Answer:
Different surfaces
You can see that dull surfaces are good absorbers and emitters of infrared radiation. Shiny surfaces are poor absorbers and emitters (but they are good reflectors of infrared radiationIn young Goodman’s Brown hawthornes reveals his feelings about his Puritan ancestors when
Answer:
In "Young Goodman Brown," Hawthorne reveals his feelings about his Puritan ancestors when the dark man reveals that he helped Brown's forebears persecute others.
Explanation:
hey mate i know it is right so don't worry this is the correct
Answer:
In "Young Goodman Brown," Hawthorne reveals his feelings about his Puritan ancestors when the dark man reveals that he helped Brown's forebears persecute others. so that means the answer is D.
Explanation:
A. brown strives to resist his dark mission
B. faith expresses her anxieties about young brown's departure
C. brown discovers his catechism teacher is on speaking terms with the devil
D. the dark man reveals that he helped brown's forebears persecute others(correct answer)
A student working in the physics laboratory connects a parallel-plate capacitor to a battery, so that the potential difference between the plates is 270 V. Assume a plate separation of d = 1.40 cm and a plate area of A = 25.0 cm^2. When the battery is removed, the capacitor is plunged into a container of distilled water. Assume distilled water is an insulator with a dielectric constant of 80.0
Required:
a. Calculate the charge on the plates (in pC) before and after the capacitor is submerged.
b. Determine the capacitance (in F) and potential difference (in V) after immersion
c. Determine the change in energy (in nJ) of the capacitor
Answer:
a) Q₀ = Q = 4.27 10² pC, b) C = 1.26 10⁻¹⁰ F, ΔV = 3.375 V, c) ΔE = 56.88 nJ
Explanation:
a) the capacitance of a parallel plate capacitor is
C = ε₀ A / d
let's reduce the magnitudes to the SI system
d = 1.40 cm = 0.0140 m
A = 25.0 cm² (1 m / 100 cm) ² = 25.0 10⁻⁴ m²
C = 8.85 10⁻¹² 25.0 10⁻⁴ / 0.0140
C = 1.58 10⁻¹² F
the capacitance is also
C = Q / ΔV
Q = V ΔV
Q = 270 1.58 10⁻¹²
Q = 4.27 10⁻¹⁰ C
When the battery is removed and the capacitor is inserted into the dielectric, the charge should remain the same,
Q₀ = Q = 4.27 10⁻¹⁰ C
let's reduce to pC
Q₀ = 4.27 10⁻¹⁰ C (10¹² pC / 1C)
Q₀ = Q = 4.27 10² pC
b) After being immersed in the fluid of constant k = 80.0
capacitance is
C = k Co
C = 80 1.58 10⁻¹²
C = 1.26 10⁻¹⁰ F
the voltage difference is
ΔV = ΔV₀ / k
ΔV = 270/80
ΔV = 3.375 V
c) We stop the energy before the dive
E₀ = ½ C₀ ΔV₀²
after the dive
E = ½ (k C₀) (ΔV₀/k)²
E = ½ C₀ ΔV₀² / k
E = E₀ / k
the change in energy is
ΔE = E -E₀
ΔE = E₀ / k - E₀
ΔE = E₀ ( [tex]\frac{1}{k} - 1[/tex] )
we calculate
E₀ = ½ 1.58 10⁻¹² 270²
E₀ = 5.76 10⁻⁸ J
ΔE = 5.76 10⁻⁸ ([tex]\frac{1}{80} -1[/tex])
ΔE = 5.688 10⁻⁸ J
we reduce to nJ
ΔE = 5.688 10⁻⁸ J (10⁹ nJ / 1J)
ΔE = 56.88 nJ
How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole
Answer: hello your question is incomplete below is the complete question
Water stands at a depth H in a large open tank whose side walls are vertical . A hole is made in one of the walls at a depth h below the water surface. Part B How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole
answer :
At Height ( h ) from the bottom of Tank
Explanation:
Determine how far above the bottom of the tank a second hole be cut
For the second hole to have the same range as the first hole
Range of first hole = Velocity of efflux of water * time of fall of water
= √ (2gh) * √( 2g (H - h) / g)
= √ ( 4(H-h) h)
Hence the Height at which the second hole should be placed to exercise same range of stream emerging = h from the bottom of the Tank
A ship is flying away from Earth at 0.9c (where c is the speed of light). A missile is fired that moves toward the Earth at a speed of 0.5c relative to the ship. How fast does the missile move relative to the Earth
Answer:
the required speed with which the missile move relative to the Earth is -0.727c
Explanation:
Given the data in the question;
relative velocity relation;
u' = u-v / 1 - [tex]\frac{uv}{c^2}[/tex]
so let V[tex]_B[/tex] represent the velocity as seen by an external reference frame; u=V[tex]_B[/tex]
and let V[tex]_A[/tex] represent the speed of the secondary reference frame; v=V[tex]_A[/tex]
hence, u' is the speed of B as seen by A
so
u' = V[tex]_B[/tex]-V[tex]_A[/tex] / 1 - [tex]\frac{V_BV_A}{c^2}[/tex]
now, given that; V[tex]_A[/tex] = 0.9c and V[tex]_B[/tex] = 0.5c
we substitute
u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{(0.5c)(0.9c)}{c^2}[/tex]
u' = ( 0.5c - 0.9c ) / 1 - [tex]\frac{c^2(0.5)(0.9)}{c^2}[/tex]
u' = ( 0.5c - 0.9c ) / 1 - (0.5 × 0.9)
u' = ( -0.4c ) / 1 - 0.45
u' = -0.4c / 0.55
u' = -0.727c
Therefore, the required speed with which the missile move relative to the Earth is -0.727c
why the walls of tyres becomes warm as the car moves
Answer:
the particles vibrate inside the tyre
Explanation:
as the car moves kinetic energy is transfered in the tyres which causes the particles to vibrate inside the tyre so the kinetic store is. transferred into thermal
A car moves at a constant speed of 90km/h from a starting point. Another car moves at 70km/h after 2hours from the same starting point. if both cars moves in the same direction, after how many hours will the distance between the first car and the second car to be 40 km.
Answer:
400
Explanation:
Three children are riding on the edge of a merry-go-round that is 182 kg, has a 1.60 m radius, and is spinning at 15.3 rpm. The children have masses of 17.4, 28.5, and 32.8 kg. If the child who has a mass of 28.5 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm
Answer:
The new angular velocity of the merry-go-round is 18.388 revolutions per minute.
Explanation:
The merry-go-round can be represented by a solid disk, whereas the three children can be considered as particles. Since there is no external force acting on the system, we can apply the principle of angular momentum conservation:
[tex]\left(\frac{1}{2}\cdot M + m_{1}+m_{2} + m_{3} \right)\cdot R^{2}\cdot \dot n_{o} = \left(\frac{1}{2}\cdot M + m_{1} + m_{3})\cdot R^{2}\cdot \dot n_{f}[/tex] (1)
Where:
[tex]M[/tex] - Mass of the merry-go-round, in kilograms.
[tex]m_{1}[/tex], [tex]m_{2}[/tex], [tex]m_{3}[/tex] - Masses of the three children, in kilograms.
[tex]R[/tex] - Radius of the merry-go-round/Distance of the children with respect to the center of the merry-go-round, in meters.
[tex]\dot n_{o}[/tex], [tex]\dot n_{f}[/tex] - Initial and final angular speed, in revolutions per minute.
If we know that [tex]M = 182\,kg[/tex], [tex]m_{1} = 17.4\,kg[/tex], [tex]m_{2} = 28.5\,kg[/tex], [tex]m_{3} = 32.8\,kg[/tex], [tex]R = 1.60\,m[/tex] and [tex]\dot n_{o} = 15.3\,\frac{rev}{min}[/tex], then the final angular speed of the system is:
[tex]\dot n_{f} = \dot n_{o}\cdot \left(\frac{\frac{1}{2}\cdot M + m_{1} + m_{2} + m_{3} }{\frac{1}{2}\cdot M + m_{1} + m_{3} } \right)[/tex]
[tex]\dot n_{f} = \left(15.3\,\frac{rev}{min} \right)\cdot \left[\frac{\frac{1}{2}\cdot (182\,kg) + 17.4\,kg +28.5\,kg + 32.8\,kg }{\frac{1}{2}\cdot (182\,kg) + 17.4\,kg + 32.8\,kg } \right][/tex]
[tex]\dot n_{f} = 18.388\,\frac{rev}{min}[/tex]
The new angular velocity of the merry-go-round is 18.388 revolutions per minute.
On the basis of following diagram/picture answer the questions given: The diagram shows a vacuum flask and an enlarged view of a section through the flask wall. The main reason for the silvering is to reduce heat transfer by
(a) conduction only
(b) conduction and convection
(c) radiation only
(d) radiation and convection
A digital signal differs from an analog signal because it a.consists of a current that changes smoothly. b. consists of a current that changes in pulses. c.carries information. d. is used in electronic devices.
Answer:
d.it is used in electronic devices
Object A is 8 kg and at rest and Object B is 16 kg and moving at 10 m/s to the Right. If Object B hits Object A and Object B is at rest after the collision, what is the velocity of Object A after the collision? Why?
Answer:
Initially...
Object A is at rest(v=0)
With Mass =8kg
Object B moving at 10ms- and Mass of 16kg.
Now After collision(Elastic)... B Came to rest...Meaning its final vel =0
But its mass remains the same
What velocity would A be moving with?
From Conversation of Momentum
M'U' + M"U"= M'V' + M"V"
8(0) + (16)(10) = 8V' + 16(0)
0 + 160 = 8V' + 0
8V'= 160
V'= 160/8
V'=20ms-¹.
A would be Moving at 20ms-¹ after collision.
N 4. Which of the following can cause a short circuit?
Answer:
pretty sure its A
Explanation:
please give brainliest if i'm correct
Answer:
A.
Explanation:
I have done this and that was correct
hope this helps
What is 3*10^-6 divided by 2.5*10^6 expressed in standard notation?
Answer:
1.2 x 10^-12
Explanation:
3/2.5 x 10^-6/10^6
1.2 x 10^-6 x 10^-6
1.2 x 10^-12
A firecracker breaks up into three pieces, one of which has a mass of 200 g and flies off along the x-axis with a speed of 82.0 m/s. A second piece has a mass of 300 g and flies off along the y-axis with a speed of 45.0 m/s. (a) What is the magnitude of the momentum for the third piece
Answer:
[tex]|P_3|=21.242m/s[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=200g=0.2[/tex]
Speed [tex]v=82.0m/s[/tex]
Mass [tex]M_2=300g=0.3kg[/tex]
Speed [tex]V_2=45.0m/s[/tex]
Generally the equation for Magnitude of the Third piece is mathematically given by
[tex]P_1+P_2+P_3=0[/tex]
[tex]|P_3|=|P_1+P_2|[/tex]
Where
[tex]P_1=m_1v_1[/tex]
[tex]P_1=0.2*82[/tex]
[tex]P_1=16.4kgm/s[/tex]
And
[tex]P_2=0.3*45[/tex]
[tex]P_2=13.5kgm/s[/tex]
Therefore
[tex]|P_3|=\sqrt{16.4^2+13.5^2}[/tex]
[tex]|P_3|=21.242m/s[/tex]
The tendon from the flexor hallucis longus muscle in the hindlimb of a turkey (Meleagris gallopavo) has a resilience of 0.93. If 9.00 J of work are done on the tendon to stretch it out, how many Joules of work does the tendon do as it is relaxing?
Answer:
8.37 Joules
Explanation:
The amount of energy that a substance, such as animal tissue or rubber can take and yet return to its original condition is known as resilience. When we stress such a substance and then let it restore to its previous state, we are referring to the substance to maintain its elastic area of the stress-strain curve.
From the given information:
the resilence is 0.93
The amount of work done during stretching of the tendon = 9.00 J
Thus,
the work done when relaxing = 0.93 × 9.00 J
= 8.37 Joules
What is Magnetic field
The area around a magnet in which it's magnetic force can ne experienced.
➛ Magnetic field has both magnitude and direction.
➛ Magnetic field can be described with help of a magnetic compass.
➛ It's SI unit is Tesla( T ) .
Explanation:
An area around a magnet or materials that behave like a magnet, where there is a force that will attract some metals towards it.
[tex] \: [/tex]
As shown in Fig. 4, an ideal gas of monatomic molecules expands from its initial state A to a state B through an isobaric process and then further expands to a volume C. Find the work done by the gas, increase in internal energy, and the energy transferred by heat to the gas over the whole process
Why do Hubble Space Telescope images show what objects looked like long ago?
Answer:
when we get images from the telescope, we're not seeing those object like they are right now. we're seeing them like they looked when that light left those distant objects...... the animation cuts to a visualization of distant galaxies imaged by Hubble.
A man weighing 720 N and a woman weighing 500 N have the same momentum. What is the ratio of the man's kinetic energy Km to that of the woman Kw?
Answer:
Because weight W = M g, the ratio of weights equals the ratio of masses.
(M_m g)/ (M_w g) = [ (p^2 Man )/ (2 K_man)] / [ (p^2 Woman )/ (2 K_woman)
but p's are equal, so
K_m/K_m = (M_w g)/(M_m g) = W_woman / W_man = 450/680 = 0.662Explanation:
When hydrogen peroxide antiseptic is tightly capped and left on the shelf for a long period of time, the bottle is observed to swell. When opened, it “hisses” noticeably. Explain these observations. Is the antiseptic likely to be as effective?
Answer:
The hydrogen peroxide decomposes as: [tex]H_2O_2\rightarrow 2H_2O+O_2[/tex]
Explanation:
Hydrogen peroxide antiseptic although tightly capped but left on the shelf for a long time leads to swollen bottle and opens with a hissing sound because hydrogen peroxide reacts with air in presence of light to dissociate into water and oxygen molecules that leads to high volume of content in the tightly closed bottle and formation of water molecules from hydrogen peroxide.
Once the hydrogen peroxide has turned into water molecules it is no longer effective as an antiseptic.
the unit of area is called a derived unit.why?
Explanation:
the unit of area is called a derived unit because it is made of two fundamental unit metre and metre.
The efficiency of a machine can be increased by
Explanation:
the efficiency of a machine can be increased by reducing the friction
please mark the brainliest
Necesito ayudaaaaaa por favor
MAnswer:
Explanation:
Three people are trying to move a box. Which set of forces will result in a net
force on the box of 20 N to the left?
Answer:
yes the answer is D as the two people if we calculate
Answer:
A
Explanation:
correct on A P E X
A light year is a unit that measures time
Answer:
Light-year is the distance light travels in one year. Light zips through interstellar space at 186,000 miles (300,000 kilometers) per second and 5.88 trillion miles (9.46 trillion kilometers) per year.
Explanation:
Question 1 of 15
Tamia goes on a run every moming in the summer to stay in shape for
soccer in the fall She runs 25 miles in 20. minutes. What is her average
speed in miles per hour
O A 7 mph
B. 8 mon
O c. 5.5 mph
OD 75 mpin
Answer:
d 75 miles per hour
Explanation:
60 ÷ 20
3 × 25 = 75
hoped that helped!!
Un recipiente contiene 224 dm3 de Ozono de masa 4.561 Kg a 51.09 grados celsius. Calcula la presión del Ozono
Answer:
Por lo tanto, la presión del ozono es:
[tex]P=0.011\: atm[/tex]
Explanation:
Podemos usar la ecuacion de los gases ideales;
[tex]PV=nRT[/tex] (1)
Tenemos:
El volumen V = 224 dm³ = 224 L
La temperatura T = 51.09 C = 324.09 K
La masa es m = 4.561 kg
Lo necesitamos ahora es calvular n que es el numero de moles;
recordemos que el peso molecular del ozono M = 48 g/mol.
[tex]n=\frac{m}{M}=\frac{4.561}{48}=0.095\: mol[/tex]
Finalmente, usando la ecuacion 1 despejamos la presion P
[tex]P=\frac{nRT}{V}[/tex]
[tex]P=\frac{0.095*0.082*324.09}{224}[/tex]
Por lo tanto, la presion del ozono es:
[tex]P=0.011\: atm[/tex]
Espero te haya ayudado!
A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying pan is only 0.150 N. Knowing the coefficient of kinetic friction between the two materials (0.04), he quickly calculates the normal force. What is it (in N)
Answer:
[tex]f_n=3.75N[/tex]
Explanation:
From the question we are told that:
Frictional force [tex]F=0.150N[/tex]
Coefficient of kinetic friction [tex]\mu=0.04[/tex]
Generally the equation for Normal for is mathematically given by
[tex]f_n=\frac{F}{\mu}[/tex]
Therefore
[tex]f_n=\frac{0.150}{0.04}[/tex]
[tex]f_n=3.75N[/tex]
A 2 kg object traveling at 5 m s on a frictionless horizontal surface collides head-on with and sticks to a 3 kg object initially at rest. Which of the following correctly identifies the change in total kinetic energy and the resulting speed of the objects after the collision? Kinetic Energy Speed
(A) Increases 2 m/s 3.2 m/s
(B) Increases Soold 2 m/s
(C) Decreases 3.2 m/s
(D) Decreases
Answer: (d)
Explanation:
Given
Mass of object [tex]m=2\ kg[/tex]
Speed of object [tex]u=5\ m/s[/tex]
Mass of object at rest [tex]M=3\ kg[/tex]
Suppose after collision, speed is v
conserving momentum
[tex]\Rightarrow mu+0=(m+M)v\\\\\Rightarrow v=\dfrac{2\times 5}{2+3}\\\\\Rightarrow v=2\ m/s[/tex]
Initial kinetic energy
[tex]k_1=\dfrac{1}{2}\times 2\times 5^2\\\\k_1=25\ J[/tex]
Final kinetic energy
[tex]k_2=\dfrac{1}{2}\times (2+3)\times 2^2\\\\k_2=10\ J[/tex]
So, it is clear there is decrease in kinetic energy . Thus, energy decreases and velocity becomes 2 m/s.
There are two substances, 1 and 2, that diffuse through a medium. The substances have diffusion constants D1 and D2, and D1 > D2. Which substance will spread out farther over a 10 s time interval?
Answer:
D1
Explanation:
Higher diffusion constants mean that the substance diffuses more in a given time. Because D1>D2, D1 should diffuse more.
A strong trough in a Rossby wave occurs when the jet stream A. bends towards the Equator. B. bends toward the poles. C. does not bend but maintains an east to west flow. D. does not bend but maintains a west to east flow.
Answer:
A. bends towards the Equator.
Explanation:
Rossby waves are inertial waves that are naturally occurring in a rotating fluids. These waves are also called as the planetary waves.
The Rossby waves are undulated that occur in the polar front jet stream when there is a significant differences in the temperatures between the polar and the tropical air masses.
It occurs when the polar air masses moves towards the equator and when the tropical air masses moves towards the pole. It is formed when the air bends away from the poles and bends towards the equator.
Hence the correct option is (A).
