Explanation:
The relation between frequency and wavelength is shown below as:
c = frequency × Wavelength
Where, c is the speed of light having value = 3×10⁸ m/s
So, Wavelength is:
Wavelength = c / Frequency
Given:
Frequency = 5.5 × 10¹⁴ Hz
So,
Wavelength = 3×10⁸ m/s / 5.5 × 10¹⁴ Hz = 545 × 10⁻⁹ m = 545 nm (1 nm = 10⁻⁹ m )
This wavelength corresponds to green color.
Frequency = 7 × 10¹⁴ Hz
So,
Wavelength = 3×10⁸ m/s / 7 × 10¹⁴ Hz = 428 × 10⁻⁹ m = 428 nm (1 nm = 10⁻⁹ m )
This wavelength corresponds to Violet color.
Frequency = 8 × 10¹⁴ Hz
So,
Wavelength = 3×10⁸ m/s / 8 × 10¹⁴ Hz = 375 × 10⁻⁹ m = 375 nm (1 nm = 10⁻⁹ m )
It does not fall in visible region. So no color. It is in UV region.
An elevator moves downward in a tall building at a constant speed of 5.70 m/s. Exactly 4.95 s after the top of the elevator car passes a bolt loosely attached to the wall of the elevator shaft, the bolt falls from rest. (a) At what time does the bolt hit the top of the still-descending elevator? (Assume the bolt is dropped at t = 0 s.)(b) Estimate the highest floor from which the bolt can fall if the elevator reaches the ground floor before the bolt hits the top of the elevator. (Assume 1 floor congruent 3 m.)
Answer:
a) t = 3.01s
b) 15th floor
Explanation:
First we need to know the distance the elevator has descended before the bolt fell.
[tex]\Delta Y_{e} = -V_{e}*t = -5.7 * 4.95 = -28.215m[/tex]
Now we can calculate the time that passed before both elevator and bolt had the same position:
[tex]Y_{b}=Y_{e}[/tex]
[tex]Y_{ob}+V_{ob}*t-g*\frac{t^{2}}{2} = Y_{oe} - V_{e}*t[/tex]
[tex]0+0-5*t^{2} = -28.215 - 5.7*t[/tex] Solving for t:
t1 = -1.87s t2 = 3.01s
In order to know how the amount of floors, we need the distance the bolt has fallen:
[tex]Y_{b}=-g*\frac{t^{2}}{2}=-45.3m[/tex] Since every floor is 3m:
Floors = Yb / 3 = 15 floors.
A spelunker is surveying a cave. She follows a passage 180 m straight west, then 230 m in a direction 45° east of south, and then 280 m at 30° east of north. After a fourth unmeasured displacement, she finds herself back where she started. A: Use a scale drawing to determine the magnitude of the fourth displacement. Express your answer using two significant figures.
B: Determine the direction of the fourth displacement. Express your answer using two significant figures.
The problem requires vector operations in two dimensions. Displacement is broken down into x and y-components which follow the east-west and north-south directions respectively. Total displacement being zero means the sum of the x and y components of displacement will also be zero. The fourth displacement is determined by negating the total x and y components of the first three displacements. The magnitude and direction are then obtained using Pythagorean theorem and arctan function respectively.
Explanation:To solve this problem, we need to deal with the changes in displacement in terms of vector operations. Given the different directions, we need to break down the vectors into their x and y components, where x represents east-west direction, and y north-south direction. Since our spelunker starts and ends in the same place, the sum of the displacements in each dimension will also be zero.
For displacement 1, moving 180m west, the x component would be -180, and y component would be 0. For displacement 2, moving 230m in a direction 45° east of south, the x would be -230×sin(45) and y would be -230×cos(45). For displacement 3, moving 280m at 30° east of north, the x would be 280×cos(30) and y would be 280×sin(30).
To determine the fourth displacement, we sum up the x and y components for displacement 1,2 and 3 and then negate them to get the x and y component of the fourth displacement. We then use the Pythagorean theorem to calculate the magnitude of the 4th displacement which is square root of (sum x² + sum y²). The direction can be obtained by calculating the arctan of the total y component / total x component.
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A quantity of 14.1 cm^3 of water at 8.4°C is placed in a freezer compartment and allowed to freeze to solid ice at -7.2°C. How many joules of energy must be withdrawn from the water by the refrigerator?
Answer:920.31 J
Explanation:
Given
Volume of water (V)[tex]=14.1 cm^3 [/tex]
mass(m)[tex]=\rho \times V=1000\times 14.1\times 10^{-6}=14.1 gm[/tex]
Temperature [tex]=8.4^{\circ} C[/tex]
Final Temperature [tex]=-7.2 ^{\circ}C[/tex]
specific heat of water(c)[tex]=4.184 J/g-^{\circ}C[/tex]
Therefore heat required to removed is
[tex]Q=mc(\Delta T)[/tex]
[tex]Q=14.1\times 4.184\times (8.4-(-7.2))[/tex]
[tex]Q=920.31 J[/tex]
A uniform electric field of magnitude 4.9 ✕ 10^4 N/C passes through the plane of a square sheet with sides 8.0 m long. Calculate the flux (in N · m^2/C) through the sheet if the plane of the sheet is at an angle of 30° to the field. Find the flux for both directions of the unit normal to the sheet.
1)unit normal with component parallel to electric field (N · m^2/C)
2)unit normal with component antiparallel to electric field (N · m^2/C)
Answer:
1. 1.568 x 10^6 N m^2 / C
2. - 1.568 x 10^6 N m^2 / C
Explanation:
E = 4.9 x 10^4 N/C
Side of square, a = 8 m
Area, A = side x side = 8 x 8 = 64 m^2
Angle between lane of sheet and electric field = 30°
Angle between the normal of plane of sheet and electric field,
θ = 90°- 30° = 60°
The formula for the electric flux is given by
[tex]\phi = E A Cos\theta[/tex]
(1) [tex]\phi = E A Cos\theta[/tex]
By substituting the values, we get
Ф = 4.9 x 10^4 x 64 x Cos 60 = 1.568 x 10^6 N m^2 / C
(2) [tex]\phi = E A Cos\theta[/tex]
By substituting the values, we get
Ф = - 4.9 x 10^4 x 64 x Cos 60 = - 1.568 x 10^6 N m^2 / C
A stone is thrown vertically upward from ground level at t = 0. At t=2.50 s, it passes the top of a tall building, and 1.50 s later, it reaches its maximum height. What is the height of the tall building? We assume an answer in meters.
Answer:67.45 m
Explanation:
Given
at t=2.5 s it passes the top of a tall building and after 1.5 s it reaches maximum height
let u is the initial velocity of stone
v=u+at
0=u-gt
[tex]u=9.81\times 4=39.24 m/s[/tex]
Let us take h be the height of building
[tex]h=ut+\frac{-1}{2}gt^2[/tex]
[tex]h=39.24\times 2.5-\frac{1}{2}\times 9.81\times 2.5^2[/tex]
h=67.45 m
A mass m = 550 g is hung from a spring with spring constant k = 2.8 N/m and set into oscillation at time t = 0. A second, identical mass and spring next to the first set is also set into motion. At what time t should the second system be set into motion so that the phase difference in oscillations between the two systems is pi/2?
Answer:
The second system must be set in motion [tex]t=0.70s[/tex] seconds later
Explanation:
The oscillation time, T, for a mass, m, attached to spring with Hooke's constant, k, is:
[tex]T=2\pi\sqrt(\frac{m}{k} )[/tex]
One oscillation takes T secondes, and that is equivalent to a 2π phase. Then, a difference phase of π/2=2π/4, is equivalent to a time t=T/4.
If the phase difference π/2 of the second system relative to the first oscillator. The second system must be set in motion [tex]t=\frac{\pi}{2}\sqrt(\frac{m}{k})=\frac{\pi}{2}\sqrt(\frac{0.55}{2.8}= 0.70s)[/tex] seconds later
Suppose you're on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the buzzer over the side while the balloon is rising at constant speed, what can you conclude about the sound you hear as the buzzer falls toward the ground?
(A) The frequency and intensity increase
(B) The frequency decrease and intensity increase
(C) The frequency decrease and intensity decrease
(D) The frequency remains the same, but the intensity decreases.
Answer:
(C) The frequency decrease and intensity decrease
Explanation:
The Doppler effect describes the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source, or the wave source is moving relative to the observer, or both.
if the observer and the source move away from each other as is the case for this problem, the wavelength heard by the observer is bigger.
The frequency is the inverse from the wavelength, so the frequency heard will increase.
The sound intensity depends inversely on the area in which the sound propagates. When the buzzer is close, the area is from a small sphere, but as the buzzer moves further away, the wave area will be from a larger sphere and therefore the intensity will decrease.
Oppositely charged parallel plates are separated by 4.67 mm. A potential difference of 600 V exists between the plates. (a) What is the magnitude of the electric field between the plates? ________ N/C (b) What is the magnitude of the force on an electron between the plates? ________ N (c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.00 mm from the positive plate?_________ J
Answer:
a) 1.28 *10^5 N/C
b)2.05 *10^{-14} N
c) 4.83 *10^{-17} J
Explanation:
Given Data:
Distance between the plates, d = 4.67 mm
[tex]= (4.67) *10^{-3} m[/tex]
[/tex]= 4.67 *10^{-3} m[/tex]
Potential difference, V = 600 V
Solution:
(a) The magnitude of the electric field between the plates is,
[tex]E = \frac{V}{d}[/tex]
[tex]= \frac{600 V}{4.67 *10^{-3}} m[/tex]
[tex]= 1.28 *10^5 V/m or 1.28 *10^5N/C[/tex]
(b) Force on electron btwn the plates is,
F = q E
[tex]= (1.6 *10^{-19} C) (1.28 *10^5N/C[/tex]
[tex]= 2.05 *10^{-14} N[/tex]
(c) Work done on the electron is
W = F * s
[tex]= (2.05 *10^{-14} N) * (5.31 *10^{-3} m - 2.95 *10^{-3} m)[/tex]
[tex]= 4.83 *10^{-17} J[/tex]
Suppose a Southwest Airlines passenger plane took three hours to fly 1800 miles in the direction of the Jetstream. The return trip against the Jetstream took four hours. What was the plane’s speed (as read on the plane’s speedometer) in still air and the Jetstream’s speed? How can applying matrices and linear systems help solve this problem?
Answer:
plane speed: 525mph, jetstream speed=75mph, in explanation it is solved with a linear equations system
Explanation:
First lets name each speed
vs:=speed of the jetstream
vp:=speed of the plane
Now when in the jetstream direction the speeds are added and on the opposite direction are subtracted, then we get these equations, that are linear.
1800 mi=(vp+vs)*3h
1800 mi=(vp-vs)*4h
which is a linear equation system equivalent to:
600 mph=vp+vs (1)
450 mph=vp-vs (2)
Now from (2) vp= 450mph+vs (3), replacing this in (1) we get:
600mph=(450mph+vs)+vs=450mph+2*vs, then 2*vs=150mph or vs=*75mph, this is the jetstream speed, replacing this in (3) we get the plane speed too vp=450 mph +75mph = 525 mph
A super snail initially traveling at 2 m/s accelerates at 1 m/s^2 for 5 seconds. How fast will it be going at the end of the 5 seconds? How far did the snail travel?
Answer:
The snail travel at the end of 5 s with a velocity of 12 m/s and the distance of the snail is 22.5 m.
Explanation:
Given that, the initial velocity of the snail is,
[tex]u=2m/s[/tex]
And the acceleration of the snail is,
[tex]a=1m/s^{2}[/tex]
And the time taken by the snail is,
[tex]t=5 sec[/tex]
Now according to first equation of motion,
[tex]v=u+at[/tex]
Here, u is the initial velocity, t is the time, v is the final velocity and a is the acceleration.
Now substitute all the variables
[tex]v=2m/s+ 1 \times 5 sec\\v=7m/s[/tex]
Therefore, the snail travel at the end of 5 s with a velocity of 7 m/s.
Now according to third equation of motion.
[tex]v^{2}- u^{2}=2as\\ s=\frac{v^{2}- u^{2}}{2a} \\[/tex]
Here, u is the initial velocity, a is the acceleration, s is the displacement, v is the final velocity.
Substitute all the variables in above equation.
[tex]s=\dfrac{7^{2}- 2^{2}}{2(1)}\\s=\dfrac{45}{2}\\ s=22.5m[/tex]
Therefore the distance of the snail is 22.5 m.
What is the electric flux through one side of a cube that has a single point charge of-2.80μC placed at its center? Hint: You do not need to integrate any equations to get the answer
Answer:
The flux through one face of a cube is [tex]- 5.2730\times 10^{4} Wb[/tex]
Given:
Charge, Q = [tex]- 2.80\micro C = -2.80\times 10^{- 6} C[/tex]
Solution:
The electric flux, [tex]\phi_{E} = \frac{Q}{\epsilon_{o}}[/tex]
Since, there are 6 faces in a cube, the flux through one face:
[tex]\phi_{E} = \frac{Q}{6\epsilon_{o}}[/tex]
[tex]\phi_{E} = \frac{-2.80\times 10^{- 6}}{6\epsilon_{o}}[/tex]
[tex]\phi_{E} = - 5.2730\times 10^{4} Wb[/tex]
What happens to the width of the central diffraction pattern (in the single slit experiment) as the slit width is changed and why?
Answer:
width of fringes are increased
Explanation:
The width of central maxima is given by the following expression
Width = 2 x Dλ / d
D is distance of screen from source , d is slit width and λ is wavelength of light source. Here we see , on d getting decreased , width will increase because d is in denominator .Due to increased width , position of a fringe moves away from the centre.
Find the critical angle for total internal reflection for a flint glass-air boundary (you may assume that λ = 580.0 nm). Express your answer to 4 significant figures!
Answer:
the critical angle of the flint glass is 37.04⁰
Explanation:
to calculate the critical angle for total internal reflection.
given,
wavelength of the flint glass = λ = 580.0 nm
= 580 × 10⁻⁹ m
critical angle = sin^{-1}(\dfrac{\mu_a}{\mu_g})
at the wavelength of 580.0 nm the refractive index of the glass is 1.66
refractive index of air = 1
critical angle = sin^{-1}(\dfrac{1}{1.66})
= 37.04⁰
hence, the critical angle of the flint glass is 37.04⁰
How long does it a take a runner, starting from rest to reach max speed, 30 f/s given acceleration 8 f/s^2? After finding the time, calculate the distance traveled in that time.
Explanation:
Given that,
Initial sped of the runner, u = 0
Final speed of the runner, v = 30 ft/s
Acceleration of the runner, [tex]a=8\ ft/s^2[/tex]
Let t is the time taken by the runner. It can be calculated using first equation of motion as :
[tex]t=\dfrac{v-u}{a}[/tex]
[tex]t=\dfrac{30-0}{8}[/tex]
t = 3.75 seconds
Let s is the distance covered by the runner. Using the second equation of motion as :
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
[tex]s=\dfrac{1}{2}\times 8\times (3.75)^2[/tex]
s = 56.25 feet
Hence, this is the required solution.
Final answer:
To reach a maximum speed of 30 f/s from rest with an acceleration of 8 f/s², it takes 3.75 seconds. During this time, the runner travels a distance of 56.25 feet.
Explanation:
The question involves calculating the time it takes for a runner to reach a maximum speed of 30 feet per second (f/s) from rest with an acceleration of 8 feet per second squared (f/s²), and then finding the distance traveled during this time. This can be solved using the basic kinematics equations.
Calculating Time to Reach Max Speed
To find the time, we use the equation v = at, where v is the final velocity (30 f/s), a is the acceleration (8 f/s²), and t is the time. Rearranging the equation to solve for t, we get t = v/a. Plugging in the values, t = 30 f/s / 8 f/s² = 3.75 seconds.
Calculating Distance Traveled
To find the distance traveled, we use the equation d = 0.5 * a * t², where d is the distance, a is the acceleration, and t is the time. Substituting the given values, d = 0.5 * 8 f/s² * (3.75 s)² = 56.25 feet.
A piece of glass of index of refraction 1.50 is coated with a thin layer of magnesium fluoride of index of refraction 1.38. It is illuminated with light of wavelength 680 nm. Determine the minimum thickness of the coating that will result in no reflection
Answer:
Thickness = 123.19 nm
Explanation:
Given that:
The refractive index of the glass = 1.50
The refractive index of thin layer of magnesium fluoride = 1.38
The wavelength of the light = 680 nm
The thickness can be calculated by using the formula shown below as:
[tex]Thickness=\frac {\lambda}{4\times n}[/tex]
Where, n is the refractive index of thin layer of magnesium fluoride = 1.38
[tex]{\lambda}[/tex] is the wavelength
So, thickness is:
[tex]Thickness=\frac {680\ nm}{4\times 1.38}[/tex]
Thickness = 123.19 nm
Assume everyone in the United States consumes one soft drink in an aluminum can every two days. if there are 280 million americans, how many tons of aluminum need to be recycles each year if each can weight 1/15 pound and once ton=2000 pounds?
Answer:
1.708*10^6 tons.
Explanation:
Number of Aluminum cans used by 1 person in 1 year = 365/2=182.5 say it as 183 cans per year.
Total number of people in US= 280,000,000
Total number of cans used by americans.
[tex] = 5.12×10^10 cans[/tex]
Weight of 1 can =1/15 pounds
Weight of all cans used in 1 year
[tex]= \frac{5.12*10^10}{15} =3.41*10^9pounds.[/tex]
we know that
1ton=2000pounds.
[tex]\frac{3.41*10^9 pounds}{2000} = 1.708*10^6 tons.[/tex]
A certain elevator cab has a total run of 218 m and a maximum speed is 319 m/min, and it accelerates from rest and then back to rest at 1.20 m/s^2. (a) How far does the cab move while accelerating to full speed from rest? (b) How long does it take to make the nonstop 218 m run, starting and ending at rest?
Answer:
a)11.6m
b)45.55s
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
X=(Vf+Vo)T/2 (4)
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
a)
for this problem
Vo=0
Vf=319m/min=5.3m/s
a=1.2m/s^2
we can use the ecuation number 1 to calculate the time
t=(Vf-Vo)/a
t=(5.3-0)/1.2=4.4s
then we use the ecuation number 3 to calculate the distance
X=0.5at^2
X=0.5x1.2x4.4^2=11.6m
b)second part
We know that when the elevator starts to accelerate and decelerate, it takes a distance of 11.6m and a time of 4.4s, which means that if the distance is subtracted 2 times this distance (once for acceleration and once for deceleration)
we will have the distance traveled in with constant speed.
With this information we will find the time, and then we will add it with the time it takes for the elevator to accelerate and decelerate
X=218-11.6x2=194.8m
X=VT
T=X/v
t=194.8/5.3=36.75s
Total time=36.75+2x4.4=45.55s
A person drives a car around a circular cloverleaf with a radius of 63 m at a uniform speed of 10 m/s. (a) What is the acceleration of the car? (b) Compare this answer with the acceleration due to gravity as a percentage?
Answer:
acceleration is 1.59 m/s²
required % is 16.22 %
Explanation:
given data
radius = 63 m
speed = 10 m/s
to find out
acceleration of the car and acceleration due to gravity
solution
we know car is moving in circular path
so acceleration will be, a = [tex]\frac{v^2}{r}[/tex] .........1
put here value
a = [tex]\frac{10^2}{63}[/tex]
a = 1.59 m/s²
so acceleration is 1.59 m/s²
so acceleration due to gravity as a percentage is 9.8 m/s²
required % = [tex]\frac{1.59}{9.8}*100[/tex]
required % = 16.22 %
Meredith walks from her house to a bus stop that is 260 yards away. If Meredith is 29 yards from her house, how far is she from the bus stop? 231 Correct yards If Meredith is 204.8 yards from her house, how far is she from the bus stop? 55.2 Correct yards Let the variable x represent Meredith's varying distance from her house (in yards). As Meredith walks from her house to the bus stop, the value of x varies from 0 Correct to 260 Correct . How many values does the variable x assume as Meredith walks from her house to the bus stop? 3 Incorrect
Answer:
a) 231 yards
b) 55.2 yards
c) 0 yards to 260 yards
d) Infinite values
Explanation:
This situation can be described as a horizontal line that begins at point [tex]P_{1}=0 yards[/tex] (Meredith's house) and ends at point [tex]P_{2}=260 yards[/tex] (Bus stop). Where [tex]x[/tex] is the varying distance from her house, which can be calculated in the following way:
x=Final Position - Initial Position
or
[tex]x=x_{f} - x_{i}[/tex]
a) For the first case Meredith is at position [tex]x_{i}=29 y[/tex] and the bus stop at position [tex]x_{f}=260 y[/tex]. So the distance Meredith is from the bus stop is:
[tex]x=260 y - 29 y=231 y[/tex]
b) For the second case the initial position is [tex]x_{i}=204.8 y[/tex] and the final position [tex]x_{f}=260 y[/tex]. Hence:
[tex]x=260 y - 204.8 y=55.2 y[/tex]
c) If we take Meredith's initial position at her house [tex]x_{i}=0 y[/tex] and her final position at the bus stop [tex]x_{f}=260 y[/tex], the value of [tex]x[/tex] varies from 0 yards to 260 yards.
d) As Meredith walks from her house to the bus stop, the variable [tex]x[/tex] assumes infinite values, since there are infinite position numbers from [tex]x=0 yards[/tex] to [tex]x=260 yards[/tex]
The answers to the possible distance covered by Meredith at the various distances from her house are;
A) distance = 231 yards
A) distance = 231 yardsB) distance = 55.2 yards
A) distance = 231 yardsB) distance = 55.2 yardsC) x will vary from 0 m to 260 m i.e 0 ≤ x ≤ 260
A) We are told that meredith walks from her house to a bus stop that is 260 yards away.
After walking, she is now 29 yards from her house. This means that she has walked a total of 29 yards from her house.
Distance left to reach bus stop = 260 - 29 = 231 yards
B) We are told that Meredith is now 204.8 yards from her house. This means that she has walked a total of 204.8 yards from here house. Thus;
Distance left to reach bus stop = 260 - 204.8 = 55.2 yards.
C) This question is basically asking for all the possible values that Meredith could have walked from her house to the bus stop.
Since she starts from her house at 0m, then it means that if the bus stop is 260 m away, then if x is the possible distance, we can say that x will vary from 0 m to 260 m i.e 0 ≤ x ≤ 260
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A golfer hits a shot to a green that is elevated 3.20 m above the point where the ball is struck. The ball leaves the club at a speed of 18.1 m/s at an angle of 49.0° above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.
Answer:
16.17 m/s
Explanation:
h = 3.2 m
u = 18.1 m/s
Angle of projection, θ = 49°
Let H be the maximum height reached by the ball.
The formula for the maximum height is given by
[tex]H=\frac{u^{2}Sin^{2}\theta }{2g}[/tex]
[tex]H=\frac{18.1^{2}\times Sin^{2}49 }{2\times 9.8}=9.52 m[/tex]
The vertical distance fall down by the ball, h' H - h = 9.52 - 3.2 = 6.32 m
Let v be the velocity of ball with which it strikes the ground.
Use third equation of motion for vertical direction
[tex]v_{y}^{2}=u_{y}^{2}+2gh'[/tex]
here, uy = 0
So,
[tex]v_{y}^{2}=2\times 9.8 \times 6.32[/tex]
vy = 11.13 m/s
vx = u Cos 49 = 18.1 x 0.656 = 11.87 m/s
The resultant velocity is given by
[tex]v=\sqrt{v_{x}^{2}+v_{y}^{2}}[/tex]
[tex]v=\sqrt{11.87^{2}+11.13^{2}}[/tex]
v = 16.27 m/s
How do resistors in series affect the total resistance?
Answer:
Explanation:
Resistance in series is given by the sum of all the resistor in series
value of Total Resistance is given by
[tex]R_{th}=R_1+R_2+R_3+R_4+..............R_n[/tex]
Where [tex]R_{th}[/tex] is the total resistance
[tex]R_1,R_2[/tex] are the resistance in series
Current in series remains same while potential drop is different for different resistor
The value of net resistor is always greater than the value of individual resistor.
If a there is a defect in a single resistor then it affects the whole circuit in series.
Which of the following is not a unit of torque? O pound-foot 0 kilogram-newton Newton-meter O pound-inch
Answer: Kilogram- newton is wrong unit for Torque
Idem pound-inch is also wrong unit for Torque
Explanation: As it is well known the torque is defined as :
Τorque= F x R
so its UNITS are Newton*meter (SI)
or in the Imperial System is often use Pound-foot
If a marathon runner averages 9.39 mi/h, how long does it take him or her to run a 26.22-mi marathon? Express your answers in h, min and s.
Answer:
[tex]t=2.8h[/tex]
[tex]t=10080s[/tex]
[tex]t=168 min[/tex]
Explanation:
From this exercise we have velocity and distance. Using the following formula, we can calculate time:
[tex]v=\frac{d}{t}[/tex]
Solving for t
[tex]t=\frac{d}{v}=\frac{26.22mi}{9.39mi/h} =2.8h[/tex]
[tex]t=2.8h*\frac{3600s}{1h} =10080s[/tex]
[tex]t=2.8h*\frac{60min}{1h} =168min[/tex]
Recent findings on the topic of brain based research indicate all of the following except
Recent findings on the topic of brain-based research indicate all of the following except
A. early environments matter.
B. all children are born ready to learn.
C. society isn't addressing the needs of young children. ?
D. the brain stops growing at around age two
A package is dropped from an airplane flying horizontally with constant speed V in the positive xdirection. The package is released at time t = 0 from a height H above the origin. In addition to the vertical component of acceleration due to gravity, a strong wind blowing from the right gives the package a horizontal component of acceleration of magnitude ¼g to the left. Derive an expression for the horizontal distance D from the origin where the package hits the ground.
Answer:
[tex]D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}[/tex]
Explanation:
From the vertical movement, we know that initial speed is 0, and initial height is H, so:
[tex]Y_{f}=Y_{o}-g*\frac{t^{2}}{2}[/tex]
[tex]0=H-g*\frac{t^{2}}{2}[/tex] solving for t:
[tex]t=\sqrt{\frac{2H}{g} }[/tex]
Now, from the horizontal movement, we know that initial speed is V and the acceleration is -g/4:
[tex]X_{f}=X_{o}+V*t+a*\frac{t^{2}}{2}[/tex] Replacing values:
[tex]D=V*\sqrt{\frac{2H}{g} }-\frac{g}{4}*\frac{1}{2} *(\sqrt{\frac{2H}{g} })^{2}[/tex]
Simplifying:
[tex]D=V*\sqrt{\frac{2H}{g} } -\frac{H}{4}[/tex]
If the car’s speed decreases at a constant rate from 64 mi/h to 30 mi/h in 3.0 s, what is the magnitude of its acceleration, assuming that it continues to move in a straight line? What distance does the car travel during the braking period?
Answer:[tex]3.874 m/s^2[/tex]
Explanation:
Given
Car speed decreases at a constant rate from 64 mi/h to 30 mi/h
in 3 sec
[tex]60mi/h \approx 26.8224m/s[/tex]
[tex]34mi/h \approx 15.1994 m/s[/tex]
we know acceleration is given by [tex]=\frac{velocity}{Time}[/tex]
[tex]a=\frac{15.1994-26.8224}{3}[/tex]
[tex]a=-3.874 m/s^2[/tex]
negative indicates that it is stopping the car
Distance traveled
[tex]v^2-u^2=2as[/tex]
[tex]\left ( 15.1994\right )^2-\left ( 26.8224\right )^2=2\left ( -3.874\right )s[/tex]
[tex]s=\frac{488.419}{2\times 3.874}[/tex]
s=63.038 m
Comment on why the acceleration due to gravity is less for the plastic ball. Why do the other two balls (steel ball and golf ball) not have such a low value for the acceleration?
Answer and Explanation:
The gravitational acceleration 'g' depend directly on the mass of the object or body and inversely on the distance or radius squared:
[tex]g = \frac{Gm_{o}}{R^{2}}[/tex]
where
[tex]m_{o}[/tex] = mass of the object
G = Gravitational constt
Thus the plastic ball is lighter and have low mass as compared to the steel and golf balls.
This is the reason that a plastic ball have a low value of acceleration as compared to that of steel and golf balls with higher values of acceleration.
Final answer:
The acceleration due to gravity is a constant (g) for all objects in the same gravitational field, and any observed difference in falling rates is likely due to air resistance, not gravitational pull. Experiments have confirmed that objects fall at the same rate regardless of their mass or composition, assuming no air resistance.
Explanation:
The acceleration due to gravity should not vary with the material of the object if we ignore effects such as air resistance. This is because, according to Newton's second law, the force acting on an object is the product of its mass and the acceleration (F = ma). Here, the force due to gravity is mg, where m is the mass and g is the acceleration due to gravity. Since a = F/m, the mass cancels out, leaving a = g, which is constant for all objects in the same gravitational field.
The question assumes that the plastic ball has a lower acceleration due to gravity, which contradicts known physics principles. All objects, regardless of their mass or composition, feel the same acceleration due to gravity near the Earth's surface, assuming no other forces, like air resistance, play a significant role. Historical experiments by scientists such as Galileo and Eötvös have confirmed the equality of gravitational acceleration (g) for different substances within exceptionally high precision.
If you observe differing acceleration rates, this can be often attributed to air resistance, not a difference in gravitational pull. Objects with a larger surface area or less aerodynamic shape, like the plastic ball, may experience greater air resistance and thus appear to fall slower, even though their acceleration due to gravity is the same as denser objects like the steel or golf ball.
A car is making a 40 mi trip. It travels the first half of the total distance 20.0 mi at 18.00 mph and the last half of the total distance 20.0 mi at 56.00 mph. What is the car’s average speed in mph for the entire second trip?
Answer: The average speed is 27,24 mph (exactly 1008/37 mph)
Explanation:
This is solved using a three rule: We know the speeds and the distances, what we can obtain from it is the time used. It is done like this:
1h--->18mi
X ---->20 mi, then X=20mi*1h/18mi= 10/9 h=1,111 h
1h--->56mi
X ---->20 mi, then X=20mi*1h/56mi= 5/14 h=0,35714 h
Then the average speed is calculated by taking into account that it was traveled 40mi and the time used was 185/126 h=1,468 h and since speed is distance over time we get the answer. Average speed= 40mi/(185/126 h)=1008/37 mph=27,24 mph.
Two tiny conducting spheres are identical and carry charges of -19.8μC and +40.7μC. They are separated by a distance of 3.59 cm. (a) What is the magnitude of the force that each sphere experiences? (b) The spheres are brought into contact and then separated to a distance of 3.59 cm. Determine the magnitude of the force that each sphere now experiences.
Answer:
(a): [tex]\rm -5.627\times 10^3\ N.[/tex]
(b): [tex]\rm 7.626\times 10^2\ N.[/tex]
Explanation:
Given:
Charge on one sphere, [tex]\rm q_1 = -19.8\ \mu C = -19.8\times 10^{-6}\ C.[/tex]Charge on second sphere, [tex]\rm q_2 = +40.7\ \mu C = +40.7\times 10^{-6}\ C.[/tex]Separation between the spheres, [tex]\rm r=3.59\ cm = 3.59\times 10^{-2}\ m.[/tex]Part (a):
According to Coulomb's law, the magnitude of the electrostatic force of interaction between two static point charges is given by
[tex]\rm F=k\cdot\dfrac{q_1q_2}{r^2}[/tex]
where,
k is called the Coulomb's constant, whose value is [tex]\rm 9\times 10^9\ Nm^2/C^2.[/tex]
From Newton's third law of motion, both the spheres experience same force.
Therefore, the magnitude of the force that each sphere experiences is given by
[tex]\rm F=k\cdot\dfrac{q_1q_2}{r^2}\\=9\times 10^9\times \dfrac{(-19.8\times 10^{-6})\times (+40.7\times 10^{-6})}{(3.59\times 10^{-2})^2}\\=-5.627\times 10^3\ N.[/tex]
The negative sign shows that the force is attractive in nature.
Part (b):
The spheres are identical in size. When the spheres are brought in contact with each other then the charge on both the spheres redistributes in such a way that the net charge on both the spheres distributed equally on both.
Total charge on both the spheres, [tex]\rm Q=q_1+q_2=-19.8\ \mu C+40.7\ \mu C = 20.9\ \mu C.[/tex]
The new charges on both the spheres are equal and given by
[tex]\rm q_1'=q_2'=\dfrac Q2 = \dfrac{20.9}{2}\ \mu C=10.45\ \mu C = 10.45\times 10^{-6}\ C.[/tex]
The magnitude of the force that each sphere now experiences is given by
[tex]\rm F'=k\cdot \dfrac{q_1'q_2'}{r^2}'\\=9\times 10^9\times \dfrac{10.45\times 10^{-6}\times 10.45\times 10^{-6}}{(3.59\times 10^{-2})^2}\\=7.626\times 10^2\ N.[/tex]
Lightning can be studied with a Van de Graaff generator, which consists of a spherical dome on which charge is continuously deposited by a moving belt. Charge can be added until the electric field at the surface of the dome becomes equal to the dielectric strength of air. Any more charge leaks off in sparks. Assume the dome has a diameter of 43.0 cm and is surrounded by dry air with a "breakdown" electric field of 3.00 10^6 V/m. (a) What is the maximum potential of the dome?
(b) What is the maximum charge on the dome?
The maximum potential is [tex]6.45 \times 10^5\ V[/tex] and the maximum charge is [tex]71 \mu C[/tex].
The electric field is the gradient of the potential. So, the electric field in terms of the potential is given as:
[tex]V=E \times r[/tex]
Here, V is the potential, E is the electric field and r is the distance.
Given:
Diameter of the dome, [tex]d=43\ cm[/tex]
Maximum electric field, [tex]E=3.00 \times10^6\ V/m[/tex]
(a) The maximum potential is computed as:
[tex]V_{max}={E}{d/2}\\V_{max}= {3.00 \times 10^6} \times \frac{0.43}{2}\\V_{max}= 6.45 \times 10^5\ V[/tex]
(b) The maximum charge is computed as:
[tex]Q= \frac{Er}{k}\\Q= \frac{3.00 \times 10^6 \times 0.215}{9 \times 10^9}\\Q=71 \mu C[/tex]
Therefore, the maximum potential is [tex]6.45 \times 10^5\ V[/tex] and the maximum charge is [tex]71 \mu C[/tex].
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The maximum potential of the Van de Graaff generator with a diameter of 43.0 cm is 645 kV, and the maximum charge it can hold, before sparks occur due to air breakdown, is 5.11 μC.
Explanation:The maximum potential of a Van de Graaff generator can be calculated using the formula V = (kQ)/r, where V is the potential, k is Coulomb's constant, Q is the charge, and r is the radius of the sphere. Given that the Van de Graaff generator has a diameter of 43.0 cm (which means the radius r is 21.5 cm or 0.215 m) and it is surrounded by air with a breakdown electric field of 3.00 x 106 V/m, the maximum surface potential (V) equals the breakdown electric field (E) times the radius (r), which gives V = E × r = 3.00 x 106 V/m × 0.215 m = 645,000 V or 645 kV.
The maximum charge on the dome can be calculated using the relationship between the electric field and charge on a sphere, which is E = (kQ)/(r2). Re-arranging this for Q gives Q = Er2/k. Inserting the known values, we get Q = (3.00 x 106 V/m) × (0.215 m)2 / (8.988 x 109 N·m2/C2) = 5.11 x 10−5 C or 5.11 μC, where k is Coulomb's constant.