Answer:
so initial speed of the rock is 30.32 m/s
correct answer is b. 30.3 m/s
Explanation:
given data
h = 15.0m
v = 25m/s
weight of the rock m = 3.00N
solution
we use here work-energy theorem that is express as here
work = change in the kinetic energy ..............................1
so it can be written as
work = force × distance ...................2
and
KE is express as
K.E = 0.5 × m × v²
and it can be written as
F × d = 0.5 × m × (vf)² - (vi)² ......................3
here
m is mass and vi and vf is initial and final velocity
F = mg = m (-9.8) , d = 15 m and v{f} = 25 m/s
so put value in equation 3 we get
m (-9.8) × 15 = 0.5 × m × (25)² - (vi)²
solve it we get
(vi)² = 919
vi = 30.32 m/s
so initial speed of the rock is 30.32 m/s
An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about _____ years.
Answer:
An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about 2 years.
Explanation:
Given;
orbital period of 3 years, P = 3 years
To calculate the years of an orbital with a semi-major axis, we apply Kepler's third law.
Kepler's third law;
P² = a³
where;
P is the orbital period
a is the orbital semi-major axis
(3)² = a³
9 = a³
a = [tex]a = \sqrt[3]{9} \\\\a = 2.08 \ years[/tex]
Therefore, An asteroid that has an orbital period of 3 years will have an orbital with a semi-major axis of about 2 years.
In the child's game of tetherball, a rope attached to the top of a tall pole is tied to a ball. Players hit the ball in opposite directions in an attempt to wrap the ball and rope around the pole. Assume the rope has negligible mass and that resistive forces, such as air resistance and friction, can be neglected. As the ball wraps around the pole between hits, how does the angular speed of the ball change
Answer:
The angular speed of the ball will increase
Explanation:
the angular speed of the ball will increase because the force of hit by the players will sum up in opposite direction to increase the angular speed
Two charged concentric spheres have radii of 0.008 m and 0.018 m. The charge on the inner sphere is 3.62 10-8 C and that on the outer sphere is 1.62 10-8 C. Find the magnitude of the electric field (in N/C) at 0.012 m.
Answer:
The electric field is [tex]E = 2.2625 *10^{6} \ N/C[/tex]
Explanation:
From the question we are told that
The radius of the inner sphere is [tex]r_1 = 0.008\ m[/tex]
The radius of the outer sphere is [tex]r _2 = 0.018 \ m[/tex]
The charge on the inner sphere is [tex]q_1 = 3.62 *10^{-8} \ C[/tex]
The charge on the outer sphere is [tex]q_2 = 1.62 *10^{-8} \ C[/tex]
The position from the origin is [tex]d = 0.012 \ m[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{k (q_1 )}{ r^2}[/tex]
The reason for using [tex]q_1[/tex] for the calculation is due to the fact that the position considered is greater than the [tex]r_1[/tex] but less than [tex]r_2[/tex]
Here k is the Coulomb constant with value [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A{-2}[/tex]
So
[tex]E = \frac{9*10^9 (3.62 *10^{-8}}{0.012^2}[/tex]
[tex]E = 2.2625 *10^{6} \ N/C[/tex]
Suppose you take two non-zero displacements represented by vectors A & B.The magnitude of A is 5 m and the magnitude of B is 6 m. Under what circumstances can you end up back at your starting point? What is the magnitude of the largest displacement you can end up from the starting point?
Suppose you take two non-zero displacements represented by vectors A & B which are perpendicular to each other. The magnitude of A is 5 m and the magnitude of B is 6 m. What is the component of vector B along the direction of vector A measured in m?
Answer:
a. When the total displacement is -(A + B)
b. A + B = 1 m or -(A + B) = -11 m
c. 0 m
Explanation:
a. Under what circumstances can you end up back at your starting point?
If we have the displacement A and displacement B. The total displacement is A + B. We would end up at the starting point if we take a displacement -(A + B) from point B
b. What is the magnitude of the largest displacement you can end up from the starting point?
The maximum displacement we can obtain is when A and B are in the same direction. So A + B = 5 m + 6 m = 11 m or -A - B = -(A + B) = -11 m.
c. When A and B are perpendicular, what is the component of B in the direction of A?
Since A is perpendicular to B, the angle between A and B is 90°
So the component of B in A,s direction is Bcos90° = B × 0 = 0 m
A 5.0-Ω resistor and a 9.0-Ω resistor are connected in parallel. A 4.0-Ω resistor is then connected in series with this parallel combination. An ideal 6.0-V battery is then connected across the series-parallel combination of the three resistors. What is the current through (a) the 4.0-Ω resistor? (b) the 5.0-Ω resistor? (c) the 9.0-Ω resistor?
Answer:
Explanation:
The current through the resistor is 0.83 A
.
Part b
The current through resistor is 0.53 A
.
Part c
The current through resistor is 0.30 A
A diver running at 2.5 m/s dives out horizontally from the edge of a vertical cliff and 3.0 seconds later reaches the water below. How far from its base did the diver hit the water
Explanation:
u = 2.5 m/s
v = 0
t = 3sec
s = ?
s = (u+v)/t
s = (0+2.5)/3
s = 2.5/3 = 0.83 m
In a two-slit experiment, monochromatic coherent light of wavelength 500 nm passes through a pair of slits separated by 1.30 x 10-5 m. At what angle away from the centerline does the first bright fringe occur
Answer:
2.20°
Explanation:
For the central bright spot, we will use the constructive pattern for a double slit interference,
[tex]m\times w = d \times Sin\beta[/tex]
where w indicates the wavelength
and [tex]\beta[/tex] indicates the angle between the bright spot and center line.
now we will use the given values,
1 × 500 × 10^-9 = 1.3 × 10^-5 × Sin [tex]\beta[/tex]
Solving for [tex]\beta[/tex],
[tex]\beta[/tex] = 2.204° ~ 2.20°
Therefore the correct answer is 2.20°
A 975-kg pickup comes to rest from a speed of 87.5 km/h in a distance of 125 m. Suppose the pickup is initially traveling in the positive direction.
(a) If the brakes are the only thing making the car come to a stop, calculate the force (in newtons, in a component along the direction of motion of the car) that the brakes apply on the car .
(b) Suppose instead of braking that the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force, in newtons, exerted on the car in this case 33%
(c) What is the ratio of the force on the car from the concrete to the braking force?
Answer:
A) Force = 2303.925 N in the negative x-direction
B) F ≈ 143998.28 N
C) Ratio = 62.5
Explanation:
A) Since the brakes are the only thing making the van to come to a stop, then first of all, we will calculate the force (in a component along the direction of motion of the car) that the brakes will apply on the van.
Let's find the deceleration using Newton's law of motion formula;
v² = u² + 2as
where;
v = final velocity,
u = initial velocity,
s = displacement
a = acceleration
We are given;
u = 87.5 km/h = 24.3056 m/s
s = 125 m
v = 0 m/s
Thus;
0 = (24.3056)² + 2a(125)
- (24.3056)²= 250a
a = - 24.3056²/250
a = - 2.363 m/s²
Now, force = mass × acceleration
We are given mass = 975 kg
Thus;
Force = 975 x (-2.363)
Force = 2303.925 N in the negative x-direction
B) formula for kinetic energy is
KE = ½mv²
KE = ½(975)(24.3056)²
= 287996.568288 J
Work done on impact = F x 2
Thus;
2F = 287996.568288
F = 287996.568288/2
F ≈ 143998.28 N
C) Ratio = Force on car/braking force = 143998.284/2303.925 = 62.5
A particle confined to a motion along the x axis moves with a constant acceleration of 2.5m/s2. Its velocity at t=0s is 6m/s. Find its velocity at t=4s.
Answer:
v = 16 m/s
Explanation:
It is given that,
Acceleration of a particle along x -axis is [tex]2.5\ m/s^2[/tex]
At t = 0s, its velocity is 6 m/s
We need to find the velocity at t = 4 s
It means that the initial velocity of the particle is 6 m/s
Let v is the velocity at t = 4 s
So,
v = u + at
[tex]v=6+2.5\times 4\\\\v=16\ m/s[/tex]
So, the velocity at t = 4 s is 16 m/s.
Answer:
v = 16 m/s
Explanation:
It is given that,
Acceleration of a particle along x -axis is
At t = 0s, its velocity is 6 m/s
We need to find the velocity at t = 4 s
It means that the initial velocity of the particle is 6 m/s
Let v is the velocity at t = 4 s
So,
v = u + at
So, the velocity at t = 4 s is 16 m/s.
A boat that has a speed of 6km / h must cross a 200m wide river perpendicular to the current that carries a speed of 1m / s. Calculate a) the final speed of the boat b) displacement experienced by the boat in the direction of the current when making the journey
Answer:
a) 1.94 m/s
b) 120 m
Explanation:
Convert km/h to m/s:
6 km/h = 1.67 m/s
a) The final speed is found with Pythagorean theorem:
v = √((1.67 m/s)² + (1 m/s)²)
v = 1.94 m/s
b) The time it takes the boat to cross the river is:
t = (200 m) / (1.67 m/s)
t = 120 s
The displacement in the direction of the current is:
x = (1 m/s) (120 s)
x = 120 m
How did the magnet’s density measurement using the Archimedes’ Principle compare to the density measurement using the calculated volume? Which method might be more accurate? Why?
Answer:
The two methods will yield different results as one is subject to experimental errors that us the Archimedes method of measurement, the the density measurement method will be more accurate
Explanation:
This is because the density method using the calculated volume will huve room for less errors that's occur in practical method i.e Archimedes method due to human error
A horizontal clothesline is tied between 2 poles, 12 meters apart. When a mass of 1 kilograms is tied to the middle of the clothesline, it sags a distance of 4 meters. What is the magnitude of the tension on the ends of the clothesline
Answer:
The tension on the clotheslines is [tex]T = 8.83 \ N[/tex]
Explanation:
The diagram illustrating this question is shown on the first uploaded image
From the question we are told that
The distance between the two poles is [tex]d = 12 \ m[/tex]
The mass tie to the middle of the clotheslines [tex]m = 1 \ kg[/tex]
The length at which the clotheslines sags is [tex]l = 4 \ m[/tex]
Generally the weight due to gravity at the middle of the clotheslines is mathematically represented as
[tex]W = mg[/tex]
let the angle which the tension on the clotheslines makes with the horizontal be [tex]\theta[/tex] which mathematically evaluated using the SOHCAHTOA as follows
[tex]Tan \theta = \frac{ 4}{6}[/tex]
=> [tex]\theta = tan^{-1}[\frac{4}{6} ][/tex]
=> [tex]\theta = 33.70^o[/tex]
So the vertical component of this tension is mathematically represented a
[tex]T_y = 2* Tsin \theta[/tex]
Now at equilibrium the net horizontal force is zero which implies that
[tex]T_y - mg = 0[/tex]
=> [tex]T sin \theta - mg = 0[/tex]
substituting values
[tex]T = \frac{m*g}{sin (\theta )}[/tex]
substituting values
[tex]T = \frac{1 *9.8}{2 * sin (33.70 )}[/tex]
[tex]T = 8.83 \ N[/tex]
Five identical cylinders are each acted on by forces of equal magnitude. Which force exerts the biggest torque about the central axes of the cylinders
Answer:
From the image, the force as shown in option A will exert the biggest torque on the cylinder about its central axes.
Explanation:
The image is shown below.
Torque is the product of a force about the center of rotation of a body, and the radius through which the force acts. For a given case such as this, in which the cylinders are identical, and the forces are of equal magnitude, the torque at the maximum radius away from the center will exert the maximum torque. Also, the direction of the force also matters. To generate the maximum torque, the force must be directed tangentially away from the circle formed by the radius through which the force acts away from the center. Option A satisfies both condition and hence will exert the most torque on the cylinder.
A proton of mass and a charge of is moving through vacuum at a constant velocity of 10000 directly to the east when it enters a region of uniform electric field that points to the south with a magnitude of E =3.62e+3 N/C . The region of uniform electric field is 5 mm wide in the east-west direction.
Required:
How far (in meters) will the proton have been deflected towards the south by the time it exits the region of uniform electric field.
Complete Question
A proton of mass mp= 1.67×10−27 kg and a charge of qp= 1.60×10−19 C is moving through vacuum at a constant velocity of 10,000 m/s directly to the east when it enters a region of uniform electric field that points to the south with a magnitude of E = 3.62e+3 N/C . The region of uniform electric field is 5 mm wide in the east-west direction. How far (in meters) will the proton have been deflected towards the south by the time it exits the region of uniform electric field. You may neglect the effects of friction and gravity, and assume that the electric field is zero outside the specified region. Answer is to be in units of meters
Answer:
[tex]s = 0.039 \ m[/tex]
Explanation:
From the question we are told that
The mass of the proton is [tex]m = 1.67 *10^{-27} \ g[/tex]
The charge of on the proton is [tex]q = 1.60 *10^{-19} \ C[/tex]
The speed of the proton is [tex]v = 10000 \ m/s[/tex]
The magnitude of the electric field is [tex]E = 3.62*10^{3 } \ N/C[/tex]
The width covered by the electric field [tex]d = 5mm = 5 *10^{-3} \ m[/tex]
Generally the acceleration of the proton due to the electric toward the south (at the point where the force on the proton is equal to the electric force due to the electric field) is mathematically represented as
[tex]a = \frac{q* E}{m}[/tex]
Substituting values
[tex]a = \frac{1.60*10^{-19 } * 3.26 *10^{3}}{ 1.67*10^{-27}}[/tex]
[tex]a = 3.12*10^{11} \ m/s^2[/tex]
Generally the time it will take the proton to cross the electric field is mathematically represented as
[tex]t = \frac{d}{v}[/tex]
Substituting values
[tex]t = \frac{5 *10^{-3}}{10000}[/tex]
[tex]t = 5 *10^{-7} \ s[/tex]
Generally the the distance covered by the proton toward the south is
[tex]s = ut + \frac{1}{2} * a*t^2[/tex]
Here u = 0 m/s this because before the proton entered the electric field region the it velocity towards the south is zero
So
[tex]s = \frac{1}{2} * a*t^2[/tex]
Substituting values
[tex]s = \frac{1}{2} * 3.12 *10^{11}*(5 *10^{-7})^2[/tex]
[tex]s = 0.039 \ m[/tex]
A very large sheet of a conductor carries a uniform charge density of on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?
Complete Question
A very large sheet of a conductor carries a uniform charge density of [tex]4.00\ pC/mm^2[/tex] on its surfaces. What is the electric field strength 3.00 mm outside the surface of the conductor?
Answer:
The electric field is [tex]E = 4.5198 *10^{5} \ N/C[/tex]
Explanation:
From the question we are told that
The charge density is [tex]\sigma = 4.00pC /mm^2 = 4.00 * 10^{-12 } * 10^{6} = 4.00 *10^{-6}C/m[/tex]
The position outside the surface is [tex]a = 3.00 \ mm = 0.003 \ m[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{\sigma}{\epsilon _o }[/tex]
Where [tex]\epsilon_o[/tex] is the permitivity of free space with values [tex]\epsilon _o = 8.85 *10^{-12} F/m[/tex]
substituting values
[tex]E = \frac{4.0*10^{-6}}{8.85 *10^{-12} }[/tex]
[tex]E = 4.5198 *10^{5} \ N/C[/tex]
When a nerve cell fires, charge is transferred across the cell membrane to change the cell's potential from negative to positive. For a typical nerve cell, 9.2pC of charge flows in a time of 0.52ms .What is the average current through the cell membrane?
Answer:
The average current will be "17.69 nA".
Explanation:
The given values are:
Charge,
q = 9.2 pC
Time,
t = 0.52ms
The equivalent circuit of the cell surface is provided by:
⇒ [tex]i_{avg}=\frac{charge}{t}[/tex]
Or,
⇒ [tex]i_{avg}=\frac{q}{t}[/tex]
On substituting the given values, we get
⇒ [tex]=\frac{9.2\times 10^{-12}}{0.52\times 10^{-3}}[/tex]
⇒ [tex]=17.69^{-9}[/tex]
⇒ [tex]=17.69 \ nA[/tex]
a car slows down from - 27.7 m/s to -10.9 m/s while undergoing a displacement of -105 m .what is its acceleration?
Answer:
3.09 m/s²
Explanation:
Given:
Δx = -105 m
v₀ = -27.7 m/s
v = -10.9 m/s
Find: a
v² = v₀² + 2aΔx
(-10.9 m/s)² = (-27.7 m/s)² + 2a (-105 m)
a = 3.09 m/s²
Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. When each of the spheres has lost half its initial charge, the magnitude of the electrostatic force will be
Answer:
F' = F/4
Thus, the magnitude of electrostatic force will become one-fourth.
Explanation:
The magnitude of force applied by each charge on one another can be given by Coulomb's Law:
F = kq₁q₂/r² -------------- equation 1
where,
F = Force applied by charges
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of 2nd charge
r = distance between the charges
Now, in the final state the charges on both spheres are halved. Therefore,
q₁' = q₁/2
q₂' = q₂/2
Hence, the new force will be:
F' = kq₁'q₂'/r²
F' = k(q₁/2)(q₂/2)/r²
F' = (kq₁q₂/r²)(1/4)
using equation 1:
F' = F/4
Thus, the magnitude of electrostatic force will become one-fourth.
The magnitude of the electrostatic force will be F' = F/4
The magnitude of the electrostatic force:Here we used Coulomb's Law:
F = kq₁q₂/r² -------------- equation 1
Here
F = Force applied by charges
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of 2nd charge
r = distance between the charges
Now
q₁' = q₁/2
q₂' = q₂/2
So, the new force should be
F' = kq₁'q₂'/r²
F' = k(q₁/2)(q₂/2)/r²
F' = (kq₁q₂/r²)(1/4)
So,
F' = F/4
Learn more about force here: https://brainly.com/question/14282312
Inexperienced physics teachers often demonstrate the use of the electroscope by touching it with a charged glass rod at a single point. More experienced teachers typically drag the length of the rod across the top of the electroscope to increase the desired effect. Why does this help
Answer:
This is because The glass rod of the electroscope is an insulator therefore only charge transferred to the ball is at the point of contact on the rod. Thus, When the charge rod is dragged across the top of the electroscope, by the experienced teacher the more charge is transferred to electroscope thereby producing a greater effect
A certain lightning bolt moves 40.0 C of charge. How many units ???? of fundamental charge e is this?
q = 40 C
e = 1.6×10^-19 C
n = ?
n = q/e
n = 40/1.6×10^-19 C
= 2.6×10^20
How does an atom of rubidium-85 become a rubidium ion with a +1 charge?
Answer:
C. The atom loses 1 electron to have a total of 36.
Explanation:
Cations have a positive charge. Cations lose electrons.
The number of electrons in a Rubidium atom is 37. If the atom loses 1 electron, then it has 36 left.
g power output of 87 W. At what distance will the decibel reading be 120 dB, which is noise level of a loud indoor rock concert
Given that,
Output power = 87 W
Decibel reading = 120 dB
We need to calculate the intensity of sound
Using formula of intensity of sound
[tex]dB=10\log(\dfrac{I}{I_{0}})[/tex]
Put the value into the formula
[tex]120=10\log(\dfrac{I}{1\times10^{-12}})[/tex]
[tex]12=log(\dfrac{I}{1\times10^{-12}})[/tex]
[tex]10^{12}=\dfrac{I}{1\times10^{-12}}[/tex]
[tex]I=10^{12}\times1\times10^{-12}[/tex]
[tex]I=1\ W/m^2[/tex]
We need to calculate the distance
Using relation of power out[ut and intensity
[tex]I=\dfrac{P}{4\pi r^2}[/tex]
Put the value into the formula
[tex]1=\dfrac{87}{4\pi r^2}[/tex]
[tex]r=\sqrt{\dfrac{87}{4\pi}}[/tex]
[tex]r=2.63\ m[/tex]
Hence, The distance is 2.63 m
Will an object present on the surface of the earth possess any energy?
Answer:
An object lying on the surface of the earth has energy like it maybe kinetic or potential energy
Explanation:
Hope it will help you :)
You want to produce three 2.00-mm-diametercylindrical wires, each with a resistance of 1.00 Ω at room temperature. One wire is gold (rhog=2.44×10^−8Ω⋅m), one is copper (rhoc=1.72×10^−8Ω⋅m), and one is aluminum (rhoa=2.75×10−8Ω⋅m).
Required:
a. What will be the length of the gold wire?
b. What will be the length of the copper wire?
c. What will be the length of the aluminum wire?
d. Gold has a density of 1.93 × 10^4 kg/m^3. What will be the mass of the gold wire?
e. If gold is currently worth $40 per gram, what is the cost of the gold wire?
Answer:
(a) L = 128.75 m
(b) L = 182.56 m
(c) L = 114.28 m
(d) Mass of Gold = 7.68 kg = 7680 gram
(e) Cost of Gold Wire = $ 307040
Explanation:
The resistance of the wire is given as:
R = ρL/A
where,
R = Resistance
ρ = resistivity
L = Length
A = cross-sectional area
(a)
For Gold Wire:
ρ = 2.44 x 10⁻⁸ Ω.m
A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²
R = 1 Ω
Therefore,
1 Ω = (2.44 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)
L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.44 x 10⁻⁸ Ω.m)
L = 128.75 m
(b)
For Copper Wire:
ρ = 1.72 x 10⁻⁸ Ω.m
A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²
R = 1 Ω
Therefore,
1 Ω = (1.72 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)
L = (1 Ω)(3.14 x 10⁻⁶ m²)/(1.72 x 10⁻⁸ Ω.m)
L = 182.56 m
(c)
For Aluminum Wire:
ρ = 2.75 x 10⁻⁸ Ω.m
A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²
R = 1 Ω
Therefore,
1 Ω = (2.75 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)
L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.75 x 10⁻⁸ Ω.m)
L = 114.28 m
(d)
Density = Mass/Volume
Mass = (Density)(Volume)
Volume of Gold = AL = (3.14 x 10⁻⁶ m²)(128.75 m) = 4.04 x 10⁻⁴ m³
Therefore,
Mass of Gold = (1.9 x 10⁴ kg/m³)(4.04 x 10⁻⁴ m³)
Mass of Gold = 7.68 kg = 7680 gram
(e)
Cost of Gold Wire = (Unit Price of Gold)(Mass of Gold)
Cost of Gold Wire = ($ 40/gram)(7680 grams)
Cost of Gold Wire = $ 307040
(a) L is = 128.75 m
(b) L is = 182.56 m
(c) L is = 114.28 m
(d) Mass of Gold is = 7.68 kg = 7680 gram
(e) Cost of Gold Wire is = $307040
Calculation of Diameter cylindricalWhen The resistance of the wire is given as:
R is = ρL/A
Now, where
R is = Resistance
ρ is = resistivity
L is = Length
A is = cross-sectional area
(a) For Gold Wire is:
ρ is = 2.44 x 10⁻⁸ Ω.m
A is = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²
R is = 1 Ω
Thus,
1 Ω is = (2.44 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)
L is = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.44 x 10⁻⁸ Ω.m)
L is = 128.75 m
(b) For Copper Wire is:
ρ is = 1.72 x 10⁻⁸ Ω.m
Then A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²
R is = 1 Ω
Thus,
After that 1 Ω = (1.72 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)
Now, L = (1 Ω)(3.14 x 10⁻⁶ m²)/(1.72 x 10⁻⁸ Ω.m)
Therefore, L = 182.56 m
(c) For Aluminum Wire is:
ρ is = 2.75 x 10⁻⁸ Ω.m
A is = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²
R is = 1 Ω
Thus,
Then 1 Ω = (2.75 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)
After that L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.75 x 10⁻⁸ Ω.m)
L = 114.28 m
(d) Density is = Mass/Volume
Mass is = (Density)(Volume)
Then Volume of Gold = AL = (3.14 x 10⁻⁶ m²)(128.75 m) = 4.04 x 10⁻⁴ m³
Therefore,
Now, Mass of Gold = (1.9 x 10⁴ kg/m³)(4.04 x 10⁻⁴ m³)
Then Mass of Gold = 7.68 kg = 7680 gram
(e) The Cost of Gold Wire is = (Unit Price of Gold)(Mass of Gold)
Than Cost of Gold Wire = ($ 40/gram)(7680 grams)
Therefore, The Cost of Gold Wire is = $ 307040
Find more information about Diameter cylindrical here:
https://brainly.com/question/26988752
A 30.0-g object moving to the right at 19.5 cm/s overtakes and collides elastically with a 13.0-g object moving in the same direction at 15.0 cm/s. Find the velocity of each object after the collision. (Take the positive direction to be to the right. Indicate the direction with the sign of your answer.) 30.0-g object Incorrect: Your answer is incorrect. seenKey 16.8 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. cm/s 13.0-g object
Answer:
[tex]V(30g)=1.4\frac{cm}{s} \\V(13g)=5.9\frac{cm}{s}[/tex]
Explanation:
So my calculations may be off (the final step is plugging in a bunch of things and getting a value, which creates room for error [and they honestly seem too small]) but I'm confident that the process is correct. I'll upload my work shortly but here is the method:
Use the Principle of Conservation of Momentum (PCoM) to set up a statement between the momentum of each object before and after the collision. It should be [before]=[after] because the collision is said to be elastic.Do the same thing using the Principle of Conservation of Energy (PCoE). Note that you can start with either one, the problem can't really be done without both anyways.You can chose to either divide the PCoM expression by the PCoE expression or do a series of substitutions. If you do the sub., you'll find that it's the same thing as just dividing them outright. (This is a little time saving trick I picked up from one of my Physics professors, I definitely didn't come up with it on my own.)After that, you should be able to reduce the expressions into something with only the two initial velocities and the two final velocities. Note that in order to reduce, you'll need to use the "difference of squares" relationship in the PCoE expression. Choose either one of the final velocities to solve for first via substitution. In my example, I chose to solve for the 30g mass's final velocty first by subbing out the 13g mass's final velocity. (If you do it the other way around, your work might look different from mine at the end but it should give the same answer either way.)After substituting, you should have an expression where the final velocity equals some combination of the masses and initial velocities of both objects (it's not a "nice" looking equation).With one final velocity found, you can plug it into the expression you found in step 4. to find the other mass's final velocity.And with that, you've got it! This whole process is kinda long and involved so I would try practicing it a lot before any tests/quizzes so it doesn't eat up your time.
Edit: In my work I made u to be the initial velocities and v to be the final velocities because it was easier to keep track of
Answer:
The final velocity of the 30 g object is 16.8 cm/s
The final velocity of the 13 g object is 21.3 cm/s
Explanation:
Let's study the elastic collision with conservation of linear momentum, assigning object 1 to the 30 g object, and object 2 to the 13 gr object:
[tex]p_{1\,i}+p_{2\,i}=p_{1\.f}+p_{2\,f}\\(30)\,(19.5)+(13)\,(15) = (30)\,v_{1\,f} +(13)\,v_{2\,f}\\780 = (30)\,v_{1\,f} +(13)\,v_{2\,f}[/tex]
so we can write one of the unknowns in terms of the other one:
[tex]v_{1\,f}=(780-13\,v_{2\,f})/30[/tex]
Now we analyze the equation for conservation of kinetic energy that verifies in elastic collisions:
[tex]\frac{30}{2} (19.5)^2+\frac{13}{2} \,(15)^2=\frac{30}{2} (v_{1\,f})^2+\frac{13}{2} \,(v_{2\,f})^2\\7166.25=15\, (v_{1\,f})^2+6.5\,(v_{2\,f})^2[/tex]
now we can write this quadratic equation replacing [tex]v_{1\,f}[/tex] with its expression in terms of [tex]v_{2\,f}[/tex] and solve it (with the help of a graphing calculator is simpler by looking for the roots).
We get two answers for [tex]v_{2\,f}[/tex] : one 15 cm/s, and the other one 21.28 cm/s.
We select the 21.28 cm/s answer since otherwise, the situation is the same as the initial one at which the second object was moving at 15 cm/s.
This velocity can be rounded to one decimal to: 21.3 cm/s
Given the value 21,28 for [tex]v_{2\,f}[/tex] , then:
[tex]v_{1\,f}=(780-13\,v_{2\,f})/30\\v_{1\,f}=(780-13\,(21.28))/30=16.78 \,\,cm/s[/tex]
which can be rounded to 16.8 cm/s
A 1,470-N force pushes a 500-kg piano up along a ramp. What is the work done by the 1,470-N pushing force on the piano as it moves 10 m up the ramp
Answer:
W = 14700 J
Explanation:
This is an exercise on Newton's second law.
To solve it we must fix a coordinate system, the most common is an axis parallel to the ramp and the other perpendicular axis, we write Newton's second law
Y Axis . Perpendicular to the ramp
N - Wy = 0
X axis. Parallel to the ramp, we assume it is positive when the ramp is going up
F - Wx = m a
in this case F = 1470 N and it is parallel to the plane.
Work is defined by
W = F .d
boldface indicates vectors
W = F d cos θ
let's calculate
W = 1470 10 cos 0
W = 14700 J
A commercial diffraction grating has 500 lines per mm. Part A When a student shines a 480 nm laser through this grating, how many bright spots could be seen on a screen behind the grating
Answer:
The number of bright spot is m =4
Explanation:
From the question we are told that
The number of lines is [tex]s = 500 \ lines / mm = 500 \ lines / 10^{-3} m[/tex]
The wavelength of the laser is [tex]\lambda = 480 nm = 480 *10^{-9} \ m[/tex]
Now the the slit is mathematically evaluated as
[tex]d = \frac{1}{s} = \frac{1}{500} * 10^{-3} \ m[/tex]
Generally the diffraction grating is mathematically represented as
[tex]dsin\theta = m \lambda[/tex]
Here m is the order of fringes (bright fringes) and at maximum m [tex]\theta = 90^o[/tex]
So
[tex]\frac{1}{500} * sin (90) = m * (480 *10^{-3})[/tex]
=> [tex]m = 4[/tex]
This implies that the number of bright spot is m =4
A dumbbell-shaped object is composed by two equal masses, m, connected by a rod of negligible mass and length r. If I1 is the moment of inertia of this object with respect to an axis passing through the center of the rod and perpendicular to it and I2 is the moment of inertia with respect to an axis passing through one of the masses, it follows that:
a. I1 > I2
b. I2 > I1.
c. I1 = I2.
Answer:
B: I2>I1
Explanation:
See attached file
Which best describes the relationship between heat, internal energy, and thermal energy?
Internal energy is heat that flows, and heat is the part of thermal energy that can be transferred.
Internal energy is thermal energy that flows, and thermal energy is the part of heat that can be transferred.
Thermal energy is heat that flows, and heat is the part of internal energy that can be transferred.
Heat is thermal energy that flows, and thermal energy is the part of internal energy that can be transferred.
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Answer:
I think it is the 4th answer choice
Explanation:
Heat is thermal energy that flows in the direction of high temp to low temp, and internal energy is the "energy contained in a system", and thermal energy is a part of that.
A projectile is launched with V0 = 7.6 m/s and initial angle = 1.27 radians above the horizontal. What is the initial horizontal component of the projectile velocity in miles per hour?
Answer:
The horizontal component is [tex]v_h = 1.7096 \ m/s[/tex]
Explanation:
A diagram illustrating the projection is shown on the first uploaded image (from IB Maths Resources from British international school Phuket )
From the question we are told that
The initial velocity is [tex]v_o = 7.6 \ m/s[/tex]
The angle of projection is [tex]\theta = 1.27 \ rad = 72.77^o[/tex]
The horizontal component of this projectile velocity is mathematically represented as
[tex]v_h = v_o * cos (\theta )[/tex]
substituting values
[tex]v_h = 7.6 * cos (72.77 )[/tex]
[tex]v_h = 1.7096 \ m/s[/tex]