Answer:
ΔV = ±0.175 cm
Explanation:
The equation for volume is
V = π/4 * d^2 * h
All the measurements are multiplied. To propagate uncertainties in multiplication we add the relative uncertainties together.
The relative uncertainty of the diameter is:
εd = Δd/d
εd = 0.0005/5.1 = 0.000098
The relative uncertainty of the height is:
εh = Δh/h
εh = 0.005/37.6 = 0.00013
Then, the relative uncertainty of the volume is:
εV = 2 * εd + εh
εV = 2 * 0.000098 + 0.00013 = 0.000228
Then we get the absolute uncertainty of the volume, for that we need the volume:
V = π/4 * 5.1^2 * 37.6 = 768.1 cm^3
So:
ΔV = ±εV * V
ΔV = ±0.000228 * 768.1 = ±0.175 cm
What is the magnitude of the electric field at a distance of 89 cm from a 27 μC charge, in units of N/C?
Answer:
306500 N/C
Explanation:
The magnitude of an electric field around a single charge is calculated with this equation:
[tex]E(r) = \frac{1}{4 \pi *\epsilon 0} \frac{q}{r^2}[/tex]
With ε0 = 8.85*10^-12 C^2/(N*m^2)
Then:
[tex]E(0.89) = \frac{1}{4 \pi *8.85*10^-12} \frac{27*10^-6}{0.89^2}[/tex]
E(0.89) = 306500 N/C
A typical radio wave has a period of 1.4 microseconds. Express this period in seconds. Answer in units of s.
Answer:
In second time period will be [tex]1.4\times 10^{-6}sec[/tex]
Explanation:
We have given the time period of wave [tex]T=1.4microsecond[/tex]
We have to change this time period in unit of second
We know that 1 micro sec [tex]10^{-6}sec[/tex]
We have to change 1.4 micro second
To change the time period from micro second to second we have to multiply with [tex]10^{-6}[/tex]
So [tex]1.4microsecond =1.4\times 10^{-6}sec[/tex]
Two particles having charges of 0.440 nC and 11.0 nC are separated by a distance of 1.80 m . 1) At what point along the line connecting the two charges is the net electric field due to the two charges equal to zero?
2) Where would the net electric field be zero if one of the charges were negative?
Answer:
a) 0.3 m
b) r = 0.45 m
Explanation:
given,
q₁ = 0.44 n C and q₂ = 11.0 n C
assume the distance be r from q₁ where the electric field is zero.
distance of point from q₂ be equal to 1.8 -r
now,
E₁ = E₂
[tex]\dfrac{K q_2}{(1.8-r)^2} = \dfrac{K q_1}{r^2}[/tex]
[tex](\dfrac{1.8-r}{r})^2= \dfrac{q_2}{q_1}[/tex]
[tex]\dfrac{1.8-r}{r}= \sqrt{\dfrac{11}{0.44}}[/tex]
1.8 = 6 r
r = 0.3 m
b) zero when one charge is negative.let us assume q₁ be negative so, distance from q₁ be r
from charge q₂ the distance of the point be 1.8 +r
now,
E₁ = E₂
[tex]\dfrac{K q_2}{(1.8+r)^2} = \dfrac{K q_1}{r^2}[/tex]
[tex](\dfrac{1.8+r}{r})^2= \dfrac{q_2}{q_1}[/tex]
[tex]\dfrac{1.8+r}{r}= \sqrt{\dfrac{11}{0.44}}[/tex]
1.8 =4 r
r = 0.45 m
At the instant the traffic light turns green, a car starts with a constant acceleration of 3.00 ft/s^2. At the same instant a truck, traveling with a constant speed of 70.0 ft/s, overtakes and passes the car. How far from the starting point (in feet) will the car overtake the truck?
Answer:
The car overtakes the truck at a distance d = 3266.2ft from the starting point
Explanation:
Problem Analysis
When car catches truck:
dc = dt = d
dc: car displacement
dt: truck displacement
tc = tt = t
tc: car time
tt : truck time
car kinematics :
car moves with uniformly accelerated movement:
d = vi*t + (1/2)a*t²
vi = 0 : initial speed
d = (1/2)*a*t² Equation (1)
Truck kinematics:
Truck moves with constant speed:
d = v*t Equation (2)
Data
We know that the acceleration of the car is 3.00 ft / s² and the speed of the truck is 70.0 ft / s .
Development problem
Since the distance traveled by the car is equal to the distance traveled by the truck and the time elapsed is the same for both, then we equate equations (1 ) and (2)
Equation (1) = Equation (2)
(1/2)*a*t² = v*t
(1/2)*3*t² = 70*t (We divide both sides by t)
1.5*t = 70
t = 70 ÷ 1.5
t = 46.66 s
We replace t = 46.66 s in equation (2) to calculate d:
d = 70*46.66 = 3266.2ft
d = 3266.2 ft
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]
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
You are camping with two friends, Joe and Karl. Since all three of you like your privacy, you don't pitch your tents close together. Joe's tent is 19.0 m from yours, in the direction 19.0° north of east. Karl's tent is 45.0 m from yours, in the direction 39.0° south of east. What is the distance between Karl's tent and Joe's tent?
Answer:
Distance between Karl and Joe is 38.467 m
Solution:
Let us assume that you are at origin
Now, as per the question:
Joe's tent is 19 m away from yours in the direction [tex]19.0^{\circ}[/tex] north of east.
Now,
Using vector notation for Joe's location, we get:
[tex]\vec{r_{J}} = 19cos(19.0^{\circ})\hat{i} + 19sin(19.0^{\circ})\hat{j}[/tex]
[tex]\vec{r_{J}} = 17.96\hat{i} + 6.185\hat{j} m[/tex]
Now,
Karl's tent is 45 m away from yours and is in the direction [tex]39.0^{\circ}[/tex]south of east, i.e., [tex]- 39.0^{\circ}[/tex] from the positive x-axis:
Again, using vector notation for Karl's location, we get:
[tex]\vec{r_{K}} = 45cos(-319.0^{\circ})\hat{i} + 45sin(- 39.0^{\circ})\hat{j}[/tex]
[tex]\vec{r_{K}} = 34.97\hat{i} - 28.32\hat{j} m[/tex]
Now, obtain the vector difference between [tex]\vec{r_{J}}[/tex] and [tex]\vec{r_{K}}[/tex]:
[tex]\vec{r_{K}} - \vec{r_{J}} = 34.97\hat{i} - 28.32\hat{j} - (17.96\hat{i} + 6.185\hat{j}) m[/tex]
[tex]\vec{d} = \vec{r_{K}} - \vec{r_{J}} = 17.01\hat{i} - 34.51\hat{j} m[/tex]
Now, the distance between Karl and Joe, d:
|\vec{d}| = |17.01\hat{i} - 34.51\hat{j}|
[tex]d = \sqrt{(17.01)^{2} + (34.51)^{2}} m[/tex]
d = 38.469 m
The distance between Karl's and Joe's tent is:
The distance between Joe's tent and Karl's tent is approximately 36.84 m.
Explanation:To find the distance between Joe's tent and Karl's tent, we can use the concept of vector addition. We first need to break down the given distances and angles into their respective components:
Joe's tent: 19.0 m at 19.0° north of east Karl's tent: 45.0 m at 39.0° south of east
Next, we can use the components to find the displacement from Joe's tent to Karl's tent:
For Joe's tent: North component = 19.0 m * sin(19.0°) = 6.36 m, East component = 19.0 m * cos(19.0°) = 17.88 m For Karl's tent: North component = -45.0 m * sin(39.0°) = -27.10 m, East component = 45.0 m * cos(39.0°) = 34.37 m
Using the components, we can calculate the displacement from Joe's tent to Karl's tent:
North displacement = -27.10 m - 6.36 m = -33.46 m East displacement = 34.37 m - 17.88 m = 16.49 m
Finally, we can use the Pythagorean theorem to find the magnitude of the displacement:
Magnitude = sqrt((-33.46 m)^2 + (16.49 m)^2) = 36.84 m
Therefore, the distance between Joe's tent and Karl's tent is approximately 36.84 m.
You slide a coffee mug across the table. The mug slides to the east and slows down while sliding. Which of the following statements best describes the net force acting on it? O The net force is zero. O The net force is directed east. O The net force is directed west. O There is not enough information to determine the direction of the net force.
Answer:
Explanation:
The mug is sliding towards the east but its velocity is going down . That means it has negative acceleration towards east . In other words , it has positive acceleration towards west. Since it has positive acceleration towards
west , it must have positive force acting on it towards west.
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
String linear mass density is defined as mass/unit length. Calculate the linear mass density in kg/m of a string with mass 0.3g and 1.5m length?
Answer:
Linear mass density,[tex]\lambda=2\times 10^{-4}\ kg/m[/tex]
Explanation:
Given that,
Mass of the string, m = 0.3 g = 0.0003 kg
Length of the string, l = 1.5 m
The linear mass density of a string is defined as the mass of the string per unit length. Mathematically, it is given by :
[tex]\lambda=\dfrac{m}{l}[/tex]
[tex]\lambda=\dfrac{0.0003\ kg}{1.5\ m}[/tex]
[tex]\lambda=0.0002\ kg/m[/tex]
or
[tex]\lambda=2\times 10^{-4}\ kg/m[/tex]
So, the linear mass density of a string is [tex]2\times 10^{-4}\ kg/m[/tex]. Hence, this is the required solution.
To calculate the linear mass density of the string, you divide the mass in kilograms by the length in meters. For a 0.3g and 1.5m string, this yields a linear mass density of 0.0002 kg/m.
The student asks how to calculate the linear mass density (μ) of a string. The linear mass density is defined as mass per unit length. To calculate it for a given string, you divide the mass of the string by its length. The provided string has a mass of 0.3g, which should be converted to kilograms (0.0003 kg), and a length of 1.5m.
The formula to calculate linear mass density is:
μ = mass/length.
Therefore, the linear mass density of the string is:
μ = 0.0003 kg / 1.5 m = 0.0002 kg/m.
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
A baseball thrown from the outfield is released form
shoulderheight at an initial velocity of 29.4 m/s at an initial
angle of30.0 with respect to the horizontal. What is the maximum
verticaldisplacement that the ball reaches during
itstrajectory?
Answer:
[tex]y_{max}=11m[/tex]
Explanation:
The maximum vertical displacement that the ball reaches can be calculate using the following formula:
[tex]v^{2}=v^{2} _{o}+2g(y-y_{o})[/tex]
At the highest point, its velocity becomes 0 because it stop going up and starts going down.
[tex]0=(29.4sin(30))^{2} -2(9.8)y[/tex]
Solving for y
[tex]y=\frac{(29.4sin(30))^{2}}{2(9.8)} =11m[/tex]
A lens with f = +11cm is paired with a lens with f = −25cm. What is the focal length of the combination?
Answer:
19.642 cm
Explanation:
f₁ = Focal length of first lens = 11 cm
f₂ = Focal length of second lens = -25 cm
Combined focal length formula
[tex]\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}\\\Rightarrow \frac{1}{f}=\frac{1}{11}+\frac{1}{-25}\\\Rightarrow \frac{1}{f}=\frac{14}{275}\\\Rightarrow f=\frac{275}{14}\\\Rightarrow f=19.642\ cm[/tex]
Combined focal length is 19.642 cm
The focal length of the combination of lenses is approximately 19.64 cm.
To find the focal length of the combination of lenses in this scenario, we need to use the lensmaker's formula for thin lenses in combination.
Given:
Focal length of lens 1, [tex]\( f_1 = +11 \)[/tex] cm
Focal length of lens 2, [tex]\( f_2 = -25 \)[/tex] cm
The formula for the focal length of two thin lenses in contact is given by:
[tex]\[ \frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2} \][/tex]
[tex]\[ \frac{1}{f} = \frac{1}{+11 \text{ cm}} + \frac{1}{-25 \text{ cm}} \][/tex]
[tex]\[ \frac{1}{f} = \frac{1}{11} - \frac{1}{25} \][/tex]
To subtract the fractions, find a common denominator, which is 275:
[tex]\[ \frac{1}{f} = \frac{25}{275} - \frac{11}{275} \][/tex]
[tex]\[ \frac{1}{f} = \frac{14}{275} \][/tex]
Now, invert both sides to solve for (f):
[tex]\[ f = \frac{275}{14} \][/tex]
[tex]\[ f \approx 19.64 \text{ cm} \][/tex]
A car is traveling at a speed of 38.0 m/s on an interstate highway where the speed limit is 75.0 mi/h. Is the driver exceeding the speed limit? Justify your answer.
Answer: Yes, he is exceeding the speed limit
Explanation:
Hi!
This is problem about unit conversion
1 mile = 1,609.344 m
Then the speed limit v is:
v = 75 mi/h = 120,700.8 m/h
1 hour = 60 min = 60*60 s = 3,600 s
v = (120,700.8/3,600) m/s = 33.52 m/s
38 m/s is higher than the speed limit v.
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.
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]
In a Young's two-slit experiment it is found that an nth-order maximum for a wavelength of 680.0 nm coincides with the (n+1)th maximum of light of wavelength 510.0nm. Determine n.
Answer:
n = 3
Solution:
Since, the slit used is same and hence slit distance 'x' will also be same.
Also, the wavelengths coincide, [tex]\theta [/tex] will also be same.
Using Bragg's eqn for both the wavelengths:
[tex]xsin\theta = n\lambda[/tex]
[tex]xsin\theta = n\times 680.0\times 10^{- 9}[/tex] (1)
[tex]xsin\theta = (n + 1)\lambda[/tex]
[tex]xsin\theta = (n + 1)\times 510.0\times 10^{- 9}[/tex] (2)
equate eqn (1) and (2):
[tex] n\times 680.0\times 10^{- 9} = (n + 1)\times 510.0\times 10^{- 9}[/tex]
[tex]n = \frac{510.0\times 10^{- 9}}{680.0\times 10^{- 9} - 510.0\times 10^{- 9}}[/tex]
n = 3
Final answer:
Using the formula for the location of maxima in a double slit interference pattern and equating the equations for the two different wavelengths, we find that the value of n is 3 for this Young's two-slit experiment scenario.
Explanation:
To determine the value of n such that an nth-order maximum for a wavelength of 680.0 nm coincides with the (n+1)th maximum of light of wavelength 510.0 nm in a Young's two-slit experiment, we can use the formula for the location of maxima in a double slit interference pattern:
dsinθ = mλ, where d is the separation of the slits, sinθ is the sine of the angle of the maxima, m is the order of the maximum, and λ is the wavelength of the light.
For two wavelengths to coincide, we equate the two equations:
nλ1 = (n+1)λ2
Substituting the given wavelengths:
n(680.0 nm) = (n+1)(510.0 nm)
Solving for n gives:
n = 510.0 / (680.0 - 510.0) = 3
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
Read more at; https://brainly.com/question/13242055
The driver of a sports car traveling at 10.0m/s steps down hard on the accelerator for 5.0s and the velocity increases to 30.0m/s. What was the average acceleration of the car during the 5.0s time interval?
Answer:
[tex]a=4m/s^{2}[/tex]
Explanation:
From the concept of average acceleration we know that
[tex]a=\frac{v_{2}-v_{1} }{t_{2}-t_{1} }[/tex]
From the exercise we know that
[tex]v_{2}=30m/s\\v_{1}=10m/s\\t_{2}=5s\\t_{1}=0s[/tex]
So, the average acceleration of the car is:
[tex]a=\frac{30m/s-10m/s}{5s}=4m/s^{2}[/tex]
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.
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 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 ball is dropped from rest from the top of a building, which is 106 m high. The magnitude of the gravitational acceleration g = 9.8 m/s2 Keep 2 decimal places in all answers. In this problem, the following setup is convenient: Take the initial location of ball (the top of the building) as origin x0 = 0 Take DOWNWARD as +x (a) How far (in meters) does the ball fall in the first 3 s ?
Answer:
44.1 m
Explanation:
initial velocity of ball, u = 0
height of building, H = 106 m
g = 9.8 m/s^2
t = 3 second
Let the ball travels a distance of h in first 3 seconds.
Use second equation of motion
[tex]s=ut+\frac{1}{2}at^{2}[/tex]
h = 0 + 0.5 x 9.8 x 3 x 3
h = 44.1 m
Thus, the distance traveled by the ball in first 3 seconds is 44.1 m.
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.
If the Sun were to disappear or somehow radically change
itsoutput, how long would it take for us on Earth to learn
aboutit?
The answer is If the Sun were to suddenly disappear or change its output, it would take 8 minutes for us on Earth to notice, due to the time it takes light and gravitational force to travel from the Sun to Earth. The Sun is approximately 1.50×10¹¹ meters away, and light travels at about 300,000 kilometers per second. Therefore, we see the Sun as it was 8 minutes ago.
If the Sun were to suddenly disappear or change its output, it would take 8 minutes for us on Earth to notice any change. This is because the Sun is approximately 1.50×10¹¹ meters away, and light, which travels at the speed of about 300,000 kilometers per second, takes 8 minutes to travel from the Sun to Earth.
Therefore, for 8 minutes, we would continue to see the Sun as it was before the change or disappearance occurred.
Similarly, if there were an immediate change in the gravitational force due to the Sun’s disappearance, it would also take 8 minutes for Earth to experience the effect because gravitational information, like light, propagates at the speed of light.
Hence, The answer is If the Sun were to suddenly disappear or change its output, it would take 8 minutes for us on Earth to notice
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
Find the force of attraction between a proton and an electron separated by a distance equal to the radius of the smallest orbit followed by an electron (5 x 10^-11 m) in a hydrogen atom
Answer:
The answer is [tex] -9.239 \times 10^{-8}\ N = [/tex]
Explanation:
The definition of electric force between two puntual charges is
[tex]F_e = \frac{K q_1 q_2}{d^2}[/tex]
where
[tex]K = 9 \times 10^9\ Nm^2/C^2[/tex].
In this case,
[tex]q_1 = e = 1.602\times 10^{-19}\ C[/tex],
[tex]q_2 = -e = -1.602\times 10^{-19}\ C[/tex]
and
[tex]d = 5 \times 10^{-11}\ m[/tex].
So the force is
[tex]F_e = -9.239 \times 10^{-8}\ N [/tex]
where the negative sign implies force of attraction.
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.
At approximately what wavelength of the continuous spectrum will the greatest (maximum) intensity occur when 60-kV electrons strike an copper (Cu) target?
Answer: 20 pm=20*10^-12 m
Explanation: To solve this problem we have to use the relationship given by:
λmin=h*c/e*ΔV= 1240/60000 eV=20 pm
this expression is related with the bremsstrahlung radiation when a flux of energetic electrons are strongly stopped hitting to a catode. The electrons give their kinetic energy to the atoms of the catode.
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.