Answer:
The displacement is constant when the rate of change in velocity is constant, for example, if a car’s speed increases by 5 km every 5 minutes it is called constant displacement
A certain heating element is made out of Nichrome wire and used with the standard voltage source of V=120 V. Immediately after the voltage is turned on, the current running through the element is measured at I1=1.28 A and its temperature at T1=25°C. As the heating element warms up and reaches its steady-state (operating) temperature, the current becomes I2=1.229 A.
Required:
Find this steady-state temperature T2.
Answer:
T₁ = 232.5 ºC
Explanation:
For this exercise let's start by finding the value of the resistance for the two currents, using Ohm's law
V = i R
R = V / i
i₀ = 1.28 A
R₀ = 120 / 1.28
R₀ = 93.75 ohm
i₁ = 1.229 A
R₁ = 120 / 1.229
R₁ = 97.64 or
Resistance in a metal is linear with temperature
ΔR = α R₀ ΔT
where the coefficient of thermal expansion for Nichrome is α=0.0002 C⁻¹
ΔT = [tex]\frac{\Delta R}{\alpha R_o}[/tex]
ΔT = [tex]\frac{97.64 \ -93.75}{ 0.00020 \ 93.75}[/tex]
ΔT = 2,075 10² C
ΔT = T₁-T₀ = 2,075 10²
T₁ = T₀ + 207.5
T₁ = 25+ 207.5
T₁ = 232.5 ºC
2. Do you expect the resistance of a light bulb to remain constant as the current through it is increased and the filament goes from red-hot to white-hot? Explain why or why not.
Answer:
Remember that as the temperature of an object increases, the kinetic energy of the particles inside it also increases.
So, if the temperature of a wire increases, the kinetic energy of the particles inside increases. Then when the electrons try to flow through the wire, the probability of a collision is increased (then the resistance increases). Thus, if the filament goes from red-hot to white-hot (so the temperature of the wire increases) we can conclude that the resistance that the current experiences also increases.
So no, we can not expect the resistance of the light bulb to remain constant as the filament goes from red-hot to white-hot.
What is the current in milliamperes produced by the solar cells of a pocket calculator through which 9.00 C of charge passes in 8.50 h
Answer:
Current = 0.000294 A
Explanation:
Below is the given values:
Given the charge = 9.00 C
Time = 8.50 h
Use the below formula to find the current:
Current = Q / t
Now plug the values:
Current = 9 / (8.5 x 3600)
Current = 0.000294 A
20 raindrops fall on an umbrella. The following day, 10 hail stones fall on the umbrella. All drops of rain and hail are of equal mass m and are traveling at the same velocity v. Both of these (rain and hail) fall on the umbrella in 1 second. What can we say about the Force exerted on the umbrella to hold it up
Answer:
the force exerted by the umbrella against the drops is the same in both cases
Explanation:
In this exercise the total mass of the raindrops and the hail are equal, therefore the momentum that these drops create is
raindrops I₁ = ∫ F dt
hail I₂ = ∫ F dt
since the mass and the velocity is the same the force is the weight of the drops
F = mg
whereby the impulse is
I₁ = mg t
I₂ = mg t
we can see that the impulses generated by hail and raindrops are equal, therefore the force exerted by the umbrella against the drops is the same in both cases
A glass block in air has critical angle of 49. What will happen to a ray of light coming through the glass when it is incident at and angle of 50 at the glass air boundary? Illustrate with a diagram
Answer:
b
Explanation:
Reflected waves change their wavelength by ______ when reflected.
A. ¾
B. ½
C. They don’t change
D. ¼
Answer:
i think its c
Explanation:
trust me
pha của dao động làm hàm
Answer:
pha của dao động là hàm bậc nhất của thời gian.
A 900 kg car travelling at 12 m/s due east collides with a 600 kg car travelling at 24 m/s due north. As a result of the collision, the two cars lock together and move in what final direction?
45.0° N of E
53.1° N of E
63.3° N of E
69.5° N of E
Calculate force of each car:
P1 = 900 kg x 12m/s = 10,800
P2= 600 x 24 = 14,400
Degree of travel = arctan(14,300/10800)
Degree of travel = 53.1 N of E
If four students separately measure the density of a rock, and they all have very low percent
differences between their measurements, what can you say for certain about the accuracy of their
results?
Answer:
Their measured results are closer to the exact or true value. Hence, their measured value is considered to be more accurate.
Explanation:
Considering the situation described above, the accuracy of a measured value depicts how closely a measured value is to the accurate value.
Hence, since the students' measured values have very low percent differences, it shows the similarity of computations or estimates to the actual values, which in turn offers a smaller measurement error.
Therefore, their measured results are closer to the exact or true value, which implies that their measured value is considered to be more accurate.
A uniform magnetic field is directed at an angle of 30o with the plane of a circular coil of radius 4cm and 1000 turns. The magnetic field changes at a rate of 5 T per second. Calculate the following:
a. Area vector
b. Induced emf
Answer:
(a) The area vector is 0.00503 m² at 30⁰ from the magnetic field
(b) The induced emf is 12.58 V
Explanation:
Given;
angle between the magnetic field and the plane of the circular coil, = 30⁰
number of turns of the coil, N = 1000
radius of the coil, r = 4 cm = 0.04 m
change in the magnetic field with time, dB/dt = 5 T/s
(a) The area vector is calculated as;
A = πr²
A = π x (0.04)²
A = 0.00503 m²
The area vector is 0.00503 m² at 30⁰ from the magnetic field.
(b) The induced emf is calculated as;
[tex]emf = N\frac{\Delta \phi}{\Delta t} \\\\where;\\\\\phi = BAcos \theta\\\\emf = N\times \frac{dB}{dt} \times Acos (\theta)\\\\where;\\\\\theta \ is \ the \ angle \ between \ a \ perpendicular \ vector \ to \ the \ area\\ and\ the \ magnetic\ field\\\\\theta = 90 - 30 = 60^0\\\\emf = N\times \frac{dB}{dt} \times Acos (\theta)\\\\emf = (1000) \times (5 )\times (0.00503) \times cos (60)\\\\emf = 12.58 \ V[/tex]
The average 8-18 year old spends how many hours per day average in front of a screen doing little physical activity
Nearly four hours every day, doing little to no physical activity.
in an experiment the following readings were observed volume of alcohol flowing per minute equals to 10 raise to power - 5 cube volume density of alcohol is 800 kg per metre cube length of cube is 0.5 m radius of tube is 0.05 cm height of alcohol is 60 cm calculate the coefficient of viscosity of alcohol
Answer:
The viscosity is 1.30 x 10^-3 deca poise.
Explanation:
Volume per minute, V = 10^-5 m^3
Volume per second, V = 1.67 x 10^-7 m^3
density, d = 800 kg/m^3
radius, r = 0.05 cm
Length, L = 0.5 m
Height, h = 60 cm
Pressure, P = h d g = 0.6 x 800 x 9.8 = 4704 Pa
Use the formula of rate of flow
[tex]V = \frac{\pi p r^4}{8\eta L}\\\\1.67\times 10^{-7}\times8\times \eta\times 0.5 = 3.14\times 4707\times (0.05\times 10^{-2})^4\\\\\eta = 1.38\times 10^{-3} deca poise[/tex]
Wind instruments like trumpets and saxophones work on the same principle as the "tube closed on one end" that we examined in our last experiment. What effect would it have on the pitch of a saxophone if you take it from inside your house (76 degrees F) to the outside on a cold day when the outside temperature is 45 degrees F?
Answer:
The correct answer is - low pitch
Explanation:
Now for the case it is mentioned that the tube closed on one end frequency is:
f = v/2l
Where,
l = length of the tube
v = velocity of longitudinal wave of gas filled in the tube
if frequency increases then pitch will be increase as well as pitch depends on frequency.
Now increase with the temperature the density of the gas decreases and velocity v is inversely proportional to density of gas so velocity increases. So if there is an increase in frequency so pitch also increases.
As the temperature inside the house is at 750 F more than outsideat 450 Fso pitch is more inside and the pitch is low outside.
An inductor has an inductance of 0.053 H. The voltage across this inductor is 48 V and has a frequency of 270 Hz. What is the current in the inductor?
Answer:
3.004A
Explanation:
The voltage across an inductor is expressed as:
[tex]V=IX_L\\V=I(2\pi fL)[/tex]
Given
f = 270Hz
V = 48V
L =0.053 H
Get the current
270 = I (2*3.14*270*0.053)
270 = 89.8668I
I = 270/89.8668
I = 3.004A
Hence the current in the inductor is 3.004A
In order to create a larger induced electric field, one could... change the magnetic field more slowly. use a larger static magnetic field. change the magnetic field more quickly.
Answer:
change the magnetic field more quickly.
Explanation:
An electric field is defined as a region or an field which surrounds an [tex]\text{electrically-charged particle}[/tex] where it exerts force on all other particles that are charged present in the field or the region.
A magnetic field is defined as the physical field created by a magnet or by a moving charge which creates a force on another moving charge.
The strength of the induced electric field depends on the magnetic field. The change in the magnetic flux induces the electric field. As a result, for creating a larger induced electric field one have change the magnetic field quickly.
The index of refraction for a vacuum is 1.00000. The index of refraction for air is 1.00029. 1) Determine the ratio of time required for light to travel through 1000 m of air to the time required for light to travel through 1000 m of vacuum. (Express your answer to six significant figures.)
Answer:
[tex]\frac{t_{air}}{t_{vaccum}}[/tex] = 1.00029
Explanation:
The refractive index is defined
n = c / v
v = c / n
the speed of light per se wave is constant, so we can use the relations of uniform motion
v = x / t
t = x / v
we substitute
t = x n / c
let's calculate the time
vacuum
t₁ = 1000 1/3 10⁸
t₁ = 3.333333 10⁻⁶ s
air
t₂ = 1000 1.00029 / 3 10⁸
t2 = 3.3343 10⁻⁶ s
the relationship between these times is
t₂ / t₁ = 3.3343 / 3.3333333
t₂ / t₁ = 1.00029
hi can anyone pls answer this question. i will mark brainliest
Answer:
a.work done by man A is zero
D is best option
Explanation:
D none of them because as we know that to do work some distance should be cover with some load as both picture they are covering some distance with carrying some load so A and B option are absolutely wrong and remaining C they are not doing same amount of work because distance cover by them and load carrying by them is different so how can work be same so D is best option none of A B C is correct answer
A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.500 rev/s. What is its angular velocity after a 22.0-kg child gets onto it by grabbing its outer edge
Answer:
I think it is of science is it true na i knew it bro dont take tension
Because of the internal resistance of a real battery, we know that the potential difference between the terminals:___________
a. decreases as more current is drawn.
b. increases as more current is drawn.
c. is greater than the emf of the battery
d. increases as the internal resistance increases.
Answer: I think the ans is A but I’m not sure.
Can someone check it and tell me I would like to know
Which has a total mass of 1612 kg. If she accelerates from rest to a speed of 12.87 m/s in 3.47 s, what is the minimum power required of the engine?
Answer:
76969.29 W
Explanation:
Applying,
P = F×v............. Equation 1
Where P = Power, F = force, v = velocity
But,
F = ma.......... Equation 2
Where m = mass, a = acceleration
Also,
a = (v-u)/t......... Equation 3
Given: u = 0 m/s ( from rest), v = 12.87 m/s, t = 3.47 s
Substitute these values into equation 3
a = (12.87-0)/3.47
a = 3.71 m/s²
Also Given: m = 1612 kg
Substitute into equation 2
F = 1612(3.71)
F = 5980.52 N.
Finally,
Substitute into equation 1
P = 5980.52×12.87
P = 76969.29 W
An eagle is flying horizontally at a speed of 2.60 m/s when the fish in her talons wiggles loose and falls into the lake 4.70 m below. Calculate the velocity (in m/s) of the fish just before it hits the water. (Assume that the eagle is flying in the x direction and that the y direction is up.)
Answer:
Explanation:
The fish will have horizontal velocity of 2.6 m/s which is also the velocity of eagle. Additionally , he will have vertical velocity due to fall under gravity .
v² = u² + 2 g H .
v² = 0 + 2 x 9.8 x 4.7 m
= 92.12
v = 9.6 m /s
The fish's final velocity will have two components
vertical component = 9.6 m/s downwards
Horizontal component = 2.6 m /s .
Resultant velocity = √ ( 9.6² + 2.6² )
= √ ( 92.16 + 6.76 )
= 9.9 m /s
Answer:
The speed of fish at the time of hitting the surface is 9.95 m/s.
Explanation:
Horizontal speed, u = 2.6 m/s
height, h = 4.7 m
Let the vertical velocity at the time of hitting to water is v.
Use third equation of motion
[tex]v^2 = u^2 - 2 gh \\\\v^2 = 0 + 2 \times 9.8\times 4.7\\\\v = 9.6 m/s[/tex]
The net velocity with which the fish strikes to the water is
[tex]v' = \sqrt{9.6^2 + 2.6^2 }\\\\v' = 9.95 m/s[/tex]
Help!!! I need it today
Thank you in advance
Answer:
F = - k (x-xo) a graph of the weight or applied force against the elongation obtaining a line already proves Hooke's law.
Explanation:
The student wants to prove hooke's law which has the form
F = - k (x-xo)
To do this we hang the spring in a vertical position and mark the equilibrium position on a tape measure, to simplify the calculations we can make this point zero by placing our reference system in this position.
Now for a series of known masses let's get them one by one and measure the spring elongation, building a table of weight vs elongation,
we must be careful when hanging the weights so as not to create oscillations in the spring
we look for the mass of each weight
W = mg
m = W / g
and we write them in a new column, we make a graph of the weight or applied force against the elongation and it should give a straight line; the slope of this line is sought, which is the spring constant.
The fact of obtaining a line already proves Hooke's law.
A 70-turn coil has a diameter of 11 cm. Find the magnitude of the emf induced in the coil (in V) if it is placed in a spatially uniform magnetic field of magnitude 0.70 T so that the face of the coil makes the following angles with the magnetic field, and the magnetic field is reduced to zero uniformly in 0.2 s.
This question is incomplete, the complete question is;
A 70-turn coil has a diameter of 11 cm. Find the magnitude of the emf induced in the coil (in V) if it is placed in a spatially uniform magnetic field of magnitude 0.70 T so that the face of the coil makes the following angles with the magnetic field, and the magnetic field is reduced to zero uniformly in 0.2 s. a) θ = 30° V b) θ = 60° V c) θ = 90° V
Answer:
the magnitude of the emf induced in the coil are;
a)- For θ = 30°, e = 1.16 V
b)- For θ = 60°, e = 2.01 V
c)- For θ = 90°, e = 2.33 V
Explanation:
Given the data in the question;
number of turns N = 70
Diameter of coil D = 11 cm
Radius r = D/2 = 11/2 = 5.5 cm = 0.055 m
magnitude of magnetic ΔB = 0.7T
Δt = 0.2 seconds
Now,
a)
For θ = 30°,
Angle of with area of vector θ' = 90° - 30° = 60°
so
emf e = N( Δ∅ / Δt ) = N( ΔBAcosθ / Δt )
hence
e = NAcosθ'(ΔB / Δt )
where A is area ( πr² )
so we substitute
e = 70 × πr² × cos(60°) × ( 0.7 / 0.2 )
e = 70 × π(0.055)² × cos(60°) × ( 0.7 / 0.2 )
e = 1.16 V
b)
For θ = 60°,
Angle of with area of vector θ' = 90° - 60° = 30°
so
e = NAcosθ'(ΔB / Δt )
we substitute
e = 70 × πr² × cos(30°) × ( 0.7 / 0.2 )
e = 70 × π(0.055)² × cos(30°) × ( 0.7 / 0.2 )
e = 2.01 V
c)
For θ = 90°,
Angle of with area of vector θ' = 90° - 90° = 0°
so
e = NAcosθ'(ΔB / Δt )
we substitute
e = 70 × πr² × cos(0°) × ( 0.7 / 0.2 )
e = 70 × π(0.055)² × cos(30°) × ( 0.7 / 0.2 )
e = 2.33 V
Therefore, the magnitude of the emf induced in the coil are;
a)- For θ = 30°, e = 1.16 V
b)- For θ = 60°, e = 2.01 V
c)- For θ = 90°, e = 2.33 V
Two resistors with resistance values of 4.5 Ω and 2.3 Ω are connected in series or parallel
across a 30 V potential difference to a light bulb.
a. Calculate the current delivered through the light bulb in the two cases.
b. Draw the circuit connection that will achieve the brightest light bulb.
Explanation:
Given that,
Two resistors 4.5 Ω and 2.3 Ω .
Potential difference = 30 V
When they are in series, the current through each resistor remains the same. First find the equivalent resistance.
R' = 4.5 + 2.3
= 6.8 Ω
Current,
[tex]I=\dfrac{V}{R'}\\\\I=\dfrac{30}{6.8}\\\\=4.41\ A[/tex]
So, the current through both lightbulb is the same i.e. 4.41 A.
When they are in parallel, the current divides.
Current flowing in 4.5 resistor,
[tex]I_1=\dfrac{V}{R_1}\\\\=\dfrac{30}{4.5}\\\\I_1=6.7\ A[/tex]
Current flowing in 2.3 ohm resistor,
[tex]I_2=\dfrac{V}{R_2}\\\\=\dfrac{30}{2.3}\\\\I_2=13.04[/tex]
In parallel combination, are brighter than bulbs in series.
A 61.1-kg person, running horizontally with a velocity of +2.16 m/s, jumps onto a 16.1-kg sled that is initially at rest. (a) Ignoring the effects of friction during the collision, find the velocity of the sled and person as they move away. (b) The sled and person coast 30.0 m on level snow before coming to rest. What is the coefficient of kinetic friction between the sled and the snow?
Answer:
(a) v = 1.71 m/s
(b) μ = 0.005
Explanation:
(a)
Using the law of conservation of the momentum:
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\[/tex]
where,
m₁ = mass of person = 61.1 kg
m₂ = mass of sled = 16.1 kg
u₁ = initial speed of the person = 2.16 m/s
u₂ = initial speed of the sled = 0 m/s
v₁ = v₂ = v = final speeds of both the person and the sled = ?
Therefore,
[tex](61.1\ kg)(2.16\ m/s)+(16.1\ kg)(0\ m/s)=(61.1\ kg)(v)+(16.1\ kg)(v)\\\\v = \frac{131.976\ kgm/s}{77.2\ kg}[/tex]
v = 1.71 m/s
(b)
The kinetic energy lost by the sled must be equal to the frictional energy:
K.E = fd
[tex]\frac{1}{2}mv^2=\mu Rd = \mu Wd\\\\\frac{1}{2}mv^2=\mu mgd\\\\\frac{1}{2}v^2=\mu g\\\\\mu = \frac{v^2}{2gd}[/tex]
where,
μ = coefficient of kinetic friction = ?
d = distance covered = 30 m
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]\mu = \frac{(1.71\ m/s)^2}{(2)(9.81\ m/s^2)(30\ m)}[/tex]
μ = 0.005
Consider an electromagnetic wave propagating through a region of empty space. How is the energy density of the wave partitioned between the electric and magnetic fields?
1. The energy density of an electromagnetic wave is 25% in the magnetic field and 75% in the electric field.
2. The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
3. The energy density of an electromagnetic wave is entirely in the magnetic field.
4. The energy density of an electromagnetic wave is 25% in the electric field and 75% in the magnetic field.
5. The energy density of an electromagnetic wave is entirely in the electric field
Answer:
Option (2) is correct.
The energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
Explanation:
An electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.
The energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields.
So, the correct option is (2).
The energy density is equally distributed among the magnetic field and electric field. Hence, option (2) is correct.
The given problem is based on the concept and fundamentals of electromagnetic waves. The waves created as a result of vibrations between an electric field and a magnetic field is known as Electromagnetic waves.
In other words, an electromagnetic waves are the waves which are produced when the oscillating electric and magnetic field are interact each other perpendicular to each other. The direction of propagation of electro magnetic waves is perpendicular to each electric and magnetic fields.
Also, the energy associated with the electromagnetic waves is equally distributed in form of electric and magnetic fields. So, the energy density of an electromagnetic wave is equally divided between the magnetic and electric fields.
Thus, we can conclude that the energy density is equally distributed among the magnetic field and electric field.
Learn more about the electromagnetic waves here:
https://brainly.com/question/25559554
When 93 is written in simplest radical form, which value remains under the radical?
Explanation:
plz brainliest me
Answer:
I think it's 9 but edge says it's not
Explanation:
it should be 3√9^2
omparing Technological Design and Scientific Investigation
Explain how the stages of technological design are similar to the process of scientific investigation. Use specific stages from each process to support your answer
Answer:
杰恩斯克克斯克奇沃伊斯克克斯克什德布德克什恩克恩德恩克恩茨克杰兹姆克斯恩斯姆斯姆德恩德姆德武伊乔奥斯克斯杰德布德赫德夫赫富伊什杰吉迪赫德赫夫赫德
Answer:
Scientists learn more about the natural world through investigation by identifying a problem, researching related information, designing and conducting an investigation, analyzing the results, evaluating the conclusion, and communicating the findings. Engineers follow similar steps when creating new products or solutions through technological design. The four stages of technological design include identifying a need, designing and implementing a solution, and evaluating the solution.
NEED AN ANSWER QUICKLY PLEASE!!
If the length and number of turns of a solenoid are doubled strength of magnetic field will :
(a) Be doubled (b) become half (c) not change d) be four time
Answer:
c). It wouldn't change.
Explanation:
[tex]{ \bf{F = \frac{ \ \gamma _{o}NI }{l} }}[/tex]
A tall cylinder contains 25 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until the total liquid depth is 40 cm. What is the gauge pressure at the bottom of the cylinder
Answer: [tex]377.3\ kPa[/tex]
Explanation:
Given
Water column height [tex]h=25\ cm[/tex]
After oil is poured, the total height becomes [tex]h'=40\ cm[/tex]
Pressure at the bottom will be the sum due to the water and oil column
Suppose the density of the oil is [tex]\rho=900\ kg/m^3[/tex]
Pressure at the bottom
[tex]\Rightarrow P=10^3\times g\times 25+900\times g\times 15\\\Rightarrow P=100g[250+135]\\\Rightarrow P=3773\times 100\ Pa\\\Rightarrow P=377.3\ kPa[/tex]