Answer:
R = 20.21 ohms
Explanation:
Given that,
The length of the wire, l = 200 ft
The diameter of the wire, d = 0.01 in
Radius, r = 0.005 in
200 ft = 60.96 m
0.005 in = 0.000127 m
The resistivity of copper is, [tex]\rho=1.68\times 10^{-8}\ \Omega-m[/tex]
So, the resistance of a wire is given by :
[tex]R=\rho \dfrac{l}{A}\\\\R=\rho \dfrac{l}{\pi r^2}\\\\R=1.68\times 10^{-8}\times \dfrac{60.96 }{\pi \times (0.000127 )^2}\\\\R=20.21\ \Omega[/tex]
So, the resistance of the copper wire is 20.21 ohms.
Physics question plz help ASAP
Liquid plastic is frozen in a physical change that increases its volume. What can be known about the plastic after the change?
(A) Its mass will increase.
(B) Its density will increase.
(C) Its mass will remain the same.
(D) Its density will remain the same.
Answer:
c
Explanation:
Liquid plastic is frozen in a physical change that increases its volume,it can be known about the plastic that Its mass will remain the same, therefore the correct answer is option C.
What is the matter?Anything which has mass and occupies space is known as matter ,mainly there are four states of matter solid liquid gases, and plasma.
These different states of matter have different characteristics according to which they vary their volume and shape.
It is known about plastic that its mass will remain the same when liquid plastic is frozen, by increasing its volume.
Liquid plastic is frozen in a physical change that increases its volume,it can be known about the plastic that Its mass will remain the same, therefore the correct answer is C.
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A bicyclist moves along a straight line with an initial velocity vo and slows downs. Which of the following the best describes the signs set for the initial position, initial velocity and the acceleration ?
The sign set after the slowdown of the bicycle will be positive for the position, negative for velocity, and negative for acceleration.
What is velocity?The rate at which an object's position changes when observed from a specific point of view and when measured against a specific unit of time is known as its velocity.
According to Que, when a bicyclist moves in a straight line and slows down, then the velocity decrease as displacement is decreasing, and the acceleration also decreases only displacement increases.
Therefore, the sign set for the position is +ve, for velocity it is -ve, and for acceleration also -ve
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A biker slows down after traveling in a long, straight line at initial velocity v0. Which of the following the best \sdescribes the signs set for the initial position, initial velocity and the acceleration? Initial position Initial velocity Acceleration
A. Positive Negative Negative
B. Positive Positive Negative
C. Negative Positive Negative
D. Negative Negative Positive
E. Negative Negative Negative
A gymnast weighs 450 N. She stands on a balance beam of uniform construction which weighs 250 N. The balance beam is 3.0 m long and is supported at each end. If the support force at the right end is four times the force at the left end, how far from the right end is the gymnast
Answer:
x = 9.32 cm
Explanation:
For this exercise we have an applied torque and the bar is in equilibrium, which is why we use the endowment equilibrium equation
Suppose the counterclockwise turn is positive, let's set our reference frame at the left end of the bar
- W l / 2 - W_{child} x + N₂ l = 0
x = [tex]\frac{-W l/2 + n_2 l}{W_{child}}[/tex] 1)
now let's use the expression for translational equilibrium
N₁ - W - W_(child) + N₂ = 0
indicate that N₂ = 4 N₁
we substitute
N₁ - W - W_child + 4 N₁ = 0
5 N₁ -W - W_{child} = 0
N₁ = ( W + W_{child}) / 5
we calculate
N₁ = (450 + 250) / 5
N₁ = 140 N
we calculate with equation 1
x = -250 1.50 + 4 140 3) / 140
x = 9.32 cm
During normal beating, the heart creates a maximum 4.10-mV potential across 0.350 m of a person's chest, creating a 1.00-Hz electromagnetic wave. (a) What is the maximum electric field strength created? V/m (b) What is the corresponding maximum magnetic field strength in the electromagnetic wave? T (c) What is the wavelength of the electromagnetic wave?
Explanation:
Given that,
Maximum potential, V = 4. mV
Distance, d = 0.350 m
Frequency of the wave, f = 100 Hz
(a) The maximum electric field strength created is given by:
[tex]E=\dfrac{V}{d}\\\\E=\dfrac{4.1\times 10^{-3}}{0.350 }\\\\E=0.0117\ V/m[/tex]
(b) The corresponding maximum magnetic field strength in the electromagnetic wave is given by :
[tex]B=\dfrac{E}{c}\\\\B=\dfrac{0.0117}{3\times 10^8}\\\\B=3.9\times 10^{-11}\ T[/tex]
(c) The wavelength of the electromagnetic wave can be calculated as :
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{100}\\\\=3\times 10^6\ m[/tex]
So, the wavelength of the electromagnetic wave is [tex]3\times 10^6\ m[/tex].
The potential difference between the plates of a capacitor is 234 V. Midway between the plates, a proton and an electron are released. The electron is released from rest. The proton is projected perpendicularly toward the negative plate with an initial speed. The proton strikes the negative plate at the same instant the electron strikes the positive plate. Ignore the attraction between the two particles, and find the initial speed of the proton.
I have tried looking at the cramster.com solution manual and do not like the way it is explained. Simply put, I cannot follow what is going on and I am looking for someone who can explain it in plain man's terms and help me understand and get the correct answer. I am willing to give MAX karma points to anyone who can help me through this. Thank you kindly.
Answer:
The speed of proton is 2.1 x 10^5 m/s .
Explanation:
potential difference, V = 234 V
let the initial speed of the proton is v.
The kinetic energy of proton is
KE = q V
[tex]0.5 mv^2 = e V \\\\0.5\times 1.67\times 10^{-27} v^2 = 1.6\times 10^{-19} \times 234\\\\v=2.1\times 10^5 m/s[/tex]
The temperature of a body falls from 30°C to 20°C in 5 minutes. The air
temperature is 13°C. Find the temperature after a further 5 minutes.
Answer:
15.88°C I am not 100% sure this is right but I am 98% sure this IS right
A chair of weight 85.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 40.0 N directed at an angle of 35.0deg below the horizontal and the chair slides along the floor.
Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair.
Answer:
N = 107.94 N
Explanation:
For this exercise we must use Newton's second law.
Let's set a reference system with the x-axis parallel to the ground and the y-axis vertical
X axis
Fₓ = ma
ej and
N -F_y - W = 0
let's use trigonometry to decompose the applied force
cos -35 = Fₓ / F
sin -35 = F_y / F
Fₓ = F cos -35
F_y = F sin -35
Fₓ = 40.0 cos -35 = 32.766 N
F_y = 40.0 sin -35 = -22.94 N
we substitute
N = Fy + W
N = 22.94 + 85
N = 107.94 N
Condensation is the process of ____________________.
a. planetesimals accumulating to form protoplanets.
b. planets gaining atmospheres from the collisions of comets.
c. clumps of matter adding material a small bit at a time.
d. clumps of matter sticking to other clumps.
e. clouds formed from volcanic eruptions.
Flag question
Consider the pressure and force acting on the
dam retaining a reservoir of water. Suppose the
dam is 500-m wide
and the water is 80.0-m
deep at the dam, as illustrated below. What is
the average pressure on the dam due to the
water?
Answer:
P = density (p) * g * h
P = 1000 kg/m^3 * 9.8 m/s^2 * 40 m = 392,000 N/m^2
since kg m / s^2 = Newtons
The average pressure is 1/2 (pressure at 0m + pressure 80 m) for liquid of uniform density
a concave mirror has a radius of curvature of 60cm. How close to the mirror should an object be placed so that the rays travel parallel to each other after reflection
Answer:
Answer:30 cm
Answer:30 cmExplanation:
Answer:30 cmExplanation:Given=ROC= 60cm
Answer:30 cmExplanation:Given=ROC= 60cmObject be placed so that the rays that came from the object to them mirror are reflected from the mirror, and, then travel parallel to each other= 30cm at focus.
A student graphs power (p) on the vertical axis and time (t) on the horizontal axis. The graph appears to be a hyperbola.
a) What should the student graph on each axis to test whether the relationship is actually
hyperbolic?
b) If the relationship is actually hyperbolic, what is the general equation for the relationship between power and time?
Answer: it would be daddy
Explanation:
Because I’m daddy
which of the following cannot be increased by using a machine of some kind? work, force, speed, torque
Explanation:
Work cannot be increased by using a machine of some kind.
Work cannot be increased by using a machine of some kind.
A machine is any device in which the effort applied at one end overcomes a load at the other end.
Machines are generally used to perform different tasks faster.
However, a simple machine can not be used to increase the amount of work done at any time.
Force, speed and torque can all be increased using machines.
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A 771.0-kg copper bar is melted in a smelter. The initial temperature of the copper is 300.0 K. How much heat must the smelter produce to completely melt the copper bar? For solid copper, the specific heat is 386 J/kg • K, the heat of fusion is 205 kJ/kg, and the melting point is 1357 K.
Answer:
4.73 × 10^5
Explanation:
A single force acts on a particle situated on the positive x axis. The torque about the origin is in the negative z direction. The force might be:_______.
A. in the positive y direction
B. in the negative y direction
C. in the positive x direction
D. in the negative x direction
A single force acts on a particle situated on the positive x axis. The torque about the origin is in the negative z direction. The force might be in the negative y direction. Thus, option B is correct.
To determine the force that could generate a torque in the negative z direction, we need to consider the right-hand rule for cross products. The torque vector, denoted by τ, is given by the cross product of the position vector, r, and the force vector, F:
[tex]τ = r × F[/tex]
In this case, the position vector, r, points along the positive x-axis. The negative z-direction torque indicates that the force vector must be perpendicular to both the position vector and the negative z-axis.
Using the right-hand rule, we can determine that the force vector must be in the negative y-direction, which is option B.
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A 2.0 kg puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. How far does the puck move from rest in 2.25 s?
Answer:
d = 6.32 m
Explanation:
Given that,
The mass of a puck, m = 2 kg
It is pushed straight north with a constant force of 5N for 1.50 s and then let go.
We need to find the distance covered by the puck when move from rest in 2.25 s.
We know that,
F = ma
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2[/tex]
Let d is the distance moved in 2.25 s. Using second equation of motion,
[tex]d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m[/tex]
So, it will move 6.32 m from rest in 2.25 seconds.
A 55kg bungee jumper has fallen far enough that her bungee cord is beginning to stretch and resist her downward motion . Find the ( magnitude and direction ) exerted on her by the bungee cord at an instant when her downward acceleration has a magnitude of 7.1m/s2
Answer:
148.5 N
Explanation:
Given that,
The mass of a bungee jumper, m = 55 kg
The downward acceleration, a = 7.1 m/s²
We need to find the net force acting on the jumper. As it is moving in downward direction, net force is :
T = m(g-a)
Put all the values,
T = 55(9.8 - 7.1)
= 148.5 N
So, the force exerted on the bungee cord is 148.5 N.
Answer:
The downward force is 148.5 N.
Explanation:
mass, m = 55 kg
downwards acceleration, a = 7.1 m/s^2
Let the force is F.
According to the newton's second law
m g - F = m a
F = m (g - a)
F = 55 (9.8 - 7.1)
F = 148.5 N
given A=4i-10j and B= 7i+5j find b such that A+bB is a vector pointing along the x-axis (i.e has no y component)
Answer:
-4/7
Explanation:
Given the following
A=4i-10j and B= 7i+5j
A+ bB = 4i-10j + (7i+5j)b
A+ bB = 4i-10j + 7ib+5jb
A+ bB =
The vector along the x-axis is expressed as i + 0j
If the vector A+ bB is pointing in the direction of the x-axis then;
[tex]A+ bB * \frac{i+0j}{|i+0j|} = 0 \\ (4+7b)i-(10-5b)j* \frac{i+0j}{\sqrt{1^2+0^2} } = 0\\(4+7b)i-(10-5b)j *(i+0j) = 0\\4+7b-0 =0\\7b=-4\\b = -4/7[/tex]
Hence the value of b is -4/7
The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.
According to the statement, we have following system of vectorial equations:
[tex]\vec A = 4\,\hat {i} - 10\,\hat{j}[/tex] (1)
[tex]\vec {B} = 7\,\hat{i} + 5\,\hat{j}[/tex] (2)
[tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] (3)
By applying (1) and (2) in (3):
[tex](4\,\hat{i}-10\,\hat{j}) + \beta\cdot (7\,\hat{i}+5\,\hat{j}) = c\,\hat{i}[/tex]
[tex](4+7\cdot \beta)\,\hat{i} +(-10+5\cdot \beta)\,\hat{j} = c\,\hat{i}[/tex]
And we get two scalar equations after analyzing each component:
[tex]4+7\cdot \beta = c[/tex] (4)
[tex]-10+5\cdot \beta = 0[/tex] (5)
We solve for [tex]\beta[/tex] in (5):
[tex]\beta = 2[/tex]
And for [tex]c[/tex] in (4):
[tex]c = 4+7\cdot (2)[/tex]
[tex]c = 18[/tex]
The value of [tex]\beta[/tex] such that [tex]\vec C = \vec A + \beta \cdot \vec B = c\,\hat{i}[/tex] is 2.
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what additional load will be required to cause the extension of 2.0cm when an elastic wire extend by 1.0cm when a load of 20g range from it
Answer:
The additional load is 20g
An object is acted upon by two and only two forces that are equal magnitude and oppositely directed. Is the objected necessarily in static equilibrium? Explain. You can draw a picture if that helps explain.
Answer:
the body is subjected to a continuous rotation and the body is not in rotational equilibrium
Explanation:
For an object to have a static equilibrium, it must meet two relationships
∑ F = 0
∑ τ =0
force acting on a body fulfills the relation of
sum F = F - F = 0
when two forces do not move from position.
To find the torque we assume that the counterclockwise rotations are positive
Σ τ = - F r - F r
Στ = -2 Fr <> 0
consequently the body is subjected to a continuous rotation and the body is not in rotational equilibrium
Megan accelerates her skateboard from 0 m/s to 8 m/s in 2 seconds. What is the magnitude of the acceleration of the skateboard?
O 8 m/s^2
O 16 m/s^2
O 2 m/s^2
O 4 m/s^2
Answer:
chk picture for eqn
Explanation:
What is not one of the main uses of springs?
A. Car suspension
B. Bike suspension
C. The seasons
D. Clock making
Our system is a block attached to a horizontal spring on a frictionless table. The spring has a spring constant of 4.0 N/m and a rest length of 1.0 m, and the block has a mass of 0.25 kg.
Compute the PE when the spring is compressed by 0.50 m.
Answer
E - 1/2 K x^2 potential energy of compressed spring
E = 1/2 * 4 N / m * (.5 m)^2 = 2 * .5^2 N-m = .5 N-m
Four identical balls are thrown from the top of a cliff, each with the same speed. The
first is thrown straight up, the second is thrown at 30° above the horizontal, the third
at 30° below the horizontal, and the fourth straight down. How do the speeds and
kinetic energies of the balls compare as they strike the ground? Ignore the effects of
air resistance. Explain fully using the concepts from this unit.
The comparison of the speeds and kinetic energy of the identical balls are as follows
The speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal
The reason for the above comparison results areas follows;
Known parameters;
First ball is thrown straight up
Second ball is thrown 30° above the horizontal
Third ball it thrown 30° below the horizontal
The fourth ball is thrown straight down
Unknown:
Comparison of the speed and kinetic energy of the four balls
Method:
The kinetic energy, K.E. = (1/2) × m × v²
The velocity of the ball, v = u × sin(θ)
Where;
u = The initial velocity of the ball
θ = The reference angle
For the first ball thrown straight up, we have;
θ = 90°
∴ [tex]v_y[/tex] = u
The final velocity of the ball as it strikes the ground is v₂ = u² + 2gh
Where;
h = The height of the cliff
∴ Kinetic energy of first ball, K.E.₁ = (1/2) × m × (u₁² + 2gh)²
For the second ball thrown 30° to the horizontal, we have;
K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²
Kinetic energy K.E.₂ = (1/2) × m × ((0.5·u₂)² + 2·g·h)²
For the third ball thrown at 30° below the horizontal, we have;K.E. = (1/2) × m × ((u×sin30)² + 2·g·h)² = K.E. = (1/2) × m × ((0.5·u)² + 2·g·h)²
Kinetic energy K.E.₃ = (1/2) × m × ((0.5·u₃)² + 2·g·h)²
For the fourth ball thrown straight down, we have;Kinetic energy K.E.₄ = (1/2) × m × (u₄² + 2gh)²
Therefore, as the ball strike the ground, the speed and the kinetic energy of the first and fourth ball are equal, while the speed and kinetic energy of the second and third balls are equal
u₁ = u₄, K.E₁ = K.E.₄, u₂ = u₃, K.E₂ = K.E.₃
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Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B
[tex]\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
Explanation:
Given:
[tex]\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}[/tex]
[tex]\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
The cross product [tex]\textbf{A}×\textbf{B}[/tex] is given by
[tex]\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|[/tex]
[tex]= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}[/tex]
[tex]= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}[/tex]
3. A microscope is focused on a black dot. When a 1.30 cm -thick piece of plastic is placed over the dot, the microscope objective has to be raised 0.410 cm to bring the dot back into focus. What is the index of refraction of the plastic
The index of refraction of the plastic is approximately 1.461
The known values in the question are;
The thickness of the piece of plastic placed on the dot = 1.30 cm
The height to which the microscope objective is raised to bring the dot back to focus = 0.410 cm
The unknown values in the question are;
The index of refraction
Strategy;
Calculate the refractive index by making use of the apparent height and real height method for the black dot under the thick piece of plastic
[tex]\mathbf{ Refractive \ index, n = \dfrac{Real \ depth}{Apparent \ depth}}[/tex]
The real depth of the dot below the piece of plastic, d₁ = 1.30 cm
The apparent depth of the dot, d₂ = The actual depth - The height to which the microscope is raised
Therefore;
The apparent depth of the dot, d₂ = 1.30 cm - 0.410 cm = 0.89 cm
[tex]The \ refractive \ index, \ n = \dfrac{d_1}{d_2}[/tex]
Therefore, n = 1.30/0.89 ≈ 1.461
The refractive index of the plastic block, n ≈ 1.461
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Although human beings have been able to fly hundreds of thousands of miles into outer space, getting inside the earth has proven much more difficult. The deepest mines ever drilled are only about 10 miles deep. To illustrate the difficulties associated with such drilling, consider the following: The density of steel is about 7900 kilograms per cubic meter, and its breaking stress, defined as the maximum stress the material can bear without deteriorating, is about 2.0×1092.0×109 pascals. What is the maximum length of a steel cable that can be lowered into a mine? Assume that the magnitude of the acceleration due to gravity remains constant at 9.8 meters per second per second.
Use two significant figures in your answer, expressed in kilometers.
Answer:
26 km
Explanation:
Let's say our "cable" has a cross section of 1 m²
Then each meter of cable would weight 7900(9.8) = 77420 N
A Pascal is a Newton per square meter
2 x 10⁹ / 77420 = 25840 m or about 26 km or about 16 miles
If the mass of an object is 15 kg and the velocity is -4 m/s, what is the momentum?
momentum p= m x v = 15 x -4 = -60 N.s
Consider two closely spaced and oppositely charged parallel metal plates. The plates are square with sides of length L and carry charges Q and -Q on their facing surfaces. What is the magnitude of the electric field in the region between the plates
Answer:
E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]
Explanation:
In this exercise you are asked to calculate the electric field between two plates, the electric field is a vector
E_ {total} = E₁ + E₂
E_ {total} = 2 E
where E₁ and E₂ are the fields of each plate, we have used that for the positively charged plate the field is outgoing and for the negatively charged plate the field is incoming, therefore in the space between the plates for a test charge the two fields point in the same direction
to calculate the field created by a plate let's use Gauss's law
Ф = ∫ E . dA = q_{int} /ε₀
As a Gaussian surface we use a cylinder with the base parallel to the plate, therefore the direction of the electric field and the normal to the surface are parallel, therefore the scalar product is reduced to the algebraic product.
E 2A = q_{int} / ε₀
where the 2 is due to the surface has two faces
indicate that the surface has a uniform charge for which we can define a surface density
σ = q_{int} / A
q_{int} = σ A
we substitute
E 2A = σ A /ε₀
E = σ / 2ε₀
therefore the total field is
E_ {total} = σ /ε₀
let's substitute the density for the charge of the whole plate
σ= Q / L²
E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]
a person lifts 60kg on the surface of the earth, how much mass can he lift on the surface of the moon if he applies same magnitude of force
Explanation:
Hey there!
According to the question;
A person can lift mass of 60 kg on earth.
mass(m1) = 60kg
acceleration due to gravity on earth (a) = 9.8m/s²
Now;
force (f) = m.a
= 60*9.8
= 588 N
Since, there is application of same magnitude of force on moon,
mass(m) =?
acceleration due to gravity on moon (a) = 1.67m/s²
Now;
force (f) = m.a
588 = m*1.67
m = 352.09 kg
Therefore, the person who can lift the mass of 60 kg on earth can lift mass of 352 kg on moon.
Hope it helps!
