Answer:
maximum shear stress = τ(max) = 95.49 × 10⁶N/m²
Explanation:
given
outer diameter at one end(D₁) = 3.0cm
outer diameter at the other end(D₂) = 4.0cm
inner diameter(d₁) = 1.0cm
torque applied(T) = 500Nm
maximum shear stress will occur at lower outer diameter
the formula is τ/r = T/J
τ= T × r/J
where r is radius
T is the torque
J is the polar 2nd M of area
attached is the calculation of the question
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.
(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
Which reverses the flow of current through
an electric motor?
Answer:
a commutator
Explanation:
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]
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:
How does increasing frequency affect the crests of a wave?
They get higher.
They get closer together.
They get lower.
They get farther apart.
Answer:
they get closer together
Explanation:
Bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving +1.72 m/s. What is the velocity of car B after the collision?
The velocity of car B after the calculation can be obtained as 5.5 m/s.
What is a collision?We have to look at the principle of the conservation of the linear momentum and that is what we need so as to be able to deal with the problem that we have in this question as we have it here.
We know that the total momentum of the system would have to be constant and the implication of this is that the momentum before collision must be the same as the momentum after collision if we have been able to designate the system as a closed system.
Using the formula as we have shown it, we know that;
Momentum before Collison must be equal to momentum after collision
As such we have the following as shown below;
(281 * 2.82) + (209 * 1.72) = (281 * 0) + (209 * v)
792.42 + 359.48 = 209 v
v = 5.5 m/s
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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.
A small child has a wagon with a mass of 10 kilograms. The child pulls on the wagon with a force of 2 newtons. What is the acceleration of the wagon?
Answer:
0.2 m/s²Explanation:
Given,
Mass ( m ) = 10 kg
Force ( f ) = 2 Newtons
Acceleration ( a ) = ?
Now, let's find the acceleration :
We know that,
[tex]f = ma[/tex]
Plug the values
[tex]2 = 10a[/tex]
Swap the sides of the equation
[tex]10a = 2[/tex]
Divide both sides of the equation by 10
[tex] \frac{10a}{10} = \frac{2}{10} [/tex]
Calculate
[tex]a = 0.2 \: {metre \: per \: second \: }^{2} [/tex]
Hope this helps...
Best regards!!
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
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.
Which is the best example of muscular endurance
Answer:
personally I'd say C by do not know if that is the exact answer
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.
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:
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.
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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 .
calculate the magnitude of the electric field intensity in vacuum at a distance of 20 cm from a charge of 5 * 10 raise to power - 3 column
Answer:
1.1259*10^9 Newton per Columb
Explanation:
the magnitude of the electric field intensity can be calculated using the expresion below;
E=Kq/r^2
Where k= constant
q= electric charge
r=distance= 2cm= 20*10^-2m( we convert to m for unit consistency
:,K=59*10^9 Columb
If we substitute the value into above formula we have
E=(9*10^9)*(5*10^-3)/(20*10^-2)^2
=1.1259*10^9 Newton per Columb
Therefore,the magnitude of the electric field intensity in vacuum at a distance of 20 cm is 1.1259*10^9 Newton per Columb
a skydiver jumped out of a plane and fell 9 miles to the ground. his average speed while falling was 174 miles/hour. how much time did the dive last?
Answer:
t = 0.051 seconds
Explanation:
Given that,
A skydiver jumped out of a plane and fell 9 miles to the ground. It means 9 miles is the distance covered.
The average speed of the skydiver is 174 miles/hour
We need to find the time for which he dive last. It means t is the time taken. Total distance covered divided by time taken is called average speed. So,
[tex]v=\dfrac{d}{t}\\\\t=\dfrac{d}{v}\\\\t=\dfrac{9}{174}\\\\t=0.051\ s[/tex]
So, a skydiver dive for 0.051 seconds.
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
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]
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!
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
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
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
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.
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).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!
Calculating heat gained, need help!!!
Q=(m)(c)(triangle T)
trial 1:
Q=(0.4)(4.186 J/kg)(78)=?
trial 2:
Q=(0.6)(4.186 J/kg)(78)=?
trial 3:
Q=(0.45)(4.186 J/kg)(75)=?
Explanation:
q = mCΔT
where m is mass,
C is specific heat capacity,
and ΔT is change in temperature.
Make sure all your units match. Assuming the substance is water, C = 4186 J/kg/°C.
Trial 1:
q = (0.4 kg) (4186 J/kg/°C) (78°C)
q = 130,600 J
Trial 2:
q = (0.6 kg) (4186 J/kg/°C) (78°C)
q = 195,900 J
Trial 3:
q = (0.45 kg) (4186 J/kg/°C) (75°C)
q = 141,300 J
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.
Which of the following object is in dynamic equilibrium?
Answer:
A car driving in a straight line 20 m/s
Explanation:
ayepecks silly