Sam moves an 800 N wheelbarrow 5 meters in 15 seconds. How much work did he do?
Answer:
work done= force × displacement
=800×5
=4000J
Explanation:
The amount of work done is the result of the magnitude of force applied and the displacement of the body due to the force applied. Therefore, work done is defined as the product of the applied force and the displacement of the body.
In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the first light source has a wavelength of 640 nm. Two different interference patterns are observed. If the 10th order bright fringe from the first light source coincides with the 12th order bright fringe from the second light source, what is the wavelength of the light coming from the second monochromatic light source
Answer:
533.33 nm
Explanation:
Since dsinθ = mλ for each slit, where m = order of slit and λ = wavelength of light. Let m' = 10 th order fringe of the first slit of wavelength of light, λ = 640 nm and m"= 12 th order fringe of the second slight of wavelength of light, λ'.
Since the fringes coincide,
m'λ = m"λ'
λ' = m'λ/m"
= 10 × 640 nm/12
= 6400 nm/12
= 533.33 nm
an iron Tyre of diameter 50cm at 288k is to be shrank on to a wheel of diameter 50.35cm.To what temperature must the tyre be heated so that it will slip over the wheel with a radial gap of 0.5mm.Linear expansivity of iron is 0.000012k-1
Answer:
The answer should be D
Explanation:
A hair dryer draws a current of 12.8 A.
(a)How many minutes does it take for
6.8 x 10° C of charge to pass through the
hair dryer? The fundamental charge is
1.602 x 10-19 C.
Answer in units of min.
(b)How many electrons does this amount of
charge represent?
Answer in units of electrons.
Answer:
(a) 8.85×10⁻³ minutes
(b) 4.24×10¹⁹ electrons
Explanation:
(a) Using,
Q = it............................. Equation 1
Where Q = quantity of charge, i = current, t = time.
Make t the subject of the equation
t = Q/i............................. Equation 2
Given: Q = 6.8×10⁰ C, i = 12.8 A
Substitute these values into equation 2
t = 6.8×10⁰/12.8
t = 8.85×10⁻³ minutes
(b) n = Q/(1.602×10⁻¹⁹)................. Equation 3
Where n = number of electrons.
Given: Q = 6.8×10⁰ C
Substitute into equation 2
n = 6.8×10⁰/1.602×10⁻¹⁹
n = 4.24×10¹⁹ electrons
(a) The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes
(b) Amount of the electrons in the charge will be [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons
What will be the time of the charge and number of the electrons in the charge ?As we know Q = IT
Where Q = quantity of charge, i = current, T = time.
From the above equation
T= Q/I.
Given: Q = [tex]6.8\times\d10^{0}[/tex] C, i = 12.8 A
Substitute these values
T= [tex]6.8[/tex]×[tex]\d10^{0}[/tex] /12.8
T = [tex]8.85[/tex]×[tex]\d10^{-3}[/tex] minutes
Now the number of the electrons present in the charge will be
n = Q/( [tex]1.602[/tex]×[tex]\d10^{-19}[/tex])
Where n = number of electrons.
Given: Q = [tex]6.8\times\d10^{0}[/tex] C
Substitute Value of Q
n = [tex]6.8\times\d10^{0}[/tex]/ [tex]1.602\times\d10^{-19}[/tex]
n = [tex]4.24\times\d10^{19}[/tex] electrons
Thus
(a)The time taken by the charge to flow from the drier will be [tex]\d8.85[/tex]×[tex]\d10^{-3}[/tex]minutes
(b)Amount of the electrons in the charge will be [tex]\d4.24[/tex]×[tex]\d10^{19}[/tex] electrons
To know more about electric charge follow
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How much energy would be required to move the earth into a circular orbit with a radius 2.0 kmkm larger than its current radius
Answer:
[tex]3.52\times 10^{25}\ \text{J}[/tex]
Explanation:
G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]
M = Mass of Sun = [tex]1.989\times 10^{30}\ \text{kg}[/tex]
m = Mass of Earth = [tex]5.972\times 10^{24}\ \text{kg}[/tex]
[tex]r_i[/tex] = Initial radius of orbit = [tex]1.5\times 10^{11}\ \text{m}[/tex]
[tex]r_f[/tex] = Final radius of orbit = [tex]((1.5\times 10^{11})+2\times 10^3)\ \text{m}[/tex]
Energy required is given by
[tex]E=\dfrac{1}{2}\Delta U\\\Rightarrow E=\dfrac{GMm}{2}(\dfrac{1}{r_i}-\dfrac{1}{r_f})\\\Rightarrow E=\dfrac{6.674\times 10^{-11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{2}(\dfrac{1}{1.5\times 10^{11}}-\dfrac{1}{(1.5\times 10^{11})+2\times 10^3})\\\Rightarrow E=3.52\times 10^{25}\ \text{J}[/tex]
The energy required would be [tex]3.52\times 10^{25}\ \text{J}[/tex].
Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine:
This question is incomplete, the complete question is;
Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine: the angular velocity of Bxy rotating frame (ω).
Answer:
the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s
Explanation:
Given the data in the question and image below and as illustrated in the second image;
distance S = 40 m
V[tex]_B[/tex] = 54 km/hr
V[tex]_A[/tex] = 72 km/hr
α = 100 m
now, angular velocity of Bxy will be;
ω[tex]_B[/tex] = V[tex]_B[/tex] / α
so, we substitute
ω[tex]_B[/tex] = ( 54 × 1000/3600) / 100
ω[tex]_B[/tex] = 15 / 100
ω[tex]_B[/tex] = 0.15 rad/s
Therefore, the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s
23
In order for a 12 Volt power source
to produce a current of 0.085 amps,
a resistance of...
[?] Ohms is needed.
Enter
Haven't learned this yet.
Answer:
141.18 ohms
Explanation:
From the question given above, the following data were obtained:
Voltage (V) = 12
Current (I) = 0.085 A
Resistance (R) =?
The resistance needed can be obtained as follow:
V = IR
12 = 0.085 × R
Divide both side by 0.085
R = 12 / 0.085
R = 141.18 ohms
Therefore, a resistor of resistance 141.18 ohms is needed.
Define Mechanical advantage
fe effort of 2125N is used to lift a Lead of 500N
through a Verticle high of 2.N using a buly System
if the distance Moved by the effort is 45m
Calculate 1. Work done on the load
2. work done by the effort
3. Efficiency of the System
Answer:
1) 1000Nm
2) 95,625Nm
3) 1.05%
Explanation:
Mechanical Advantage is the ratio of the load to the effort applied to an object.
MA = Load/Effort
1) Workdone on the load = Force(Load) * distance covered by the load
Workdone on the load = 500N * 2m
Workdone on the load = 1000Nm
2) work done by the effort = Effort * distance moves d by effort
work done by the effort = 2125 * 45
work done by the effort = 95,625Nm
3) Efficiency = Workdone on the load/ work done by the effort * 100
Efficiency = 1000/95625 * 100
Efficiency = 1.05%
Hence the efficiency of the system is 1.05%
The gravitational force between two objects with masses 1kg and 28kg separated by a distance 7m is ____________10-11 N.
a.
3.81
b.
26.68
c.
9151.24
d.
1307.32
Hhhhhellllppp fastt
Answer:
a. 3.81
Explanation:
F = GMm/r^2
F = (6.67 x 10^-11 x 28 x 1) / 7^2
F = 3.81 x 10^-11 N
Is there a way to see moon and the sun at once?
In this experiment, you will use a track and a toy car to explore the concept of movement. You will measure the time it takes the car to travel certain distances, and then complete some calculations. In the space below, write a scientific question that you will answer by doing this experiment.
Answer: if weight affects how fast they go?
Explanation:
Answer:
How can we change the speed of a toy car on a racetrack to describe the car’s motion?
Explanation:
thats the sample respond
If you could help me please.
1) Does a 1 kg object weight 9.8 newtons on the moon? why?
2) How much does a 3-kg object weigh (on earth) in newtons?
3) How much does a 20-kg object weigh (on earth) in newton?
4) What must happen for the mass of an object to change?
5) What are 2 ways the weight of an object can change?
1) Does a 1 kg object weight 9.8 newtons on the moon? why?
No. 1kg of mass does not weigh 9.8N on the moon.
Weight = (mass) x (gravity).
Gravity is 9.8 m/s² on Earth, but gravity is only 1.62 m/s² on the moon.
2) How much does a 3-kg object weigh (on earth) in newtons?
Weight = (mass) x (gravity)
Gravity = 9.8 m/s² on Earth.
Weight = (3 kg) x (9.8 m/s² )
Weight = 29.4 N
3) How much does a 20-kg object weigh (on earth) in newton?
Weight = (mass) x (gravity)
Gravity = 9.8 m/s² on Earth.
Weight = (20 kg) x (9.8 m/s² )
Weight = 196 N
4) What must happen for the mass of an object to change?
When an object moves, its mass increases. The faster it moves, the greater its mass gets. But this is all part of Einstein's "Relativity". The object has to move at a significant fraction of the speed of light before any change can be noticed or measured. So as far as we are concerned, in everyday life, the mass of an object doesn't change, no matter where it is, or what you do to it.
5) What are 2 ways the weight of an object can change?
First, remember that the mass of an object doesn't change, no matter where it is, what you do to it, or what else is around it.
But its weight can change, because its weight depends on the strength of gravity in the place where the object is, and that gravity is the result of what else is around it in the neighborhood. So the weight can change even though the mass doesn't.
The weight of an object changes if you take it to a place where gravity is stronger or weaker.
Let's say we have an object whose mass is 90.72 kilograms. Like me !
As long as I stay on earth, where gravity is 9.8 m/s² , I weigh 889 Newtons (200 pounds).
. . . Fly me to the moon. Gravity = 1.62 m/s² Weight = 147 Newtons (33 lbs)
. . . Drag me to Jupiter. Gravity = 24.8 m/s² Weight = 2,249 N (506 pounds)
My mass never changed, but my weight sure did.
The gravitational potential energy of an object is defined as the energy it has due to its position in a gravitational field. A ball with a weight of 50 N is lifted to a height of 1 meter. Which graph correctly represents the change in gravitational potential energy (shaded in blue) as it is lifted to this height?
Answer:
athletic
Explanation:
because internet system has been down since we were in few days
What is happening in the graph shown below?
A.
The object moves away from the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 2 m/s.
B.
The object moves toward the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves away from the origin at a speed of 2 m/s.
C.
The object moves toward the origin at a speed of 6 m/s, stands still 6 m away from the origin for 3 seconds, then moves away from the origin at a speed of 8 m/s.
D.
The object moves away from the origin at a speed of 6 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 8 m/s.
Answer:
D. The object moves away from the origin at a speed of 3 m/s, stands still 6 m away from the origin for 3 seconds, then moves toward the origin at a speed of 2 m/s.
Explanation:
I just got it right lol
An object is accelerated by a net force in which direction?
A. at an angle to the force
B. in the direction of the force
C. in the direction opposite to the force
D. Any of these is possible.
Answer:
B. in the direction of the force
Explanation:
Sana nakatulong
Which of the following changes when an unbalanced force acts on an object?
A. mass
B. motion
C. inertia
D. weight
The answer is Motion
Soap bubbles can display impressive colors, which are the result of the enhanced reflection of light of particular wavelengths from the bubbles' walls. For a soap solution with an index of refraction of 1.21, find the minimum wall thickness that will enhance the reflection of light of wavelength 711 nm in air.
Answer:
the minimum wall thickness that will enhance the reflection of light is 146.9 nm
Explanation:
Given the data in the question;
At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.
At the second interface, no shift occurs,
condition for constructive interference;
t = ( m + 1/2) × λ/2n
where m = 0, 1, 2, 3 . . . . . .
now, the condition for the constructive interference;
t = mλ/2n
where t is the thickness of the soap bubble, λ is the wavelength of light and n is the refractive index of soap bubble.
so the minimum thickness of the film which will enhance reflection of light will be;
t[tex]_{min[/tex] = ( m + 1/2) × λ/2n
we substitute
t[tex]_{min[/tex] = ( 0 + 1/2) × 711 /2(1.21)
t[tex]_{min[/tex] = 0.5 × 711/2.42
t[tex]_{min[/tex] = 0.5 × 293.80165
t[tex]_{min[/tex] = 146.9 nm
Therefore, the minimum wall thickness that will enhance the reflection of light is 146.9 nm
which one is odd copper,plastic,rubber
Answer:
It's plastic.
trust me it's plastic, i've rad it somewhere.
All of them have something that's not like the others.
-- Rubber is the only one on the list that has two repeated letters.
-- Plastic is the only one on the list thagt has no repeated letters.
-- Plastic is the only one on the list that has no 'r' in its name.
-- Copper is the only one on the list that is an element, not a compound.
-- Copper is the only good electrical conductor on the list.
-- Plastic is the only one on the list with more than six letters in its name.
-- Rubber is the only one on the list with no 'p' in its name.
-- Plastic is the only one on the list that doesn't end in "-er".
The amount of light that enters the pupil is controlled by the:
retina.
lens.
inis.
Answer: The amount of light that enters the pupil is controlled by the Iris
Explanation:
a cohesive force between the liquids molecules is responsible for the fluids is called
Answer:
static force
Explanation:
mark me brainliest
At position A within a tube containing fluid that is moving with steady laminar flow, the speed of the fluid is 12.0 m/s and the tube has a diameter 12.00 cm. At position B, the speed of the fluid is 18.0 m/s and the tube has a diameter 6.00 cm. What is the ratio of the density of the fluid at position A to the density of the fluid at position B
Answer:
0.375
Explanation:
For incompressible flow, we know that;
ρ1•v1•A1 = ρ2•v2•A2
Where;
ρ1 = density of fluid at position A
v1 = speed of fluid at position A
A1 = area of tube
ρ2 = density of fluid at position B
v2 = speed of fluid at position B
A2 = area of tube
We want to find ratio of the density of the fluid at position A to the density of the fluid at position B.
Thus;
ρ1/ρ2 = (v2•A2)/(v1•A1)
Now, the tube will have the same height.
But we are given;
diameter of A = 12.00 cm = 0.12 m
diameter of B = 6 cm = 0.06 m
Thus;
A1 = π(d²/4)h = πh(0.12²/4)
A2 = πh(0.06²/4)
We are also given;
v1 = 12 m/s
v2 = 18 m/s
Thus;
ρ1/ρ2 = (18 × πh(0.06²/4))/(12 × πh(0.12²/4))
πh/4 will cancel out to give;
ρ1/ρ2 = (18 × 0.06²)/(12 × 0.12²)
ρ1/ρ2 = 0.375
३.रात में घूमने वाला write one word substitute
Explanation:
रात में घूमने वाला arthaarat निशाचर
If you blow across the open end of a soda bottle and produce a tone of 250 Hz, what will be the frequency of the next harmonic heard if you blow much harder?
___Hz
Answer:
Generally, the lowest overtone for a pipe open at one end and closed would be at y / 4 where y represents lambda, the wavelength.
Since F (frequency) = c / y Speed/wavelength
F2 / F1 = y1 / y2 because c is the same in both cases
F2 = y1/y2 * F1
F2 = 3 F1 = 750 /sec
Note that L = y1 / 4 = 3 y2 / 4 for these wavelengths to fit in the pipe
and y1 = 3 y2
The second harmonic will be three times the first harmonic. The answer is 750 Hz
VIBRATION OF WAVES IN PIPESClosed pipes have odd multiples of frequencies or harmonics. That is,
If [tex]F_{0}[/tex] = fundamental frequency = first harmonic
[tex]F_{1}[/tex] = 3[tex]F_{0}[/tex] = second harmonic
[tex]F_{2}[/tex] = 5[tex]F_{0}[/tex] = third harmonic
[tex]F_{3}[/tex] = 7[tex]F_{0}[/tex] = fourth harmonic
Let assume that the first harmonic is 250 Hz, If you blow it much harder, second, third or fourth harmonic can be produced.
By using the formula above,
second harmonic will be 3 x 250 = 750Hz
Therefore, the frequency of the next harmonic heard if you blow much harder will be 750 Hz
Learn more about Sound waves here: https://brainly.com/question/1199084
why food cook faster with salt water than cook with pure water
Answer:
oil heats faster
Explanation:
On the Moon's surface, lunar astronauts placed a corner reflector, off which a laser beam is periodically reflected. The distance to the Moon is calculated from the round-trip time. The Earth's atmosphere slows down light. Assume the distance to the Moon is precisely 3.84×108 m, and Earth's atmosphere (which varies in density with altitude) is equivalent to a layer 30.0 km thick with a constant index of refraction n=1.000293. What is the difference in travel time for light that travels only through space to the moon and back and light that travels through the atmosphere and space?
Answer:
a) space only t = 1.28 s
b) space+ atmosphere t_ {total} = 1.28000003 s
Explanation:
The speed of light in each material medium is constant, which is why we can use the uniform motion relations
v= x / t
a) let's look for time when it only travels through space
t = x / c
t = 3.84 10⁸/3 10⁸
t = 1.28 s
b) we look for time when it travels part in space and part in the atmosphere
space
as it indicates that the atmosphere has a thickness of e = 30 10³ m
t₁ = (D-e) / c
t₁ = (3.84 10⁸ - 30.0 10³) / 3 10⁸
t₁ = 1.2799 s
atmosphere
we use the refractive index
n = c / v
v = c / n
we substitute in the equation of time
t₂ = e n / c
t₂ = 30 10³ 1,000293 /3 10⁸
t₂ = 1.000293 10⁻⁴ s
therefore the total travel time is
t_ {total} = t₁ + t₂
t_ {total} = 1.2799+ 1.000293 10⁻⁴
t_ {total} = 1.28000003 s
we can see that the time increase due to the atmosphere is very small
a sharp image is formed when light reflects from a
Answer:
Regular reflection
Explanation:
Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.
i hope this helps a bit.
According to the context, a sharp image is formed when light reflects from a regular reflection.
What is regular reflection?It is reflection without diffusion that obeys the laws of geometrical optics, as in mirrors.
This reflection of light happens when the angles that the two rays determine with the surface are equal.
Therefore, we can conclude that according to the context, a sharp image is formed when light reflects from a regular reflection.
Learn more about regular reflection here: https://brainly.com/question/3778324
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why doesn't a radio operating with two batteries function when one of the batteries is reversed?
Answer:
If you have two batteries and they have precisely the same voltage then placing one backwards will effectively cancel out the voltages and no current will flow. However, batteries aren't like that. The slightest difference in voltages mean that current will flow.
Explanation:
What factors affect the speed of a wave? Check all that apply.
the amplitude of the wave
the energy of the wave
the temperature of the medium
the type of wave
the type of medium
Answer:
the amplitude of the wave
the energy of the wave
the type of wave
the type of medium
A light year is the amount of time it takes for light from the Sun to reach the Earth.
True
False
A positive charge Q2 is uniformly distributed over a nonconducting disc of radius a which has a concentric circular hole of radius b. At the center of the hole there is another nonconducting disc of radius d where a charge Q1 is uniformly distributed.
a) Find the surface charge density of the disc with the hole σ2.
b) Find the surface charge density 01 of the disc of radius d.
c) Find the total charge enclosed by the circle of radius
Answer:
a) σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex] , b) σ = [tex]\frac{Q_2}{d^2}[/tex] , c) Q_ {total} = Q₁ + Q₂, σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]
Explanation:
a) The very useful concept of charge density is defined by
σ = Q / A
In this case we have a circular disk
The are of a circle is
A = π r²
in this case we have a hole in the center of radius r = b, so
A_net = π r² - π r_ {hollow} ²
A_ {net} = π (a² - b²)
whereby the density is
σ = [tex]\frac{Q_1}{ a^2 - b^2}[/tex]
b) The density of the other disk is
σ = Q₂ / A₂
σ = [tex]\frac{Q_2}{d^2}[/tex]
c) The total waxed load is requested by the larger circle
Q_ {total} = Q₁ + Q₂
the net charge density, in the whole system is
σ = [tex]\frac{Q_{total} }{ A_{total} }[/tex]
the area is
A_{total} = π a²
since the other circle is inside, we are ignoring the space between the two circles
σ_ {net} = [tex]\frac{Q_1 + Q_2}{\pi \ a^2}[/tex]
