¯⁴ mole of ozone will contain = 1.67×10¯⁴ x 6.02×10²³ = 1.01×10^20 molecules.
Therefore, 1.01×10^20 molecules of ozone are present in 2 L of air.
Which of the following object is in dynamic equilibrium?
Answer:
A car driving in a straight line 20 m/s
Explanation:
ayepecks silly
Which reverses the flow of current through
an electric motor?
Answer:
a commutator
Explanation:
The large-scale distribution of galaxies in the universe reveals Group of answer choices a smooth, continuous, and homogenous arrangement of clusters large voids, with most of the galaxies lying in filaments and sheets a large supercluster at the center of universe a central void with walls of galaxies at the edge of the universe
Question
The large-scale distribution of galaxies in the universe reveals
A) a smooth, continuous, and homogenous arrangement of clusters
B) large voids, with most of the galaxies lying in filaments and sheets a large supercluster at the center of the universe
C) a central void with walls of galaxies at the edge of the universe
Group of answer choices
Answer:
The correct answer is B)
Explanation:
The universe is arranged in a filamentary structure. Filamentary structures are very large. They are the largest kind of structures in the universe and comprise mostly of galaxies that are held together by gravity.
The structures found within Galaxy filaments have thread-like qualities spanning 52 to 78.7 megaparsecs h⁻¹ in lenght.
Other phenomena associated with the nature fo the universe is the existence of void spaces.
Cheers!
Fig.4.1
The switch S is closed for a period of 5.0 minutes. Calculate
(a)
the current through each resistor.
чре
(b)
the current through the battery.
rad
rad
(c)
the total charge which passes through the battery.
ma
(d)
the energy supplied by the battery.
ars
Answer:
A
Explanation:
Which statement describes one feature of a closed circuit? Charges do not flow. Bulbs will not shine. The circuit is broken. The circuit is complete.
I inferred you've referring to a close electrical circuit.
Answer:
The circuit is complete.
Explanation:
A closed electrical circuit is indeed a complete circuit. Also, it allows charges to flow, the bulbs in the circuit will shine and it is not broken.
It is termed closed circuit because there is no brokage in the series of electrical wires or the switch; which may prevent the free flow of current or charges. Thus, a feature that marks closed circuits is that they are complete.
Answer:
The circuit is complete.
Explanation:
This is a form of energy representing the motion of the molecules which make up an object. A. Thermal Energy B. Kinetic Energy C. Gravitational Potential Energy D. Chemical Potential Energy
Answer:
Kinetic energy.
Explanation:
There are many kinds of energy. Some of them are kinetic energy, potential energy, thermal energy etc. The energy that shows the motion of the object is called its kinetic energy.Also, the sum of kinetic energy and the gravitational potential energy is called mechanical energy. Out of the given options, kinetic energy is the form of energy that represents the motion of the molecules which make up an object. Hence, the correct option is (B).a solid weighs 20gf in air and 18 gf in water.Find the specific gravity of the solid. Please show your work.
Answer: It is given that A body weighs 20gf in air and 18. 0gf in water. Hence, the answer X-3 = 7.
A carpenter measured the lengeth of a small piece of timber as 24.6cm .Calculate the relative error in the measurement if the true length is 24.5cm
ANSWER:
0.4081%
Explanation:
Difference=24.6-24.5=0.1
Relative error = 0.1/24.5*100=0.4081%
Relative error is equal to the = difference between both the values/The true value *100
Suppose a 50.0 g block of silver (specific heat = 0.2350 J/g·°C) at 100.°C is placed in contact with a 50.0 g block of iron (specific heat = 0.4494 J/g·°C) at 0.00°C, and the two blocks are insulated from the rest of the universe. The final temperature of the two blocks will be:
Answer:
34.34 °C
Explanation:
From the question,
Heat lost by the silver block = heat gained by the iron block.
cm(x-y) = c'm'(y-z)................... Equation 1
Where c = specific heat capacity of the silver block, m = mass of the silver block, c' = specific heat capacity of the iron, m' = mass of the iron. x = initial temperature of the silver block, z = initial temperature of the iron, y = final temperature of the mixture.
make y the subject of the equation
y = (cmx+c'm'z)/(cm+c'm')............... Equation 2
Given: c = 50 g, c = 0.2350 J/g·°C, x = 100°C, m' = 50 g, c' = 0.4494 J/g.°C, z = 0°C
Substitute these values into equation 2
y = [(50×0.2350×100)+(50×0.4494×0)]/[(50×0.2350)+(50+0.4494)]
y = 1175/(11.75+22.47)
y = 1175/34.22
y = 34.34 °C
A uniform string of length 10.0 m and weight 0.32 N is attached to the ceiling. A weight of 1.00 kN hangs from its lower end. The lower end of the string is suddenly displaced horizontally. How long does it take the resulting wave pulse to travel to the upper end
Answer: 0.0180701 s
Explanation:
Given the following :
Length of string (L) = 10 m
Weight of string (W) = 0.32 N
Weight attached to lower end = 1kN = 1×10^3
Using the relation:
Time (t) = √ (weight of string * Length) / weight attached to lower end * acceleration due to gravity
g = acceleration due to gravity = 9.8m/s^2
Weight of string = 0.32N
Time(t) = √ (0.32 * 10) / [(1*10^3) * (9.8)]
Time = √3.2 / 9800
= √0.0003265
= 0.0180701s
2x+4y
GCF for the polynomial
Answer:
[tex]\boxed{2}[/tex]
Explanation:
The GCF is the greatest common factor of the polynomial.
[tex]2x = 1, \ 2, \ x[/tex]
[tex]4y=1, \ 2, \ 4,\ y[/tex]
The greatest common factor is 2.
Two charged objects are separated by some distance. The charge on the first object is greater than the charge on the second object. How do the forces that the two objects exert on each other compare
Even though the charge on the first object is greater, the forces that the two objects exert on each other are equal
What does Electromagnetic induction mean?
Transformers are of two types: Step up and Step down.
What is step up transformer?
What is step down transformer?
What is the difference between them?
Please I really need help.
Don't answer the question for points if you don't know what it means!
Answer:
Electromagnetic introduction is the production of an electromotive force (voltage) across an electrical conductor in a changing magnetic field.
Step up transformers is a transformer in which the output (secondary) voltage is greater than its input (primary) voltage is called a step-up transformer. The step-up transformer decreases the output current for keeping the input and output power of the system equal.
Step down transformer is a transformer in which the output (secondary) voltage is less than its input (primary) voltage is called a step-down transformer. The number of turns on the primary of the transformer is greater than the turn on the secondary of the transformer.
The difference between them:
A transformer is a static device which transfers a.c electrical power from one circuit to the other at the same frequency, but the voltage level is usually changed. For economical reasons, electric power is required to be transmitted at high voltage whereas it has to be utilized at low voltage from a safety point of view. This increase in voltage for transmission and decrease in voltage for utilization can only be achieved by using a step-up and step-down transformer.
Hopefully this helped.
Two small, identical conducting spheres repel each other with a force of 0.035 N when they are 0.35 m apart. After a conducting wire is connected between the spheres and then removed, they repel each other with a force of 0.055 N. What is the original charge on each sphere? (Enter the magnitudes in C.)
Explanation:
Given:
distance between two sphere =0.35 m
Electrical repel force =0.035 N
Electrical repel force after connecting wire =0.055 N.
The electrical force between the two spheres:
[tex]F=k \frac{q_{1} q_{2}}{r^{2}}[/tex]
The electrical force between the two spheres after the wire is attached and removed:
[tex]F=k \frac{q^{2}}{r^{2}}[/tex]
[tex]q^{2}=\frac{F r^{2}}{k}[/tex]
[tex]q=r \sqrt{\frac{F}{k}}=0.35 \times \sqrt{\frac{0.055}{8.99 \times 10^{9}}}=6.46 \times 10^{-7} \mathrm{C}[/tex]
So the total charge of the two spheres [tex]=2 q=2 \times 6.46 \times 10^{-7}=1.29 \times 10^{-6} \mathrm{C}[/tex]
Then before connecting the wire, one sphere charge was [tex]q[tex] and the charge of the other sphere was [tex]\left(1.29 \times 10^{-6}-q\right)[/tex]
The electrical force between the two spheres before connecting the wire:
[tex]F=k \frac{q\left(1.29 \times 10^{-6}-q\right)}{r^{2}}[/tex]
[tex]q\left(1.29 \times 10^{-6}-q\right)=\frac{F r^{2}}{k}=\frac{0.035 \times(0.35)^{2}}{8.99 \times 10^{9}}=0.348 \times 10^{-12}[/tex]
[tex]-q^{2}+\left(1.29 \times 10^{-6}\right) q-\left(0.348 \times 10^{-12}\right)=0[/tex]
[tex]-q^{2}+\left(1.29 \times 10^{-6}\right) q-\left(0.348 \times 10^{-12}\right)=0[/tex]
HELP me pleaseeee somebody
an object is placed 30cm from a mirror of focal length 15 cm the object is 7.5cm tall. where is the image located? how tall is the image??
Explanation:
It is given that,
Object distance from the mirror, u = -30 cm
Focal length of the mirror, f = +15 cm
Size of the object, h = 7.5 cm
We need to find the image distance and the size of the image.
Mirror's formula, [tex]\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}[/tex]
v is image distance
[tex]\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(15)}-\dfrac{1}{(-30)}\\\\v=10\ cm[/tex]
Let h' is the size of the image. So,
[tex]\dfrac{h'}{h}=\dfrac{-v}{u}\\\\h'=\dfrac{-vh}{u}\\\\h'=\dfrac{-10\times 7.5}{-30}\\\\h'=2.5\ cm[/tex]
So, the image is located at a distance of 10 cm and the size of the image is 2.5 cm.
We have seen that starlight passing through the interstellar medium is dimmed and reddened. Look at the photo of a sunset on Earth. The Sun’s light also appears reddish at sunset. Given your understanding of the reddening of starlight, why do you think sunsets appear red?
Answer:
Explanation:
Reddening of sun's rays at sunset and sunrise is due to scattering of light . The white light consisting of seven colours coming from the sun are scattered in different directions when they fall on the air particles present in atmosphere . Red coloured light scatters least and it travels straight forward to the viewer on the earth . On the other hand other colours scatter most and therefore they go out of area of vision for the viewer on the earth . Since only red colour reaches the eye of the viewer , sun's ray appear red . This happens during sunrise and sunset . It is so because during this period , sun rays travel far greater distance through atmosphere , so scattering is most pronounced .
1. Analogies exist between rotational and translational physical quantities. Identify the rotational term analogous to each of the following: acceleration, force, mass, work, translational kinetic energy, linear momentum, impulse.
2. Explain why centripetal acceleration changes the direction of velocity in circular motion but not its magnitude.
Answer:
1) a α, m I, W=F.d W =τ . θ,
2) a = v²/r
Explanation:
1) The amounts of rotational and translational motion are related
acceleration is
a = d²x / dt²
linear displacement is equivalent to angular rotation, therefore angular acceleration is
α = d²θ / dt²
force in linear motion is equivalent to moment in endowment motion
F = m a
τ = I α
the mass is the inertia of the translation, in rotational motion the moment of inertia is the rotational inertia
I = m r²
Work is defined by W = F. d
in rotation it is defined by W = τ . θ
The linear momentum is p = mv
the angular momentum L = I w
momentum the linear motion is I = F dt
in the rotation it is I = τ dt
2) The velocity is a vector therefore it has modulus and direction, linear acceleration changes the modulus of velocity, whereas circular motion changes the direction (the other element of the vector).
[tex]a_{c}[/tex]Ac = v²/r
Runner 1 has a velocity of 10 m/s west. Runner 2 has a velocity of 7 m/s east. From the frame of reference of runner 2, what is the velocity of runner 1? A.17 m/s east. B.3 m/s east. C.17 m/s west. D.3 m/s west
Answer:
17 m/s west
Explanation:
Runner 1 has velocity = 10 m/s west
runner 2 has velocity = 7 m/s east
From the frame of reference of runner 2, we can imagine runner 2 as standing still, and runner 1 moving away from him, towards the west with their combined velocity of
velocity = 10 m/s + 7 m/s = 17 m/s west
Answer:
17 m/s west
Explanation:
Hope this helps!
Did to gravity, the moon has a much smaller acceleration than earth. How do you think that would affect the period of pendulum?
I think any pendulum would swing slower on the moon than it would on Earth.
The time it takes a pendulum to go through a complete back and forth swing is:
Time period = 2 π √(length/gravity)
You can see that 'gravity' is in the denominator of the fraction, so the smaller gravity gets, the longer the period gets.
To be a little bit more technical, the period is inversely proportional to the square root of gravity.
So the period for a complete swing on the moon would be √(9.8/1.6) times as long as the complete swing of the same pendulum on Earth.
That number is roughly 2.47 .
So, for every 1 second that a pendulum takes to swing back and forth once on Earth, the same pendulum would take 2.47 seconds to do it on the moon.
Answer:
based on my opinion....
as we know that gravity in moon are less than gravity in earth.. since the force of gravity is less on the moon, the pendulum would swing slower at the same length, angle
and the frequency would be less.
I hope this helps