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
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]
mention two similarities of citizen and aliens
Answer:
The main points of difference between a citizen and alien are: (a) A citizen is a permanent resident of a state, while an alien is a temporary resident, who comes for a specific duration of time as a tourist or on diplomatic assignment. ... Aliens do not possess such rights in the state where they reside temporarily
Explanation:
A package is dropped from a helicopter moving upward at 1.5 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was released if air resistance is negligible?
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
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
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
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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
Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 25.0 m above water with an initial speed of 20.0 m/s strikes the water with a final speed of what, independent of the direction thrown.
Complete question is;
Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 25.0 m above water with an initial speed of 20.0 m/s strikes the water with a final speed of 31.1 m/s, independent of the direction thrown
Answer:
It is proved that the final speed is truly 31.1 m/s
Explanation:
From energy - conservation principle;
E_i = Initial potential energy + Initial Kinetic Energy
Or
E_i = U_i + K_i
Similarly, for final energy
E_f = U_f + K_f
So, expressing the formulas for potential and kinetic energies, we now have;
E_i = (m × g × y_i) + (½ × m × v_i²)
Similarly,
E_f = (m × g × y_f) + (½ × m × v_f²)
We are given;
y_i = 25 m
y_f = 0 m
v_i = 20 m/s
v_f = 31.1 m/s
So, plugging in relevant values;
E_i = m((9.8 × 25) + (½ × 20²))
E_i = 485m
Similarly;
E_f = m((9.8 × 0) + (½ × v_f²)
E_f ≈ ½m•v_f²
From energy conservation principle, E_i = E_f.
Thus;
485m = ½m•v_f²
m will cancel out to give;
½v_f² = 485
v_f² = 485 × 2
v_f² = 970
v_f = √970
v_f ≈ 31.1 m/s
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
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.
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]
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
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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 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
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
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.
Mark this and return
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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 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
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 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 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]
1. In a Millikan type experiment, two horizontal plates are 2.5 cm apart. A latex sphere of
mass 1.5 x 10-15 kg remains stationary when the potential difference between the
plates is 460 V, with the upper plate positive. [2+2+2+2 = 8 marks]
a. Is the sphere charged negatively or positively?
b. What is the magnitude of the electric field intensity between the plates?
C. Calculate the magnitude of the charge on the latex sphere.
d. How many excess or deficit electrons does the sphere have?
Answer:
Explanation:
a. Is the sphere charged negatively or positively?
The sphere us negatively charged. In a Millikan type experiment, there will be two forces that will be acting on the sphere which are the electric force which acts upward and also the gravity which acts downward.
Because the upper plate is positively charged, there'll what an attractive curve with an upward direction which will be felt by the negatively charged sphere.
b. What is the magnitude of the electric field intensity between the plates?
The magnitude of the electric field intensity between the plates is 18400v/m.
C. Calculate the magnitude of the charge on the latex sphere.
The magnitude of the charge on the latex sphere hae been solved and attached
d. How many excess or deficit electrons does the sphere have?
There are 5 excess electrons that the sphere has.
Check the attachment for further explanation.
The magnitude of the magnetic flux through the surface of a circular plate is 6.80 10-5 T · m2 when it is placed in a region of uniform magnetic field that is oriented at 43.0° to the vertical. The radius of the plate is 8.50 cm. Determine the strength of the magnetic field. mT A circular plate of radius r is lying flat. A field of arrows labeled vector B rising up and to the right pass through the plate.
Answer:
B = 4.1*10^-3 T = 4.1mT
Explanation:
In order to calculate the strength of the magnetic field, you use the following formula for the magnetic flux trough a surface:
[tex]\Phi_B=S\cdot B=SBcos\alpha[/tex] (1)
ФB: magnetic flux trough the circular surface = 6.80*10^-5 T.m^2
S: surface area of the circular plate = π.r^2
r: radius of the circular plate = 8.50cm = 0.085m
B: magnitude of the magnetic field = ?
α: angle between the direction of the magnetic field and the normal to the surface area of the circular plate = 43.0°
You solve the equation (1) for B, and replace the values of the other parameters:
[tex]B=\frac{\Phi_B}{Scos\alpha}=\frac{6.80*10^{-5}T.m^2}{(\pi (0.085m)^2)cos(43.0\°)}\\\\B=4.1*10^{-3}T=4.1mT[/tex]
The strength of the magntetic field is 4.1mT
An X-Ray machine delivers a radiation dose of 5mRem/hr. at 3ft from the machine. How far will the X-Ray technician have to move to reduce his exposure to 2mRem/hr.? I1/I2 = (D2)2/(D1)2 -------> I1(D1)2 = I2(D2)2
Answer:
4.7ft
Explanation:
Pls see attached file
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 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]
Please Help!!!! I WILL GIVE BRAINLIEST!!!!!!!!!!!!!
Upon using Thomas Young’s double-slit experiment to obtain measurements, the following data were obtained. Use these data to determine the wavelength of light being used to create the interference pattern. Do this using three different methods.
The angle to the eighth maximum is 1.12°.
The distance from the slits to the screen is 302.0 cm.
The distance from the central maximum to the fifth minimum is 3.33 cm.
The distance between the slits is 0.000250 m.
The 3 equations I used were 1). d sin θ_m =(m)λ 2). delta x =λL/d and 3.) d(x_n)/L=(n-1/2)λ
but all my answers are different.
DID I DO SOMETHING WRONG!!!!!!!
Given info
d = 0.000250 meters = distance between slits
L = 302 cm = 0.302 meters = distance from slits to screen
[tex]\theta_8 = 1.12^{\circ}[/tex] = angle to 8th max (note how m = 8 since we're comparing this to the form [tex]\theta_m[/tex])
[tex]x_n = x_5 = 3.33 \text{ cm} = 0.0333 \text{ meters}[/tex] (n = 5 as we're dealing with the 5th minimum )
---------------
Method 1
[tex]d\sin(\theta_m) = m\lambda\\\\0.000250\sin(\theta_8) = 8\lambda\\\\8\lambda = 0.000250\sin(1.12^{\circ})\\\\\lambda = \frac{0.000250\sin(1.12^{\circ})}{8}\\\\\lambda \approx 0.000 000 61082633\\\\\lambda \approx 6.1082633 \times 10^{-7} \text{meters}\\\\ \lambda \approx 6.11 \times 10^{-7} \text{ meters}\\\\ \lambda \approx 611 \text{ nm}[/tex]
Make sure your calculator is in degree mode.
-----------------
Method 2
[tex]\Delta x = \frac{\lambda*L*m}{d}\\\\L*\tan(\theta_m) = \frac{\lambda*L*m}{d}\\\\\tan(\theta_m) = \frac{\lambda*m}{d}\\\\\tan(\theta_8) = \frac{\lambda*8}{0.000250}\\\\\tan(1.12^{\circ}) = \frac{\lambda*8}{0.000250}\\\\\lambda = \frac{1}{8}*0.000250*\tan(1.12^{\circ})\\\\\lambda \approx 0.00000061094306 \text{ meters}\\\\\lambda \approx 6.1094306 \times 10^{-7} \text{ meters}\\\\\lambda \approx 611 \text{ nm}\\\\[/tex]
-----------------
Method 3
[tex]\frac{d*x_n}{L} = \left(n-\frac{1}{2}\right)\lambda\\\\\frac{0.000250*3.33}{302.0} = \left(5-\frac{1}{2}\right)\lambda\\\\0.00000275662251 \approx \frac{9}{2}\lambda\\\\\frac{9}{2}\lambda \approx 0.00000275662251\\\\\lambda \approx \frac{2}{9}*0.00000275662251\\\\\lambda \approx 0.00000061258279 \text{ meters}\\\\\lambda \approx 6.1258279 \times 10^{-7} \text{ meters}\\\\\lambda \approx 6.13 \times 10^{-7} \text{ meters}\\\\\lambda \approx 613 \text{ nm}\\\\[/tex]
There is a slight discrepancy (the first two results were 611 nm while this is roughly 613 nm) which could be a result of rounding error, but I'm not entirely sure.
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 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]
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.
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.