Answer:
angular acceleration = - 217.5 rad/s²
Explanation:
given data
angular speed = 3600 rev/min
rotate = 50 revolution
to find out
angular acceleration
solution
we know here no of rotation n = 3600 rev/min i.e 60 rev/s
so initial angular velocity will be ω(i) = 2π× n
ω(i) = 2π× 60 = 376.9 rad/s
and
final angular velocity will be ω(f) = 0
so
angular displacement will be = 2π × 52 = 326.56 rad
and angular acceleration calculated as
angular acceleration = [tex]\frac{\omega(f)^2-\omega(i)^2}{2*angular displacement}[/tex]
put here value
angular acceleration = [tex]\frac{-376.9^2}{326.56}[/tex]
angular acceleration = - 217.5 rad/s²
A computer monitor accelerates electrons and directs them to the screen in order to create an image. If the accelerating plates are 1.45 cm apart, and have a potential difference of 2.50 x 10^4 V , what is the magnitude of the uniform electric field between them?
Answer:
Electric field at a distance of 1.45 cm will be [tex]172.41\times 10^4N/C[/tex]
Explanation:
We have given the distance d = 1.45 cm = 0.0145 m
And the potential difference [tex]V=2.5\times 10^4volt[/tex]
There is a relation between potential difference and electric field
Electric field at a distance d due to a potential difference is given by
[tex]E=\frac{V}{d}[/tex], here E is electric field, V is potential difference and d is distance
So [tex]E=\frac{V}{d}=\frac{2.5\times 10^4}{0.0145}=172.41\times 10^4N/C[/tex]
The magnitude of the uniform electric field between the plates is approximately [tex]\( 1.724 \times 10^4 \) V/m.[/tex]
The magnitude of the uniform electric field between the plates is [tex]\( E = \frac{\Delta V}{d} \)[/tex], where [tex]\( \Delta V \)[/tex] is the potential difference and ( d ) is the distance between the plates.
Now, we can calculate the electric field ( E ):
[tex]\[ E = \frac{\Delta V}{d} = \frac{2.50 \times 10^4 \text{ V}}{1.45 \times 10^{-2} \text{ m}} \][/tex]
[tex]\[ E = \frac{2.50 \times 10^4}{1.45} \times 10^2 \text{ V/m} \][/tex]
[tex]\[ E = 1.724 \times 10^4 \text{ V/m} \][/tex]
When sound travels through the ocean, where the bulk modulus is 2.34 x 10^9 N/m^2, the wavelength associated with 1000.0 Hz waves is 1.51 m. 1) Calculate the density of seawater. (Express your answer to three significant figures.)
Answer:
1.03×10³ kg/m³
Explanation:
β = Bulk modulus = 2.34×10⁹ N/m²
f = Frequency = 1000 Hz
λ = Wavelength = 1.51 m
ρ = Density
Speed of the wave
v = fλ
⇒v = 1000×1.51
⇒v = 1510 m/s
Speed of sound
[tex]v=\sqrt{\frac{\beta}{\rho}}\\\Rightarrow \rho=\frac{\beta}{v^2}\\\Rightarrow \rho=\frac{2.34\times 10^9}{1510^2}\\\Rightarrow v=1026.27\ kg/m^3[/tex]
Density of seawater is 1.03×10³ kg/m³ or 1.03 g/cm³ (rounding)
A rubber ball is shot straight up from the ground with speed v0. Simultaneously, a second rubber ball at height h directly above the first ball is dropped from rest. a) At what height above the ground do the balls collide? Your answer will be a symbolic expression in terms of v0, h, and g .
b) What is the maximum value of h for which a collision occurs before the first ball falls back to the ground?
Express your answer in terms of the variable v0 and appropriate constants.
c) For what value of h does the collision occur at the instant when the first ball is at its highest point?
Express your answer in terms of the variable v0 and appropriate constants.
In this exercise we have to have knowledge about the horizontal launch, so we have to use the known formulas to find that:
a) [tex]h*(1 - 1/2 g * h/v_0^2)[/tex]
b)[tex]h = v_0^2/ g[/tex]
c)[tex]h = v_0^2/ g[/tex]
So we have to remember some famous equations like the position and velocity of an object moving in a constant, like this:
[tex]y = y_0 + v_0*t + 1/2 * a * t^2\\v = v_0 + a * t[/tex]
where:
y = height at time ty0 = initial heightv0 = initial velocitya = accelerationt = timev = velocitya) When the balls collide, h1 = h2. Then,
[tex]h_1 = h_2\\v_0 * t - 1/2 g * t^2 = h - 1/2 * g * t^2\\v_0 * t = h\\t = h / v_0[/tex]
Replacing in the equation of the height of the first ball:
[tex]h_1 = v_0 * h/v_0 - 1/2g * h^2/v_0^2\\h_1 = h - 1/2 g * h^2/ v_0^2\\h_1 = h*(1 - 1/2 g * h/v_0^2)[/tex]
b) that the balls collide at t = h/v0. Then:
[tex]h/ v_0 = -v_0/-g\\h = v_0^2/ g[/tex]
c) The value of h for which the collision occurs at the instant when the first ball is at its highest point is the maximum value,Then:
[tex]h = v_0^2/ g[/tex]
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An airplane is flying at altitude 2 000[m] through a thundercloud. If +40[C] is concentrated at altitude 3 000[m] , directly above the airplane, and −40[C] is concentrated at altitude 1 000[m] directly below it, what Electric field intensity (magnitude) is at the aircraft?
Answer:
The electric field intensity at the aircraft is [tex]7.18\times10^{5}\ N/C[/tex]
Explanation:
Given that,
Altitude = 2000 m
Charge = +40 C
At the position of the airplane, the electric field will be due to both the positive and negative charges and will be downwards due to both the charges.
The position is equidistant from both the charges.
We need to calculate the resultant electric field
The contribution due to the positive charge at 3000 m altitude
Using formula of electric field
[tex]E_{+}=k\dfrac{|q|}{r^2}[/tex]
Put the value into the formula
[tex]E_{+}=8.99\times10^{9}\times\dfrac{40}{(3000-2000)^2}[/tex]
[tex]E_{+}=3.59\times10^{5}\ N/C[/tex]
The contribution due to the negative charge at 1000 m altitude
Using formula of electric field
[tex]E_{-}=8.99\times10^{9}\times\dfrac{40}{(2000-1000)^2}[/tex]
[tex]E_{-}=3.59\times10^{5}\ N/C[/tex]
We need to calculate the electric field intensity at the aircraft
The resultant electric field is
[tex]E=E_{+}+E_{-}[/tex]
Put the value into the formula
[tex]E=3.59\times10^{5}+3.59\times10^{5}[/tex]
[tex]E=7.18\times10^{5}\ N/C[/tex]
Hence, The electric field intensity at the aircraft is [tex]7.18\times10^{5}\ N/C[/tex]
What is the electric potential 16.0 cm from a 4.00 μC point charge?
Answer:
the potential at a distance of 16 cm from the charge [tex]4\mu C[/tex] will be 225000 volt
Explanation:
We have given charge [tex]q=4\mu C=4\times 10^{-6}C[/tex]
Distance between the charge r = 16 cm = 0.16 m
Electric potential is given by [tex]V=\frac{1}{4\pi \varepsilon _0}\frac{q}{r}=\frac{Kq}{r}[/tex], here K is constant which value is [tex]9\times 10^9Nm^2/C^2[/tex]
So potential [tex]V=\frac{9\times 10^9\times 4\times 10^{-6}}{0.16}=225000volt[/tex]
So the potential at a distance of 16 cm from the charge [tex]4\mu C[/tex] will be 225000 volt
The electric potential, V, of a point charge, is found using the formula V = kq/r. Given the charge of 4.00 μC and the distance of 16.0 cm, the electric potential comes out to approximately 2.24 x 10⁵ V.
Explanation:The electric potential, V, of a point charge, can be calculated by the formula V = kq/r where:
k is the electrostatic constant, which is equal to 8.99 × 10⁹ Nm²/C². q is the charge in Coulombs, andr is the distance from the charge in meters.In the question, q is given as 4.00 μC, which is equivalent to 4.00 x 10⁻⁶ C. The distance r is given as 16.0 cm, which is equivalent to 0.16 m. We get by substituting the values into the formula:
V = (8.99 × 10⁹ Nm²/C² x 4.00 x 10⁻⁶ C)/0.16 m
Through calculation, the electric potential 16.0 cm from a 4.00 μC point charge is therefore approximately 2.24 x 10⁵ V.
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In straight line motion, if the velocity of an object is changing at a constant rate, then its position is _________ and its acceleration is___________: O changing: zero O changing; changing O constant and non-zero; constant and non-zero O None of the above
Answer:
None of the above
It should be position is changing and acceleration is constant.
Explanation:
Since the velocity is changing, this means the object is moving, so the position must also be changing.
Acceleration is the change in velocity in time, if this change of velocity happens at a constant rate, the acceleration must be constant too.
So, for example, if the velocity were to stay the same (not changing), acceleration would be zero, because there wouldn't be a change in time on the velocity.
So in this case the answer sould be position is changing and acceleration is constant. But this isn't in the options so the correct answer is "None of the above"
In straight line motion, if velocity changes at a constant rate, then the position is changing and the acceleration is constant and non-zero. This is defined under the principles of kinematics and implies that as the velocity alters constantly, the object is in motion, hence its position is changing.
Explanation:In straight line motion, if the velocity of an object is changing at a constant rate, then its position is changing and its acceleration is constant and non-zero. This condition is defined under the laws of physics, more specifically, under the study of kinematics.
The acceleration is constant because you're considering a situation where velocity is changing at a constant rate. In this case, the change in velocity is the acceleration, which is a constant and not zero. This situation is described by the kinematic equations for constant acceleration.
The position is changing because the object is moving. A change in position over time constitutes motion, and in this case, because the velocity (the rate of change of position) is changing, the object's position cannot be constant.
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What is the repulsive force between two pith balls that are 11.0 cm apart and have equal charges of −38.0 nC?
Answer:
[tex]F=1.07\times 10^{-3}\ N[/tex]
Explanation:
Given that,
Charge on two balls, [tex]q_1=q_2=-38\ nC=-38\times 10^{-9}\ C[/tex]
Distance between charges, r = 11 cm = 0.11 m
We need to find the repulsive force between balls. Mathematically, it is given by :
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
[tex]F=9\times 10^9\times \dfrac{(38\times 10^{-9})^2}{(0.11)^2}[/tex]
F = 0.001074 N
or
[tex]F=1.07\times 10^{-3}\ N[/tex]
So, the magnitude of repulsive force between the balls is [tex]1.07\times 10^{-3}\ N[/tex].
A paintball’s mass is 0.0032kg. A typical paintball strikes a target moving at 85.3 m/s.
A) If the paintball stops completely as a result of striking its target, what is the magnitude of the change in the paintball’s momentum?
B) If the paintball bounces off its target and afterward moves in the opposite direction with the same speed, what is the magnitude of the change in the paintball’s momentum?
C) The strength of the force an object exerts during impact is determined by the amount the object’s momentum changes. Use this idea along with your answers to (a) and (b) to explain why a paintball bouncing off your skin hurts more than a paintball exploding upon your skin.
Answer:
(A) - 0.273 kg m /s
(B) - 0.546 kg m /s
Explanation:
mass of paintball, m = 0.0032 kg
initial velocity, u = 85.3 m/s
(A) Momentum is defined as the product of mass of body and the velocity of body.
initial momentum, pi = mass x initial velocity
pi = 0.0032 x 85.3 = 0.273 kg m /s
Finally the ball stops, so final velocity, v = 0 m/s
final momentum, pf = mass x final velocity = 0.0032 x 0 = 0 m/s
change in momentum = pf - pi = 0 - 0.273 = - 0.273 kg m /s
(B) initial velocity, u = + 85.3 m/s
final velocity, v = - 85.3 m/s
initial momentum, pi = mass x initial velocity
pi = 0.0032 x 85.3 = 0.273 kg m /s
final momentum, pf = mass x final velocity = - 0.0032 x 85.3 = - 0.273 kgm/s
change in momentum = pf - pi = - 0.273 - 0.273 = - 0.546 kg m /s
(C) According to the Newton's second law, the rate of change of momentum is directly proportional to the force exerted.
As the change in momentum in case (B) is more than the change in momentum in case (A), so the force exerted on the skin is more in case (B).
Assuming that 70 percent of the Earth’s surface is covered with water at average depth of 0.95 mi, estimate the mass of the water on Earth. One mile is approximately 1.609 km and the radius of the earth is 6.37 × 10^6 m. Answer in units of kg. Your answer must be within ± 20.0%
Answer:
The mass of the water on earth is [tex]5.4537\times10^{20}\ kg[/tex]
Explanation:
Given that,
Average depth h= 0.95 mi
[tex]h=0.95\times1.609[/tex]
[tex]h =1.528\ km[/tex]
[tex]h=1.528\times10^{3}\ m[/tex]
Radius of earth [tex]r= 6.37\times10^{6}\ m[/tex]
Density = 1000 kg/m³
We need to calculate the area of surface
Using formula of area
[tex]A =4\pi r^2[/tex]
Put the value into the formula
[tex]A=4\pi\times(6.37\times10^{6})^2[/tex]
[tex]A=5.099\times10^{14}\ m^2[/tex]
We need to calculate the volume of earth
[tex]V = Area\times height[/tex]
Put the value into the formula
[tex]V=5.099\times10^{14}\times1.528\times10^{3}[/tex]
[tex]V=7.791\times10^{17}\ m^3[/tex]
Now, 70 % volume of the total volume
[tex]V= 7.791\times10^{17}\times\dfrac{70}{100}[/tex]
[tex]V=5.4537\times10^{17}\ m^3[/tex]
We need to calculate the mass of the water on earth
Using formula of density
[tex]\rho = \dfrac{m}{V}[/tex]
[tex]m = \rho\times V[/tex]
Put the value into the formula
[tex]m=1000\times5.4537\times10^{17}[/tex]
[tex]m =5.4537\times10^{20}\ kg[/tex]
Hence, The mass of the water on earth is [tex]5.4537\times10^{20}\ kg[/tex]
A man stands on top of a 25 m tall cliff. The man throws a rock upwards at a 25 degree angle above the horizontal with an initial velocity of 7.2 m/s. What will be the vertical position of the rock at t = 2 seconds? What will be the horizontal position of the rock at that time?
Answer:
a) vertical position at 2s: 50.68 m
b) horizontal position at 2s: 13.05 m
Explanation:
This situation is a good example of the projectile motion or parabolic motion, in which the travel of the rock has two components: x-component and y-component. Being the equations to find the position as follows:
x-component:
[tex]x=V_{o}cos\theta t[/tex] (1)
Where:
[tex]V_{o}=7.2 m/s[/tex] is the rock's initial speed
[tex]\theta=25\°[/tex] is the angle at which the rock was thrown
[tex]t=2 s[/tex] is the time
y-component:
[tex]y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}[/tex] (2)
Where:
[tex]y_{o}=25 m[/tex] is the initial height of the rock
[tex]y[/tex] is the height of the rock at 2 s
[tex]g=9.8m/s^{2}[/tex] is the acceleration due gravity
Knowing this, let's begin with the answers:
a) Vertical position at 2 s:
[tex]y=25 m+(7.2 m/s)sin(25\°) (2 s)-\frac{(9.8m/s^{2})(2)^{2}}{2}[/tex] (3)
[tex]y=25 m+6.085 m-19. 6 m[/tex] (4)
[tex]y=50.68 m[/tex] (5) This is the vertical position at 2 s
b) Horizontal position at 2 s:
[tex]x=(7.2 m/s) cos(25\°) (2 s)[/tex] (5)
[tex]x=13.05 m[/tex] (6) This is the horizontal position at 2 s
Explanation:
It is given that,
Position of the man, h = 25 m
Initial velocity of the rock, u = 7.2 m/s
The rock is thrown upward at an angle of 25 degrees.
(a) Let y is the vertical position of the rock at t = 2 seconds. Using the equation of kinematics to find y as :
[tex]y=-\dfrac{1}{2}gt^2+u\ sin\theta\ t+h[/tex]
[tex]y=-\dfrac{1}{2}\times 9.8(2)^2+7.2\ sin(25)\times 2+25[/tex]
y = 11.48 meters
(b) Let x is the horizontal position of the rock at that time. It can be calculated as :
[tex]x=u\ cos\theta \times t[/tex]
[tex]x=7.2\ cos(25)\times 2[/tex]
x = 13.05 meters
Hence, this is the required solution.
You have been working on a new, strong, lightweight ceramic which will be used to replace steel bearing balls. One cubic meter of the steel alloy currently in use has a mass of 8.08 ⨯ 10^3 kg, whereas a cubic meter of your new material has a mass of 3.14 ⨯ 10^3 kg. If the balls currently in use have a radius of 1.9 cm and for this application to keep balls of the same mass instead of the same size, what would the radius replacement ball of the new alloy be?
Answer:
r=2.6 cm
Explanation:
Hi!
Lets call steel material 1, and the new alloy material 2. You know their densities:
[tex]\rho_1=8.08*10^3\frac{kq}{m^3}\\\rho_2=3.14*10^3\frac{kq}{m^3}[/tex]
The volume of a sphere with radius r is given by:
[tex]V=\frac{4}{3}\pi r^3[/tex]
Then the masses of the bearings are:
[tex]m_1=\rho_1V_1=\frac{4}{3}\pi r_1^3 \\m_1=\rho_2V_2=\frac{4}{3}\pi r_2^3[/tex]
For the masses to be the same:
[tex]\rho_1 V_1 = \rho_2 V_2\\\frac{V_2}{V_1} =\frac{\rho_1}{\rho_2}\\(\frac{r_2}{r_1})^3 = \frac{8.08}{3.14}=2.57\\r_2=\sqrt[3]{2.57}\;r1 =1.37r_1=1.37*1.9cm=2.6cm[/tex]
An infinitely long line of charge has a linear charge density of 7.50×10^−12 C/m . A proton is at distance 14.5 cm from the line and is moving directly toward the line with speed 3000 m/s . How close does the proton get to the line of charge? Express your answer in meters.
Answer:
10.22 cm
Explanation:
linear charge density, λ = 7.5 x 10^-12 C/m
distance from line, r = 14.5 cm = 0.145 m
initial speed, u = 3000 m/s
final speed, v = 0 m/s
charge on proton, q = 1.6 x 10^-19 C
mass of proton, m = 1.67 x 10^-27 kg
Let the closest distance of proton is r'.
The potential due t a line charge at a distance r' is given by
[tex]V=-2K\lambda ln\left (\frac{r'}{r} \right )[/tex]
where, K = 9 x 10^9 Nm^2/C^2
W = q V
By use of work energy theorem
Work = change in kinetic energy
[tex]qV = 0.5m(u^{2}-v^{2})[/tex]
By substituting the values, we get
[tex]V=\frac{mu^{2}}{2q}[/tex]
[tex]-2K\lambda ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{2q}[/tex]
[tex]- ln\left ( \frac{r'}{r} \right )=\frac{mu^{2}}{4Kq\lambda }[/tex]
[tex]- ln\left ( \frac{r'}{r} \right )=\frac{1.67 \times 10^{-27}\times 3000\times 3000}{4\times 9\times 10^{9}\times 1.6\times 10^{-19}\times 7.5\times 10^{-12} }[/tex]
[tex]- ln\left ( \frac{r'}{r} \right )=0.35[/tex]
[tex]\frac{r'}{r} =e^{-0.35}[/tex]
[tex]\frac{r'}{r} =0.7047[/tex]
r' = 14.5 x 0.7047 = 10.22 cm
What is the electric force (with direction) on an electron in a uniform electric field of strength 3400 N/C that points due east? Take the positive direction to be east.
Answer:
So, the force its [tex] \ 5.4468 \ 10 ^{-16} [/tex] N to the west.Explanation:
The force [tex]\vec{F}[/tex] on a charge q made by an electric field [tex]\vec{E}[/tex] its
[tex]\vec{F} = q \vec{E}[/tex]
The electric charge of the electron its
[tex]q \ = \ - \ 1.602 \ 10 ^{-19} \ C[/tex].
Taking the unit vector [tex]\hat{i}[/tex] pointing towards the east, the electric field will be:
[tex]\vec{E}= 3400 \ \frac{N}{C} \ \hat{i}[/tex].
So, the force will be:
[tex]\vec{F} = \ - \ 1.602 \ 10 ^{-19} \ C \ * \ 3400 \ \frac{N}{C} \ \hat{i} [/tex]
[tex]\vec{F} = \ - \ 5446.8 \ 10 ^{-19} \ N \ \hat{i} [/tex]
[tex]\vec{F} = \ - \ 5.4468 \ 10 ^{-16} \ N \ \hat{i} [/tex]
[tex]\vec{F} = \ - \ 5.4468 \ 10 ^{-16} \ N \ \hat{i} [/tex]
[tex]\vec{F} = \ - \ 5.4468 \ 10 ^{-16} \ N \ \hat{i} [/tex]
So, the force its [tex] \ 5.4468 \ 10 ^{-16} [/tex] N to the west.
The current in a 100 watt lightbulb is 0.890 A. The filament inside the bulb is 0.280 mm in diameter. What is the current density in the filament? Express your answer to three significant figures and include the appropriate units. What is the electron current in the filament? Express your answer using three significant figures.
Answer:
Current density [tex]j=1.44\times 10^7A/m^2[/tex]
Electron density [tex]=5.55\times 10^{18}electron/sec[/tex]
Explanation:
We have given power = 100 watt
Current = 0.89 A
Diameter d = 0.280 mm
So radius [tex]r=\frac{d}{2}=\frac{0.28}{2}=0.14mm=0.14\times 10^{-3}m[/tex]
Area [tex]A=\pi r^2=3.14\times (0.14\times 10^{-3})^2=0.016\times 10^{-6}m^2[/tex]
We know that current density [tex]J=\frac{I}{A}=\frac{0.89}{0.016\times 10^{-6}}=1.44\times 10^7A/m^2[/tex]
Now we have to calculate the electron density
We have current i = 0.89 A = 0.89 J/sec
Charge on 1 electron [tex]1.6\times 10^{-19}C/electron[/tex]
So electron density [tex]=\frac{0.89j/sec}{1.6\times 10^{-19}C/electron}=5.55\times 10^{18}electron/sec[/tex]
Final answer:
The current density in the filament is 14.4 x 10^6 A/m^2. The electron current in the filament is 1.42 x 10^-19 C/s.
Explanation:
The current density in a filament can be calculated by dividing the current in the lightbulb by the cross-sectional area of the filament. To find the current density, we first need to find the cross-sectional area of the filament. The diameter of the filament is given as 0.280 mm, so the radius is 0.140 mm (0.140 mm = 0.280 mm / 2).
Using the formula for the area of a circle (A = πr^2), we can find the cross-sectional area of the filament:
A = π(0.140 mm)^2 = 0.0616 mm^2
Now, we can calculate the current density by dividing the current (0.890 A) by the cross-sectional area (0.0616 mm^2), and converting mm to m:
Current density = 0.890 A / (0.0616 mm^2) * (1 m / 1000 mm)^2 = 14.4 x 10^6 A/m^2
The electron current in the filament can be calculated using the formula:
Electron current = charge * number of charges passing per second
The charge of an electron is approximately 1.6 x 10^-19 C, and the current is given as 0.890 A, so:
Electron current = (1.6 x 10^-19 C) * (0.890 A) = 1.42 x 10^-19 C/s
Problem 1 The Van de Graaff electrostatic generator develops a charge of approximately −1 × 10−5C and a pith ball has charge of approximately 1 × 10−9C. Which of the following describes the relationship between the size of the force the Van de Graaff exerts on the pith ball and the size of the force the pith ball exerts on the Van de Graaff? Select One of the Following: (a) The Van de Graaff exerts a much larger force on the pith ball. (b) The pith ball exerts a much larger force on the Van de Graaff. (c) The forces have the same magnitude.
Answer:
(c) The forces have the same magnitude.
Explanation:
The electric Force between the two charges are proportional to the product of both charges:
[tex]F_{electric}=k*q_{1}q_{2}/r^{2}[/tex]
Both charges feels the same force but with opposite direction.
(c) The forces have the same magnitude.
A guy wire 1005 feet long is attached to the top of a tower. When pulled taut, it touches level ground 552 feet from the base of the tower. What angle does the wire make with the ground? Express your answer using degree measure rounded to one decimal place.
Answer:
56.7°
Explanation:
Imagine a rectangle triangle.
The triangle has 3 sides.
One side is the height of the tower, let's name it A.
Another side is the distance from the base of the tower to the point where the waire touches the ground. Let's name that B.
Sides A and B are perpendicular.
The other side is the length of the wire. Let's name it C.
From trigonometry we know that:
cos(a) = B / C
Where a is the angle between B anc C, between the wire and the ground.
Therefore
a = arccos(B/C)
a = arccos(552/1005) = 56.7°
[tex]56.7^\circ[/tex] angle the wire makes with the ground.
Given :
A guy wire 1005 feet long is attached to the top of a tower.
When pulled taut, it touches level ground 552 feet from the base of the tower.
Solution :
We know that,
[tex]\rm cos \theta = \dfrac{Base}{Hypotanuse}[/tex]
Given that,
base = 552 ft
hypotanuse = 1005 ft
Therefore,
[tex]\rm cos\theta = \dfrac{552}{1005}[/tex]
[tex]\rm \theta = cos^-^1 \dfrac{552}{1005}[/tex]
[tex]\theta = 56.7^\circ[/tex]
[tex]56.7^\circ[/tex] angle the wire makes with the ground.
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A 6.0-m board is leaning against a wall. A 28° angle is formed with the WALL What is the height above the ground where the board makes contact with the wall? (In Meters)
Answer:
the height above the ground where the board makes contact with the wall equals 5.297 meters above ground.
Explanation:
The wall and ladder are shown in the attached figure
In the figure we can see that
[tex]cos(28^{o})=\frac{H}{L}\\\\cos(28^{o})=\frac{H}{6}\\\\\therefore H=6.0\times cos(28^{o})==5.297meters[/tex]
You need to figure out how high it is from the 3rd floor of Keck science building to the ground. You know that the distance from the ground to the 2nd floor is 5m. Your friend drops a ball from rest from the 3 rd floor and you start to time the ball when it passes by you at the 2nd floor. You stop the timer when the ball hits the ground. The time recorded is 0.58 seconds.
Answer:1.624 m
Explanation:
Given
height of 2nd floor =5 m
time recorded is 0.58 sec
Let h be the height of third floor above 2 nd floor
[tex]h=ut+\frac{at^2}{2}[/tex]
here u=0
t=time taken to cover height h
[tex]h=\frac{gt^2}{2}[/tex] ----------1
Now time taken to complete whole building length
[tex]h+5=\frac{g(t+0.58)^2}{2}[/tex] ----------2
Subtract 1 form 2
[tex]5=\frac{g(2t+0.58)(0.58)}{2}[/tex]
Taking gravity [tex]g=1 m/s^2[/tex]
1=(2t+0.58)(0.58)
t=0.57 s
Substitute the value of t in 1
h=1.624 m
Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at y2 = +0.34 m. A third point charge q = +8.4 μC is fixed at the origin. The net electrostatic force exerted on the charge q by the other two charges has a magnitude of 25 N and points in the +y direction. Determine the magnitude of q2.
Answer:
[tex]50.91 \mu C[/tex]
Explanation:
The magnitude of the net force exerted on q is known, we have the values and positions for [tex]q_{1}[/tex] and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by[tex]q_{1}[/tex] on q. Then we can know the magnitude of the force exerted by [tex]q_{2}[/tex] about q, finally this will allow us to know the magnitude of [tex]q_{2}[/tex]
[tex]q_{1}[/tex] exerts a force on q in +y direction, and [tex]q_{2}[/tex] exerts a force on q in -y direction.
[tex]F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\[/tex]
The net force on q is:
[tex]F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}[/tex]
Rewriting for [tex]q_{2}[/tex]:
[tex]q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C[/tex]
A car travels in a straight line from a position 50 m to your right to a position 210 m to your right in 5 sec. a. What is the average velocity of the car?b. Construct a position vs. time graph and calculate the slope of the line of best fit.Please give me the exact coordinates to plot. Thank you!
Answer:
a) The average velocity is 32 m/s
b) See the attached figure. Slope of the line = 32.
Graphic: a line that passes through the points (0; 50) and (5; 210)
Explanation:
a) The average velocity is calculated as the displacement over time:
v = ΔX / Δt
where
ΔX : displacement ( final position - initial position)
Δt : time (final time - initial time)
Considering the origin of the reference system as the position where the observer is:
ΔX = 210 m - 50 m = 160 m
Δt = 5 s
v = 160 m / 5 s = 32 m/s
b) In the graphic position vs time, plot a line that passes through the points (0, 50) (because at time 0, the car is 50 m away from you, the center of the reference system) and (5, 210). The x-axis, time, is in seconds and the y-axis, position, in meters. The slope of the line is calculated as:
slope = (X₂ - X₁) /(t₂ - t₁) = (210 - 50) / (5 - 0) = 32. Then, the velocity is equal to the slope of the line. See the attached figure.
A certain field line diagram illustrates the electric field due to three particles that carry charges 5.0 μC, -3.0 μC, and -2.0 μC. 15 field lines emanate from the positively charged particle. How many field lines terminate on the -3.0 μC particle? How many field lines terminate on the -2.0 μC particle?
Answer:
6
Explanation:
Number of lines emanate from + 5 micro coulomb is 15 .
They terminates at negative charges that means at - 3 micro coulomb and - 2 micro Coulomb.
the electric field lines terminates at - 3 micro Coulomb and - 2 micro Coulomb is in the ratio of 3 : 2.
So the lines terminating at - 3 micro coulomb
= [tex]\frac{3}{5}\times 15 = 9[/tex]
So the lines terminating at - 2 micro coulomb
= [tex]\frac{2}{5}\times 15 = 6[/tex]
So, the number of filed lines terminates at - 2 micro Coulomb are 6.
The number of lines emanating from the negative charged particles are 9.09 and 6.06 respectively.
How to calculate the number of lines.In an electric field, the number of lines emanating from a charged particle is directly proportional to the magnitude of the charge. Thus, this is given by this mathematical expression:
[tex]q \;\alpha \;L\\\\q=kL\\\\5=k15\\\\k=\frac{5}{15}[/tex]
k = 0.33.
For the -3.0 μC particle:
[tex]L=\frac{q}{k} \\\\L=\frac{3.0}{0.33}[/tex]
L = 9.09.
For the -2.0 μC particle:
[tex]L=\frac{q}{k} \\\\L=\frac{2.0}{0.33}[/tex]
L = 6.06.
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A certain reaction X + Y → Z is described as being first order in [X] and third order overall. Which of the following statements is or are true?: The rate law for the reaction is: Rate = k[X][Y]2. If the concentration of X is increased by a factor of 1.5, the rate will increase by a factor of 2.25. If the concentration of Y is increased by a factor of 1.5, the rate will increase by a factor of 2.25.
Answer:
The second statement is true: If the concentration of Y is increased by a factor of 1.5, the rate will increase by a factor of 2.25.
Explanation:
Hi there!
Let´s write the rate law for the original reaction and the reaction with X increased by 1.5:
rate 1 =k [X][Y]²
rate 2 = k[1.5 X][Y]²
Now we have to demonstrate if rate 2 = 2.25 rate 1
Let´s do the cocient between the two rates:
rate 2/ rate 1
if rate 2 = 2.25 rate 1
Then,
rate 2 / rate 1 = 2.25 rate 1 / rate 1 = 2.25
Let´s see if this is true using the expressions for the rate law:
rate 2 / rate 1
k[1.5 X][Y]² / k [X][Y]² = 1.5 k [X][Y]² / k[X][Y]² = 1.5
2.25 ≠ 1.5
Then the first statement is false.
Now let´s write the two expressions of the rate law, but this time Y will be increased by 1.5:
rate 1 = k[X][Y]²
rate 2 = k[X][1.5Y]²
Again let´s divide both expressions to see if the result is 2.25
rate 2 / rate 1
k[X][1.5Y]²/ k [X][Y]²
(distributing the exponent)
(1.5)²k [X][Y]² / k [X][Y]² = (1.5)² = 2.25
Then the second statement is true!
Answer:
a) True
b)False
c) True
Explanation:
The order of reactants decides the exponents of respective reactant concentrations.
Since X is first order, exponent is = 1
Overall third order, Exponent X + Exponent Y = 3
Hence, exponent Y = 2.
Hence,
Rate = k[X][Y]^2
b)
If X conc is increased by 1.5 Rate should increase by 1.5 because k proportional to [X]. Hence, statement is False
c)
If Y conc is increased by 1.5 Rate should increase by 2.25 because k proportional to [Y]^2 = (1.5)^2 = 2.25. Hence, statement is True
A 50-V de voltage source was connected in series with a resistor and capacitor. Calculate the current in A to two significant figures) after 5.0 s if the resistance was 25 MΩ and the capacitance 0.10μF.
Answer:
Current through circuit will be [tex]0.2706\times 10^6A[/tex]
Explanation:
We have given source voltage v = 50 volt
Resistance [tex]R=25Mohm=25\times 10^6ohm[/tex]
Capacitance [tex]C=0.1\mu F=0.1\times 10^{-6}F[/tex]
Time t = 5 sec
Time constant of RC circuit is given by [tex]\tau =RC=25\times 10^6\times 0.1\times 10^{-6}=2.5sec[/tex]
We know that voltage across capacitor is given by [tex]v_c=v_s(1-e^{\frac{-t}{\tau }})[/tex]
[tex]v_c=50(1-e^{\frac{-5}{2.5 }})=43.2332v[/tex]
So current will be [tex]=\frac{v_s-v_c}{R}=\frac{50-43.2332}{25\times 10^{-6}}=0.2706\times 10^6A[/tex]
So current through circuit will be [tex]0.2706\times 10^6A[/tex]
An electron with a kinetic energy of 22.5 eV moves into a region of uniform magnetic field B of magnitude 4.55 x 104 T. The angle between the directions of the magnetic field and the electron's velocity is 65.5 degrees. What is the pitch of the helical path taken by the electron?
The pitch of the helical path followed by an electron in a magnetic field is given by the formula p = v * cos(theta) * T, where v is the velocity of the electron, theta is the angle between the direction of the magnetic field and the electron's velocity, and T is the period of the electron's motion in the magnetic field. By substituting the given values into this formula, we can calculate the pitch.
Explanation:The pitch of a helical path can be determined by considering how the charged particle reacts in a magnetic field. The helical path of an electron in a magnetic field is caused by the perpendicular and parallel components of the electron's velocity. In particular, the parallel component of the velocity, v(cos(theta)), is responsible for the linear movement of the electron along the field lines. This leads to the corkscrew-like helical path, and the pitch of this helix is equivalent to the linear distance moved by the electron in one revolution.
The pitch (p) can be calculated by the formula p = v * cos(theta) * T, where v is the velocity of the electron, theta is the angle between the direction of the magnetic field and the velocity of the electron, and T is the period of circular motion of the electron in the magnetic field. From the provided data, we can first calculate the velocity of the electron using its kinetic energy, 22.5 eV: v = sqrt((2 * KE) / m), where KE is the kinetic energy and m is the mass of the electron (9.11 * 10^-31 kg). Next, we calculate the period using the formula T = 2 * pi * m / (q * B), where q is the charge of the electron (-1.6 * 10^-19 C) and B is the magnetic field intensity (4.55 * 10^4 T).
Finally, we substitute these values into the pitch formula and calculate the pitch of the helical path taken by the electron in the magnetic field.
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The pitch of the helical path taken by the electron is approximately [tex]\( 9.20 \times 10^{-11} \)[/tex] meters.
To find the pitch of the helical path taken by the electron, we need to follow these steps:
1. Calculate the speed of the electron:
The kinetic energy (K.E.) of the electron is given by:
[tex]\[ \text{K.E.} = \frac{1}{2}mv^2 \][/tex]
where [tex]\( m \)[/tex] is the mass of the electron [tex](\( 9.11 \times 10^{-31} \) kg)[/tex] and [tex]\( v \)[/tex] is the speed of the electron.
First, we need to convert the kinetic energy from electron volts (eV) to joules (J):
[tex]\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \][/tex]
[tex]\[ 22.5 \, \text{eV} = 22.5 \times 1.602 \times 10^{-19} \, \text{J} = 3.6045 \times 10^{-18} \, \text{J} \][/tex]
Now, using the kinetic energy formula:
[tex]\[ \text{K.E.} = \frac{1}{2}mv^[/tex]
To find the pitch of the helical path taken by the electron, we need to determine the components of the electron's velocity and the motion along the magnetic field.
1. Calculate the speed of the electron:
The kinetic energy (KE) of the electron is given by:
[tex]\[ \text{KE} = \frac{1}{2} mv^2 \][/tex]
where m is the mass of the electron [tex](\( 9.11 \times 10^{-31} \) kg)[/tex] and v is the speed of the electron.
First, convert the kinetic energy from electron volts (eV) to joules (J):
[tex]\[ 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \][/tex]
[tex]\[ 22.5 \, \text{eV} = 22.5 \times 1.602 \times 10^{-19} \, \text{J} = 3.6045 \times 10^{-18} \, \text{J} \][/tex]
Now, solve for the speed \( v \) using the kinetic energy formula:
[tex]\[ \frac{1}{2} mv^2 = 3.6045 \times 10^{-18} \, \text{J} \][/tex]
[tex]\[ v^2 = \frac{2 \times 3.6045 \times 10^{-18}}{9.11 \times 10^{-31}} \][/tex]
[tex]\[ v^2 = 7.9137 \times 10^{12} \][/tex]
[tex]\[ v = \sqrt{7.9137 \times 10^{12}} \approx 2.81 \times 10^6 \, \text{m/s} \][/tex]
2. Determine the components of the velocity:
The angle between the velocity and the magnetic field is [tex]\( 65.5^\circ \)[/tex]. The component of the velocity parallel to the magnetic field [tex](\( v_{\parallel} \))[/tex] and the component perpendicular to the magnetic field[tex](\( v_{\perp} \))[/tex] can be found using trigonometry:
[tex]\[ v_{\parallel} = v \cos(65.5^\circ) \][/tex]
[tex]\[ v_{\perp} = v \sin(65.5^\circ) \][/tex]
Calculate these components:
[tex]\[ v_{\parallel} = 2.81 \times 10^6 \, \text{m/s} \times \cos(65.5^\circ) \approx 2.81 \times 10^6 \times 0.417 \approx 1.17 \times 10^6 \, \text{m/s} \][/tex]
[tex]\[ v_{\perp} = 2.81 \times 10^6 \, \text{m/s} \times \sin(65.5^\circ) \approx 2.81 \times 10^6 \times 0.909 \approx 2.55 \times 10^6 \, \text{m/s} \][/tex]
3. Find the cyclotron frequency [tex](\( f \))[/tex] and the period [tex](\( T \))[/tex] of the electron's circular motion:
The cyclotron frequency is given by:
[tex]\[ f = \frac{qB}{2\pi m} \][/tex]
where q is the charge of the electron [tex](\( 1.602 \times 10^{-19} \) C)[/tex], B is the magnetic field [tex](\( 4.55 \times 10^4 \) T)[/tex], and m is the mass of the electron.
[tex]\[ f = \frac{1.602 \times 10^{-19} \times 4.55 \times 10^4}{2\pi \times 9.11 \times 10^{-31}} \][/tex]
[tex]\[ f \approx \frac{7.29 \times 10^{-15}}{5.72 \times 10^{-31}} \approx 1.27 \times 10^{16} \, \text{Hz} \][/tex]
The period [tex](\( T \))[/tex] is the inverse of the frequency:
[tex]\[ T = \frac{1}{f} \approx \frac{1}{1.27 \times 10^{16}} \approx 7.87 \times 10^{-17} \, \text{s} \][/tex]
4. Calculate the pitch of the helical path:
The pitch is the distance the electron travels parallel to the magnetic field in one period:
[tex]\[ \text{Pitch} = v_{\parallel} \times T \][/tex]
[tex]\[ \text{Pitch} \approx 1.17 \times 10^6 \, \text{m/s} \times 7.87 \times 10^{-17} \, \text{s} \approx 9.20 \times 10^{-11} \, \text{m} \][/tex]
An airplane flies northwest for 250 mi and then west for 150 mi. What is the resultant displacement of the plane after this time?
Answer:371.564 mi
Explanation:
Given
Airplane flies northwest for 250 mi and then travels west 150 mi
That is first it travels 250cos45 in - ve x direction and simultaneously 250sin45 in y direction
after that it travels 150 mi in -ve x direction
So its position vector is given by
[tex]r=-250cos45\hat{i}-150\hat{i}+250sin45\hat{j}[/tex]
[tex]r=-\left ( 250cos45+150\right )\hat{i}+250sin45\hat{j}[/tex]
so magnitude of displacement is
[tex]|r|=\sqrt{\left ( \frac{250}{\sqrt{2}}+150\right )^2+\left ( \frac{250}{\sqrt{2}}\right )^2}[/tex]
|r|=371.564 mi
How do the significant figures in a measurement affect the significant figures in a calculation?
Answer:
Explanation:
The digits which are reliable and first uncertain digit is called significant figure.
In any measurement, if the significant figure is more than the accuracy of the measurement is also more.
If there are some numbers in a calculations and they having different numbers of significant figures, then in the final result the number of significant figures is the least number of significant figures which are in the individual number.
When performing calculations with measurements that have different numbers of significant figures, the final answer will be rounded to the same number of significant figures as the measurement with the fewest significant figures. This applies to both addition and subtraction, as well as multiplication and division.
Explanation:For addition and subtraction, the result should be rounded to the same decimal place as the least precise measured value. For example, if one measurement has two decimal places and another measurement has three decimal places, the final answer should be rounded to two decimal places.
For multiplication and division, the final answer should have the same number of significant figures as the measurement with the fewest significant figures. For example, if one measurement has three significant figures and another measurement has four significant figures, the final answer should have three significant figures.
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A positive point charge on a 2-D plane starts at the position (2 cm, 7 cm). It moves along an equipotential line to the position (10 cm, 7 cm). The plane is oriented parallel to a
uniform electric field of 120 N/C. How much work does the electric field do in this scenario? (Answer in Joules, J. You do not need to type the units, only the number.)
Answer:
W = 0
Explanation:
given data:
starting position is ( 2cm , 7 cm)
ending position is (10 cm, 7 cm)
we know thta work done is given as
[tex] W = q\Delta v[/tex]
here , q is particle charge
[tex]\Delta v[/tex] is change in potential
as we know that intial and final position are having uniform eleectric potential, therefore it have same potential i.e. V_1 = V_2, thus
one more reason of being work done equal to zero is,
Since the distance covered by the item and the direction of motion is always perpendicular to each other during the round journey, no work is therefore carried out.
W = 0
n Section 12.3 it was mentioned that temperatures are often measured with electrical resistance thermometers made of platinum wire. Suppose that the resistance of a platinum resistance thermometer is 125 Ω when its temperature is 20.0°C. The wire is then immersed in boiling chlorine, and the resistance drops to 99.6 Ω. The temperature coefficient of resistivity of platinum is α = 3.72 × 10−3(C°)−1. What is the temperature of the boiling chlorine?
The temperature of the boiling chlorine is calculated using the resistance change of a platinum wire thermometer and the temperature coefficient of resistivity. After calculations, the boiling point of chlorine is found to be -34.6°C.
Explanation:The temperature of the boiling chlorine can be deduced using the relation between temperature change and resistance change for a platinum wire thermometer. The equation we use is:
R = R0 (1 + αΔT)
Where R represents the resistance at the new temperature, R0 the resistance at a reference temperature (20°C in this case), α the temperature coefficient of resistivity, and ΔT the change in temperature. We can rearrange this equation to solve for the new temperature:
ΔT = ∂(T₂ - T₁)
Giving us:
(99.6 ohms / 125 ohms - 1) = 3.72 × 10-3°C-1 ∂T
Simplifying:
ΔT = (0.7968 - 1) / 3.72 × 10-3°C-1
ΔT = -0.2032 / 3.72 × 10-3°C-1
ΔT = -54.6°C
Thus, the boiling point of chlorine is:
20°C – 54.6°C = -34.6°C
A boy uses a slingshot to launch a pebble straight up into the air. The pebble reaches a height of 37.0 m above the launch point 2.3 seconds later. Assume air resistance is negligible (a) What was the pebble's initial speed (just after leaving the slingshot)? m/s (b) How much time did it take for the pebble to first reach a height of 18.5 m above its launch point? s
The pebble's initial speed is approximately 22 m/s. It takes approximately 1.45 seconds for the pebble to first reach a height of 18.5 m.
Explanation:To determine the pebble's initial speed, we can use the equation for projectile motion:
h = v0yt - (1/2)gt2
Where h is the height, v0y is the initial vertical speed, t is the time, and g is the acceleration due to gravity.
Since the pebble is launched straight up, the final height is equal to the initial height. Plugging in the given values, we have:
37 m = v0y(2.3 s) - (1/2)(9.8 m/s2)(2.3 s)2
Simplifying this equation gives us the value of v0y, the initial vertical speed. To find the pebble's initial speed, we can use the Pythagorean theorem:
v0 = √(v0x2 + v0y2)
Where v0 is the initial speed and v0x is the initial horizontal speed. Since the pebble is launched straight up, v0x = 0. Plugging in the calculated value of v0y, we can solve for v0.
(a) The pebble's initial speed is approximately 22 m/s.
(b) To find the time it takes for the pebble to first reach a height of 18.5 m, we can use the equation for height:
18.5 m = v0yt - (1/2)gt2
Solving for t gives the time it takes for the pebble to reach the desired height.
(b) It takes approximately 1.45 seconds for the pebble to first reach a height of 18.5 m.
How many electrons would have to be removed from a coin to leave it with a charge of +2.2 x 10^- 8 C?
Answer:
[tex]N=1.375*10^{11}[/tex] electrons
Explanation:
The total charge Q+ at the coin is equal, but with opposite sign, to the charge negative Q- removed of it. Q- is the sum of the charge of the N electrons removed from the coin:
[tex]Q_{-}=N*q_{e}[/tex]
[tex]Q_{-}=-Q_{+}[/tex]
q_{e}=-1.6*10^{-19}C charge of a electron
We solve to find N:
[tex]N=-Q_{+}/q_{e}=-2.2*10^{-8}/(-1.6*10^{-19})=1.375*10^{11}[/tex]