Answer:
♕ [tex]\large{ \red{ \tt{Step - By - Step \: Explanation}}}[/tex]
☃ [tex] \underline{ \underline{ \blue{ \large{ \tt{G \: I \: V \: E\: N}}}}} : [/tex]
Frequency ( f ) = 7 HertzWavelength ( λ ) = 42m♨ [tex] \underline {\underline{ \orange{ \large{ \tt{T \: O \: \: F \: I \: N\: D}}}} }: [/tex]
Wave velocity ( v )☄ [tex]\underline{ \underline{ \large{ \pink{ \tt{S\: O \: L \: U \: T\: I \: O \: N}}}}}: [/tex]
✧ [tex] \red{ \boxed{ \large{ \purple{ \sf{Wave \: velocity(v) = Frequency(f) \times Wavelength(λ)}}}}}[/tex]
~Plug the known values and then multiply!
↦ [tex] \large{ \tt{7 \times 42}}[/tex]
↦ [tex] \boxed{ \boxed{ \large{ \bold{ \tt{294 \: m {s}^ {- 1} }}}}}[/tex]
☥ [tex] \large{ \boxed{ \boxed{ \large{ \tt{Our \: Final \: Answer : \underline{ \large{ \tt{294 \: m {s}^{ - 1}}}}}}}}} [/tex]
---------------------------------------------------------------
❁ [tex] \underline{ \large{ \red{ \tt{D\: E\: T \: A \: I \: L\: E \: D \: \: I\: N \: F \: O}}}} : [/tex]
Frequency ( f ) : The number of complete waves , set up in a medium in one second is called frequency of the wave. The SI unit of frequency is Hertz ( Hz ). For example : if a sound wave completes 15 compressions and 15 rarefactions in one second , it's frequency is 15 Hz.Wavelength ( λ ) : The distance between two consecutive troughs or crests in a transverse wave or the distance between two consecutive compressions or rarefactions in a longitudinal wave us called wavelength. It is the distance travelled by a wave in a time equal to it's time period. It's SI unit is metre ( m ).Wave velocity ( v ) : The velocity with which a wave propagates in a medium is called wave velocity. It's SI unit is m/s.# KILL : Excuses
KISS : Opportunities
MARRY : Goals
♪ Hope I helped! ♡
☂ Have a wonderful day / night ! ツ
✎ [tex] \underbrace{ \overbrace{ \mathfrak{Carry \: On \: Learning}}}[/tex] ✔
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Which of the following means that an
image is real?
A. +do
B. +di
c. -di
D.-do
Answer:
B
Explanation:
Please help... I'm confused on what I represents in terms of solving the total current. Would variable would I be singling out?
Answer:
the researcher say hi for us the best pa the best of us are going out to eat that I can get my money toward a little bit but the best of luck to be at work by then and we will see what the status
using the image provided, which moon phase is next?
Answer:
Waning Crescent
Answer: waxing crescent
Explanation:
In a double-slit experiment, the third-order maximum for light of wavelength 510 nm is located 17 mm from the central bright spot on a screen 1.6 m from the slits. Light of wavelength 670 nm is then projected through the same slits. Part A How far from the central bright spot will the second-order maximum of this light be located
Answer:
14.9 mm
Explanation:
We know dsinθ = mλ where d = separating of slit, m = order of maximum = 3 and λ = wavelength = 510nm = 510 × 10⁻⁹ m
Also tanθ = L/D where L = distance of m order fringe from central bright spot = 17 mm = 0.017 m and D = distance of screen from slit = 1.6 m
So, sinθ = mλ/d
Since θ is small, sinθ ≅ tanθ
So,
mλ/d = L/D
d = mλD/L
Substituting the values of the variables into the equation, we have
d = 3 × 510 × 10⁻⁹ m × 1.6 m/0.017 m
d = 2448 × 10⁻⁹ m²/0.017 m
d = 144000 × 10⁻⁹ m
d = 1.44 × 10⁻⁴ m
d = 0.144 × 10⁻³ m
d = 0.144 mm
Now, for the second-order maximum, m' of the 670 nm wavelength of light,
m'λ'/d = L'/D where m' = order of maximum = 2, λ' = wavelength of light = 670 nm = 670 × 10⁻⁹ m, d = slit separation = 0.144 mm = 0.144 × 10⁻³ m, L' = distance of second order maximum from central bright spot and D = distance of screen from slit = 1.6m
So, L' = m'λ'D/d
So, substituting the values of the variables into the equation, we have
L' = 2 × 670 × 10⁻⁹ m × 1.6 m/0.144 × 10⁻³ m
L' = 2144 × 10⁻⁹ m²/0.144 × 10⁻³ m
L' = 14888.89 × 10⁻⁶ m
L' = 0.01488 m
L' ≅ 0.0149 m
L' = 14.9 mm
b. Why are the resources listed in part (a) considered "renewable"?
Answer:
I would need to see the part a to be more specific but this may help:
resources like solar, water, wind are considered renewable because they will never run out, they can be replenished over and over again. There will always be a sun (and if there wasn't then we'd have bigger conserns than renewable energy) and the wind will always blow. It can never run out because you use too much unlike coal and fossil fuels.
Hope this helps!
The diagram below shows a person swinging a hammer.
Which position has the least amount of kinetic energy?
A-1
B-2
C-3
D-4
A child moving at constant velocity carries a 2 N ice-cream cone 1 m across a level surface. What is the net work done on the ice-cream cone?
Answer:
2 Joule
Explanation:
Work=force *dISPLACMENT
2N*1M
2 JOUL
An electrical insulator is a material that:
A) contains no charge
B) does not allow electrons to flow
C) has more protons than electrons
D) must be a crystal
Answer:
Option B is appropriate for this question
Describe each of the FITT training principles. (Site 1
Answer:
The FITT principles are an exercise prescription to help participants understand how long and how hard they should exercise. FITT is acronym that stands for Frequency, Intensity, Time, and Type. FITT can be applied to exercise in general or specific components of exercise.
Explanation:
a spring has a spring constant of 330 n/m. how far is the spring compressed when 150 newtons of force are used?
a. 0.0014 meters
b. 0.45 meters
c. 2.2 meters
d. 5.0 meters
show your work
Answer:
I think its B
Explanation:
True False: Marke each statement as true or false.
shift
1. Light intensity affects the rate of photosynthesis.
in
2. Energy is required by all organisms for life.
3. The ability of a plant to repair tissue depends on respiration.
4. The gas needed for photosynthesis is carbon dioxide (CO2).
5. Plants only carry on photosynthesis, not respiration.
6. Respiration can occur without photosynthesis.
Answer:
1) true 2) true 3) true 4) true 5) false 6) false
Explanation:
After 60 days, 100g of a certain element has decayed to only 12.5g.
What is the half- life of this element?
А
30 days
B.
8 days
С
20 days
D
5 days
Answer:
8 days
Explanation:
Solve the following problem. A force of 5N is being applied to the right, and a force of
3N is being applied to the left. What is the net force?
A- 15 N
B- 8 N
C- 7N
D- 2N
Answer:
D 2N
Explanation:
these are opposite forces so one becomes negative and the other positive that is
-3+5=2 or 5-3=2
is it possible to have rainbows during the fine sunny day? Explain your answer.
Answer:
As long as there is rain, a rainbow is possible. Rain is possible on a sunny day, and is known as a sunshower.The rainbow can be observed in a sunny day if the water droplets are present in air and the sun rays pass through it reaches the eye of the observer. In this situation, the observer can see a rainbow.
or
If you happened to look up at the sky this past weekend, you might have noticed a rare and beautiful sight: iridescent rainbow clouds, but not a drop of rain in sight. This phenomenon is known, fittingly, as cloud iridescence or irisation. The effect is not unlike seeing a rainbow painted on the clouds.
please help, im having trouble can you please explain how to do it
Answer:
independent and the second one with the (-8, 1, -3)
A magnetic field of 0.27 T exists in the region enclosed by a solenoid that has 599 turns and a diameter of 9.29 cm. Within what period of time must the field be reduced to zero if the average magnitude of the induced emf within the coil during this time interval is to be 12.8 kV
Answer: [tex]8.54\times 10^{-5}\ s[/tex]
Explanation:
Given
The initial magnetic field is [tex]B=0.27\ T[/tex]
No of turns [tex]N=599\ \text{turns}[/tex]
Diameter of the solenoid [tex]d=9.29\ cm[/tex]
Induced EMF [tex]E=12.8\ kV[/tex]
Induced emf is the product of no of turns and rate of change of flux.
[tex]\Rightarrow E=-N\cdot \dfrac{\Delta \phi }{\Delta t}\\\\\Rightarrow E=-N\cdot \dfrac{\Delta (B\cdot A)}{\Delta t}\\\\\Rightarrow E=-NA\cdot \dfrac{\Delta B}{\Delta t}\\\\\text{Insert the values}\\\\\Rightarrow 12.8=-599\times \pi r^2\cdot \dfrac{(0-B)}{\Delta t}\\\\\Rightarrow \Delta t=\dfrac{599\times \pi \times (4.64\times 10^{-2})\times 0.27}{12.8\times 10^3} \\\\\Rightarrow \Delta t=854.71\times 10^{-7}\ s\\\\\text{Taking absolute value}\\\Rightarrow \Delta t=8.54\times 10^{-5}\ s[/tex]
A skydiver weighing 200 lbs with clothes that have a drag coefficient of .325 is falling in an area that has an atmospheric density of 1.225 kg/m2 (and assuming that altitude has a negligible effect on atmospheric density). The skydiver can change the body orientation from straight-erect with a cross-sectional area of 6 sqft to a belly-flat cross-sectional area of 24 sqft. Calculate the terminal velocity of the person when the body has straight and when the body has belly-flat orientations. Calculate the terminal velocity on these two different orientations.
Answer:
The right solution is:
(a) 89.455 m/s
(b) 44.73 m/s
Explanation:
The given values are:
Mass,
m = 200 lbs
or,
= [tex]\frac{200}{2.205} \ kg[/tex]
= [tex]90.7 \ kg[/tex]
Air's density,
[tex]\delta = 1.225 \ kg/m^3[/tex]
Drag coefficient,
[tex]C_d=0.325[/tex]
When body is straight, area,
[tex]A_1=6 \ ft^2[/tex]
As we know,
Terminal velocity,
⇒ [tex]V_t=\sqrt{\frac{2W}{C_d \delta A} }[/tex]
or,
⇒ [tex]=\sqrt{\frac{2mg}{C_d \delta A} }[/tex]
At straight orientation,
⇒ [tex]V_t'=\sqrt{\frac{2\times 90.7\times 9.8}{0.325\times 1.225\times 0.558} }[/tex]
⇒ [tex]=\sqrt{\frac{1777.72}{0.223}}[/tex]
⇒ [tex]=89.455 \ m/s[/tex]
When belly flat,
⇒ [tex]V_t''=\sqrt{\frac{2\times 90.7\times 9.8}{0.325\times 1.225\times 0.558\times 4} }[/tex]
⇒ [tex]=\sqrt{\frac{1777.72}{0.889} }[/tex]
⇒ [tex]=44.73 \ m/s[/tex]
A current of 5 a flows in a copper wire for 20 s. how many coulombs of charge passed through the wire in this time?
Answer:
I think so too
Explanation:
:*
During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, the bags produce a maximum acceleration of 60g, but lasting for only 32 ms (or less). How far does a person travel in coming to a complete stop in 32 ms at a constant acceleration of 60g
Answer:
s = 0.3 m
Explanation:
In order to calculate the distance traveled by the person, we will use the second equation of motion:
[tex]s = v_it + \frac{1}{2}at^2[/tex]
where,
s = distance traveled = ?
vi = initial speed = 0 m/s
t = time = 32 ms = 0.032 s
a = acceleration = 60g = (60)(9.81 m/s²) = 588.6 m/s²
Therefore,
[tex]s = (0\ m/s)(0.032\ s)+\frac{1}{2}(588.6\ m/s^2)(0.032\ s)^2[/tex]
s = 0.3 m
2. A force of magnitude 44 N compresses a spring 0.69 m from equilibrium. What is the value of the spring constant?
60 N/m
64 N/m
32 N/m
49 N/m
Answer:
Spring constant = 63.77 N/m
Explanation:
Given the following data;
Force = 44 N
Extension = 0.69 m
To find the spring constant;
Force = spring constant * extension
Substituting into the formula, we have;
44 = spring constant * 0.69
Spring constant = 44/0.69
Spring constant = 63.77 N/m
En una carga de prueba de 2 microcoulomb se sitúa en un punto en el que la intensidad del campo eléctrico es de 5 * 10 / 2 N/C cuál es la fuerza de la magnitud que actúa sobre ella?
Answer:
Fuerza, F = 0.001 Newton
Explanation:
Dados los siguientes datos;
Intensidad del campo eléctrico = 5 * 10² N/C
Carga = 2 microculombios = 2 * 10^{-6} culombios
Para encontrar la fuerza que actúa sobre la carga;
Matemáticamente, la intensidad del campo eléctrico viene dada por la fórmula;
Intensidad del campo eléctrico = fuerza/carga
F = E * Q
F = 5 * 10² * 2 * 10^{-6}
Fuerza, F = 0.001 Newton
A 1 kg flashlight is dropped from rest and falls to the floor without air resistance . At the point during its fall, when it is 0.7 meters above the floor, its potential energy exactly equals its kinetic energy. How fast is the flashlight moving at this point, in m/s 3.7
Answer:
The speed of the flashlight at that point is 3.7 m/s
Explanation:
When an object of mass M is at a height H above the ground, the potential energy of the object is:
U = M*H*g
Where g is the gravitational acceleration, g = 9.8 m/s^2
And for an object with velocity v, the kinetic energy is:
K = (M/2)*v^2
We know that when the flashlight of mass 1kg is 0.7 meters above the ground, the potential energy is equal to the kinetic energy, then:
M = 1kg
H = 0.7m
g = 9.8 m/s^2
Replacing these in the equations, we get:
U = K
(1kg)*(0.7m)*(9.8 m/s^2) = ((1kg)/2)*v^2
As the mass factor appears in both sides, we can remove it:
(0.7 m)*(9.8 m/s^2) = (v^2)/2
Now we can multiply both sides by 2:
2*(0.7 m)*(9.8 m/s^2) = v^2
Now let's apply the square root to both sides:
√(2*(0.7 m)*(9.8 m/s^2)) = v = 3.7 m/s
Select the correct answer. A body travels 10 meters during the first 5 seconds of its travel, and it travels a total of 30 meters over the first 10 seconds of its travel. What is its average speed during the time from t = 5 seconds to t = 10 seconds? A. 2 meters/second B. 3 meters/second C. 4 meters/second D. 5 meters/second E. 6 meters/second
Answer:
2 m/s
Explanation:
...........................
Answer:
A:2 m/s
Explanation:
Plato
A parallel beam of light of wavelength 4.5 x 10^-7 m is incident on a pair of slits that are 5.0 x 10^-4 m apart. The interference pattern is focused on a screen 2.0 m from the slits. The separation between two adjacent bright fringes is most nearly:_________
Answer:
1.8x10⁻³m
Explanation:
From the question above, the following information was used to solve the problem.
wavelength λ = 4.5x10⁻⁷m
Length L = 2.0 meters
distance d = 5 x 10₋⁴m
ΔY = λL/d
= 4.5x10⁻⁷m (2) / 5 x 10₋⁴m
= 0.00000045 / 0.0005
= 0.0000009/0.0005
= 0.0018
= 1.8x10⁻³m
from the solution above The separation between two adjacent bright fringes is most nearly 1.8x10⁻³m
thank you!
The power output of a car engine running at 2500 rpm is 500 kW . You may want to review (Pages 574 - 577) . Part A How much work is done per cycle if the engine's thermal efficiency is 20.0 %
Answer:
Explanation:
2500 rpm = 2500 / 60 rps ( revolution per second )
= 41.67 rps .
power output = 500 kW
thermal efficiency = 20 %
power input = (100 /20) x 500 kW
= 2500 kW .
work done per second = 2500 kJ
work done per cycle = 2500 / 41.67
= 60 kJ .
We have that the workdone per cycle is mathematically given as
Wc=12.00KJ
Workdone per cycle of engineQuestion Parameters:
The power output of a car engine running at 2500 rpm is 500 kW
Generally the work done per cycle is mathematically given as
Wc=500*1/41.666
Where
=2500rev/60sec
=41.666
Meaning one cycle takes 1/41.666
Hence
Wc=500*1/41.666
Wc=12.00KJ
For more information on work
https://brainly.com/question/756198
The temperature of a body is from 200 to 300C.The change of temp at absolute scale is
Answer:
mark me brainliest
Explanation:
The change of temperature at absolute scale is. A. 3.73 K
Answer:
373K
Explanation:
300°c - 200°c =100°c
Absolute scale means Kelvin scale so
0°c= 273°c
100°c = 100 + 273
=373K
PLEASE HELP I WILL MARK BRAINLYIST!!!!!!!!!
All of the matter in the universe formed after the big bang. Scientists
predicted that hydrogen and helium would be the most abundant elements.
What did scientists discover?
A. The universe has 73% hydrogen and 25% helium, which does not
support the big bang theory
B. Planets are made mostly of other elements, which does not
support the big bang theory
C. Stars are made mostly of other elements, which supports the big
bang theory
D. The universe has 74% hydrogen and 24% helium, which supports
the big bang theory.
Answer:
The answer is D. Here is proof . Plz mark brainliest
Explanation:
The illustration shows ?
A)open circuit
B)broken circuit
C)series circuit
D)parallel circuit
Answer:
D)parallel circuit
Explanation:
the components are placed parallel from eachother
A boat that can travel at 4.60 km/h in still water crosses a river with a current of 1.80 km/h. At what angle must the boat be pointed upstream to travel straight across the river
Answer:
The angle the boat must be pointed upstream is 69⁰
Explanation:
Check the image uploaded for the diagram;
Given;
speed of the boat on still water, = 4.6 km/h
speed of the boat on a river with current, = 1.8 km/h
The angle the boat must be pointed upstream is calculated as follows
[tex]tan(\theta) = \frac{4.6}{1.8} \\\\tan(\theta) = 2.556\\\\\theta = tan^{-1}(2.556)\\\\\theta = 68.63^0\\\\\theta = 69^0[/tex]
g hoop rolls without slipping on a horizontal surface. The hoop has a mass m and radius r. What is the ratio of the hoop's translational kinetic energy to the hoop's rotational kinetic energy, KT /KR
Answer:
[tex]\frac{K.E_t}{K.E_r} = 1[/tex]
Explanation:
The translational kinetic energy of the hoop is given as:
[tex]K.E_t = \frac{1}{2} mv^2[/tex] ---------------------- equation (1)
where,
[tex]K.E_t[/tex] = translational kinetic energy
m = mass of hoop
v = linear speed of hoop
The rotational kinetic energy of the hoop is given as:
[tex]K.E_r = \frac{1}{2} I\omega^2[/tex]
where,
[tex]K.E_r[/tex] = rotational kinetic energy of the hoop
I = Moment of Inertia of the hoop = mr²
r = radius of the hoop
ω = angular speed of hoop = [tex]\frac{v}{r}[/tex]
Therefore,
[tex]K.E_r = \frac{1}{2} (mr^2)(\frac{v}{r} )^2\\\\K.E_r = \frac{1}{2} mv^2[/tex]------------------- equation (2)
dividing equation (1) and equation (2), we get:
[tex]\frac{K.E_t}{K.E_r} = \frac{\frac{1}{2}mv^2 }{\frac{1}{2}mv^2 }\\\\\frac{K.E_t}{K.E_r} = 1[/tex]