Answers
Answer:
The total force on the charge at the top of the triangle is approximately
[tex]1.56\,\,10^{-14}\,\,N[/tex]
Explanation:
Please look at the attached image to follow the explanation.
Since all charges are of the same positive value, the force exerted on the top charge by the other two are going to be on the line that joins the top charge with each of the other vertices of the triangle, pointing away from the top charge (illustrated by green vectors of the same length in the image).
We need to find the x and y components of these force vectors in order to find the resultant force by combining the x components among themselves, and the y components among themselves. Notice that the angle needed is in all cases [tex]60^o[/tex].
The x- components include the cosine of [tex]60^o[/tex], while the y components include the sine of [tex]60^o[/tex].
Notice as well that the x-components cancel each other (they have the same magnitude but point in opposite directions), while the y-components are both of the same value but pointing in the same direction (pointing up).
Then we just need to multiply by two the y component of one of the forces to find this total force.
Now, the magnitude of the forces in question are given by Coulomb's Law:
[tex]F_C=k\,\frac{q_1\,\,q_2}{d^2} =9\,\,10^9\,\frac{2\,\,10^{-12}\,\,2\,\,10^{-12}}{2^2} =9\,\,10^{-15}\,\,N[/tex]
therefore we can calculate what the y-component of this is using:
[tex]F_y=F\,sin(60^o)=9\,\,10^{-15} \,\frac{\sqrt{3} }{2} \\[/tex]
and twice this value becomes:
[tex]2\, \,F_y=(2)\,\,9\,\,10^{-15} \,\frac{\sqrt{3} }{2}=9\,\,\sqrt{3} \,\,10^{-15}\approx 1.56\,\,10^{-14}\,\,N[/tex]
Related Questions
what is the net force on an object that is experiencing a force of 25 N north, a force of 25 N south, a force of 50 N to the east and a force of 45 N to th west?
Answers
Answer:
5 n
Explanation:
25 and 25 cancel each other out and 50-45 is 5
Define fluid flow. Write five difference between uniform and non uniform flow.
Answers
Answer:
Fluid Flow is a part of fluid mechanics and deals with fluid dynamics. Fluids such as gases and liquids in motion are called fluid flow. It involves the motion of a fluid subjected to unbalanced forces. This motion continues as long as unbalanced forces are applied.
Difference:
Whereas in real fluids velocity varies across the section. But when the size and shape of cross section are constant along the length of channels under consideration, the flow is said to be uniform. A non-uniform flow is one in which velocity is not constant at a given instant.
What is the maximum wavelength of incident light that can produce photoelectrons from silver? The work function for silver is Φ=2.93 eV. (in nm)
Answers
Answer: Wavelength of 424 nm can produce photoelectrons from silver
Explanation:
[tex]\phi=h\times \nu_o=\frac{hc}{\lambda}[/tex]
[tex]\phi[/tex] = work function = energy of photon
h = Planck's constant = [tex]6.63\times 10^{-34}Js[/tex]
[tex]\nu_0[/tex] = frequency of the metal
c = speed of light = [tex]3\times 10^8m/s[/tex]
[tex]\lambda[/tex] =longest wavelength of the radiation
Now put all the given values in the above formula, we get the wavelength of the photons.
[tex]\lambda=\frac{hc}{\phi}[/tex]
[tex]\lambda=\frac{6.63\times 10^{-34}\times 3\times 10^8m/s}{2.93\times 1.6\times 10^{-19}J}[/tex] ( as 1ev=[tex]1.6\times 10^{-19}J[/tex] )
[tex]\lambda=4.24\times 10^{-7}m[/tex]
[tex]1nm=10^{-9}m[/tex]
[tex]\lambda=424nm[/tex]
Here, wavelength of 424 nm can produce photoelectrons from silver
A tall cylinder contains 20 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until the total liquid depth is 40 cm.
Required:
What is the gauge pressure at the bottom of the cylinder?
Answers
Answer:
Pressure, P = 3724 Pa
Explanation:
Given that,
Depth of water, [tex]h_w=20\ cm =0.2\ m[/tex]
Depth of oil, [tex]h_o=40-20=20\ cm=0.2\ m[/tex]
The density of water, [tex]d_w=1000\ kg/m^3[/tex]
The densinty of oil, [tex]d_o=900\ kg/m^3[/tex]
We need to find the gauge pressure at the bottom of the cylinder. So, total pressure is equal to :
[tex]P=d_wgh_w+d_ogh_o\\\\P=(d_wh_w+d_oh_o)g\\\\P=(1000\times 0.2+900\times 0.2)\times 9.8\\\\P=3724\ Pa[/tex]
So, the gauge pressure at the bottom of the cylinder is 3724 Pa.
"Find the change in gravitational potential energy that the box undergoes as it rises to its final height."
Answers
Explanation:
Given that,
Mass of a box is 2.6 kg
(1) We need to find the work done on the box by this force as it is pushed up the 5.00-m ramp to a height of 3.00 m.
It means that the position of the object is 3 m i.e. h = 3 m
Work done = Fd
= mgh
So,
[tex]W=2.6\times 9.8\times 3\\\\W=76.44\ J[/tex]
(2) Now the gravitational potential energy that the box undergoes as it rises to its final height is equal to the work done by the box. So,
Change in potential energy = 76.44 J
Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.
Answers
Answer:
R_total = 14.57 Ω , V_C = 1.176 V
Explanation:
To solve this circuit we are going to find the equivalent resistance of each branch, let's remember
* Serial resistance
[tex]R_{eq}[/tex] = ∑ [tex]R_{i}[/tex]
* For resistance in parallel
1 / R_{eq} = ∑ 1/R_{i}
We solve the two branches of the wheatstone bridge
Series resistors
Branch B
R_B = Rb + R4
R_B = 2 + 18
R_B = 20 Ω
Branch C
R_C5 = Rc + R5
R_C5 = 3 + 12
R_C5 = 15 Ω
Resistance in parallel R_B and R_C5
1 / R_BC = 1 / R_B + 1 / R_C5
1 / R_BC = 1/20 + 1/15 = 0.116666
R_BC = 8.57 Ω
Now we have a single branch, we solve the series resistance
R_total = R_A + R_BC
R_total = 6 + 8.57
R_total = 14.57 Ω
b) they ask us for the voltage in the resistance R_C
Let's remember that the voltage in a series circuit is the sum of the voltages
10 = V_a + V_BC
10 = i R_a + i R_BC = i (R_a + R_BC)
i = 10 / (R_a + R_BC)
i = 10 / (14.57)
i = 0.6863 A
The current in the series circuit is constant
V_BC = i R_BC
V_BC = 0.6863 8.57
V_BC = 5.8819 V
This voltage is divided in the bridge, for the two branches in parallel it is the same, but the resistance is different in each branch.
Branch C
V_BC = i R_C5
i = V_BC / R_C5
i = 5.8819 / 15
i = 0.39213 A
In this branch we have two resistors in series, let's remember that the current in a series circuit is constant
V_C = i R_C
V_C = 0.39213 3
V_C = 1.176 V
what single property was the most important in jesseca's material
Answers
Answer:
Jesseca wanted to create a material that reflected most of the light that fell on it.
Explanation: The Graphite was the material in the passage that had reflected most of the light.
a. What quantum number of the hydrogen atom comes closest to giving a 61-nm-diameter electron orbit?
b. What are the electron's speed and energy in this state?
Answers
Answer:
a
[tex]n = 23[/tex]
b
[tex]v = 87377.95 \ m/s[/tex]
Explanation:
From the question we are told that
The diameter is [tex]d = 61\ nm = 61 *10^{-9} \ m[/tex]
Generally the radius electron orbit is mathematically represented as
[tex]r = \frac{61 *10^{-9}}{2}[/tex]
=> [tex]r = 3.05*10^{-8} \ m[/tex]
This radius can also be represented mathematically as
[tex]r = n^2 * a_o[/tex]
Here n is the quantum number and [tex]a_o[/tex] is the Bohr radius with a value
[tex]a_o = 0.0529 *10^{-9} \ m[/tex]
So
[tex]n = \sqrt{\frac{3.05*10^{-8}}{ 0.059*10^{-9}} }[/tex]
=> [tex]n = 23[/tex]
Generally the angular momentum of the electron is mathematically represented as
[tex]L = m * v * r = \frac{n * h }{2 \pi}[/tex]
Here h is the Planck constant and the value is [tex]h = 6.626*10^{-34} J \cdot s[/tex]
m is the mass of the electron with values [tex]m = 9.1*10^{-31} \ kg[/tex]
So
[tex]v = \frac{23 * 6.626*10^{-34} }{2\pi * 9.1 *10^{-31} * 3.05*10^{-8} }[/tex]
[tex]v = 87377.95 \ m/s[/tex]
The half-life of element X is 20 years. If there are 48 g initially a) How much is there after 80 years
Answers
Answer:
After 80 years there will be 3 g of element X remaining
Explanation:
Given;
the half life of element X = 20 year
initial mass of element X = 48 g
a) How much is there after 80 years
0 year --------------------------> 48 g
20 years -----------------------> (48g / 2) = 24 g
40 years ------------------------> 12 g
60 years ------------------------> 6 g
80 years --------------------------> 3 g
Therefore, after 80 years there will be 3 g of element X remaining.
A 11.0kg bucket is lowered vertically by a rope in which there is 164N of tension at a given instant.
Determine the magnitude of the acceleration of the bucket.
Express your answer to two significant figures and include the appropriate units.
Answers
Answer:
The magnitude of the acceleration is 5.11 m/s²
Explanation:
Given;
Tension on the rope, T = 164 N
mass of the bucket, m = 11 kg
Weight of the rope is given by;
W = mg = 11 x 9.8 = 107.8 N
According to Newton's second law of motion, the tension on the rope is given by;
T = W - ma
ma = W - T
ma = 107.8N - 164N
ma = -56.2 N
a = -56.2 / m
a = -56.2 / 11
a = -5.11 m/s²
The magnitude of the acceleration is 5.11 m/s²
What is Physics to you? What do you know about it?
Answers
Answer:
Physics is the study of matter and the way living things behave everyday....It is related to maths in how we measure the way objects or people do specific things physics has many branches under it that can be helpful too.....Physics teaches us about the world,about the mechanical things we do about air,space,matter about the solar system and about simple machines and things that we do everyday in life.... What we do everyday is related to physics....
A 50g marble is moving at 2m/s when it strikes a 20g marble at rest. Immediately after the collision, the 50g ball is moving at 1m/s. Is this an elastic collision
Answers
Answer:
Since the two marbles have different velocities after collision, it can be concluded that the collision is elastic.
Explanation:
Given;
mass of first marble, m₁ = 50g = 0.05 kg
initial velocity of the first marble, u₁ = 2 m/s
mass of second marble, m₂ = 20 g = 0.02 kg
initial velocity of the second marble, u₂ = 0
final velocity of the first marble, v₁ = 1 m/s
Let the final velocity of the second marble, = v₂
Determine the final velocity of the second marble by applying principle of momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.05 x 2 + 0.02 x 0 = 0.05 x 1 + 0.02v₂
0.1 = 0.05 + 0.02v₂
0.02v₂ = 0.1 - 0.05
0.02v₂ = 0.05
v₂ = 0.05 / 0.02
v₂ = 2.5 m/s
During inelastic collision both objects will move at the same velocity after collision.
During elastic collision both objects will move at different velocities after collision.
Since the two marbles have different velocities after collision, it can be concluded that the collision is elastic.
A part of a circuit contains an inductor. The current through the inductor is changing uniformly from 2.40A to 0.30A over the course of 1.75s. If the EMF ( voltage change ) from one side of the inductor to the other is 5.70 Volts, what is the value of the inductance
Answers
Answer:
The value of the inductance is 4.75 H
Explanation:
Given;
initial current through the inductor, I₁ = 2.4 A
final current through the inductor, I₂ = 0.3 A
duration of change of current, dt = 1.75 s
voltage change of the inductor, V = 5.7 Volts
The voltage change of the inductor is given by;
[tex]V_L = -L\frac{di}{dt}\\\\ V_L = -L(\frac{I_2-I_1}{dt} )\\\\V_L = L(\frac{I_1-I_2}{dt} )\\\\[/tex]
Where;
L is the inductance of the coil;
[tex]5.7 = L(\frac{2.4-0.3}{1.75} )\\\\5.7 = 1.2 L\\\\L = \frac{5.7}{1.2}\\\\ L = 4.75 \ H[/tex]
Therefore, the value of the inductance is 4.75 H
Calculate the density of a rod of metal in g/cm3, with a mass of 9.58g, a diameter of 8 mm and a height of 3.5cm
Answers
Answer:
5.448 g/cm³
Explanation:
Density: This is defined as the ratio of the mass of a body to its volume.
The unit of density is kg/m³ other sub units are g/cm³, mg/mm³.
From the question,
D = m/πr²h......................... Equation 1
Where D = density, m = mass, r = radius, h = height.
Given: m = 9.58 g, r = 8/2 mm = 4 mm = 0.4 cm, h = 3.5 cm
Substitute this values into equation 1
D = 9.58/(3.14×0.4²×3.5)
D = 5.448 g/cm³
Hence the density of the metal rod is 5.448 g/cm³
Suppose a big chunk of gold is submerged in water and its volume is found to be 12.5 cm?
Compute the mass of the chunk of gold in grams if you know the density is 19.3 g/cm2. Round
appropriately.
Answers
Answer:
The mass of the chunk of gold is 241.25 g
Explanation:
Since density is defined as mass divided by volume, we can solve for the mass (m) via its equation:
[tex]density=\frac{mass}{volume} \\19.3=\frac{m}{12.5} \\m=19.3\,(12.5)\, grams\\m = 241.25 \,\,g[/tex]
As the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor:___________.
A) approaches zero.
B) approaches infinity.
C) approaches unity.
D) none of the given answers
Answers
Answer:
B) approaches infinity
Explanation:
The capacitive reactance of an AC capacitor is given by;
[tex]X_C = \frac{1}{\omega C } = \frac{1}{2\pi f C}[/tex]
Where;
C is the capacitance
f is the frequency of the ac voltage
[tex]X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C} \\\\X_C = \frac{1}{2\pi f C} \\\\X_C = \frac{1}{2\pi (0) C} \\\\X_C = \frac{1}{0} \\\\X_C = \ infinity[/tex]
Therefore, as the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor approaches infinity.
The correct option is (B) approaches infinity
Plane-polarized light is incident on a single polarizing disk, withthe direction of E0 parallel to thedirection of the transmission axis. Through what angle should thedisk be rotated so that the intensity in the transmitted beam isreduced by a factor of each of the following?
(a) 2.20
(b) 5.20
(c) 12.0
Answers
Answer;
Cos²စ= I/Io
So
A. Cos²စ = 1/2.2
Cosစ= √1/2.2
စ = cos^-1 0.68
= 47.2°
B.
Cosစ = √1/5.2
စ = ,cos^-1 0.4385
= 64°
C.
Cosစ = √1/12
စ = cos^-1 0.2886
= 73.2°
A man starts from rest and accelerates at 4.00 m/s2. If he covers a distance of 525 m, how long does he accelerate?
Answers
Answer:
16.2 s
Explanation:
Given:
Δx = 525 m
v₀ = 0 m/s
a = 4.00 m/s²
Find: t
Δx = v₀ t + ½ at²
525 m = (0 m/s) t + ½ (4.00 m/s²) t²
t = 16.2 s
A narrow beam of white light is incident on a sheet of quartz. Thebeam disperses in the quartz, with red light (λË700nm)traveling at an angle of 26.3o with respect to thenormal and violet light (λË400nm) traveling at25.7o. The index of refraction of quartz for red lightis 1.45. What is the index of refraction of quartz for violetlight?
Answers
Answer:
The index of refraction of quartz for violet light is 1.47.
Explanation:
It is given that, a narrow beam of white light is incident on a sheet of quartz.
The beam disperses in the quartz, with red light at an angle, [tex]\theta_r=26.3^{\circ}[/tex] wrt to the normal and violet light traveling at an angle of [tex]\theta_v=25.7^{\circ}[/tex]
The index of refraction of quartz for red light is 1.45.
We need to find the index of refraction of quartz for violet light.
Using Snell's law of red light as follows :
[tex]\mu_a\sin\theta_i=\mu_r\sin\theta_r[/tex]
Here,
[tex]\mu_a[/tex] is the refractive index of air
[tex]\theta_i[/tex] is the angle of incidence
We can find the value of angle of incidence as follows :
[tex]\sin\theta_i=\dfrac{\mu_r \sin\theta_r}{\mu_a}\\\\\sin\theta_i=\dfrac{1.45\times \sin(26.3)}{1}\\\\\theta_i=\sin^{-1}(0.642)\\\\\theta_i=39.79^{\circ}[/tex]
Now again using Snell's law for violet light as follows :
[tex]\mu_a\sin\theta_i=\mu_v\sin\theta_v\\\\\mu_v=\dfrac{\mu_a\sin\theta_i}{\sin\theta_v}\\\\\mu_v=\dfrac{1\times \sin(39.79)}{\sin(25.7)}\\\\\mu_v=1.47[/tex]
So, the index of refraction of quartz for violet light is 1.47.
If you are pushing on a crate on a frictionless surface in one direction, and your friend is pushing on the crate in the opposite direction with an equal amount of force. Which of the following statements is the most accurate? a. The crate will not move as the forces cancel each other out b. Because the surface is frictionless, the crate will always move regardless of who is pushing c. The crate can continue to move, but it will move at a constant velocity d. The net force is towards the direction that you are pushing, as you started the crate's motion
Answers
Answer:
Its not A..
Explanation:
I chose A - was incorrect
2. A 15 kg mass fastened to the end of a steel wire of un-
stretched length 0.5 m is whirled in a vertical circle with an
angular velocity of 2 rev/s at the bottom of the circle. The cross
section of the wire is 0.02 cm2. Calculate the elongation of the
wire when the weight is at the lowest point of the path. Steel
has Y.M.= 2.0 x 1011 Pa. [1.66mm]
Answers
Explanation:
Elongation of the wire is:
ΔL = F L₀ / (E A)
where F is the force,
L₀ is the initial length,
E is Young's modulus,
and A is the cross sectional area.
ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)
ΔL = T (1.25×10⁻⁶ m/N)
T = (80,000 N/m) ΔL
Draw a free body diagram of the mass at the bottom of the circle. There are two forces: tension force T pulling up and weight force mg pulling down.
Sum of forces in the centripetal direction:
∑F = ma
T − mg = mv²/r
T − mg = mω²r
T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)
T − 147 N = (2368.7 N/m) (0.5 m + ΔL)
Substitute:
(80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)
(80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL
(797631.3 N/m) ΔL = 1331.35 N
ΔL = 0.00167 m
ΔL = 1.67 mm
While taking the stairs it takes you 10 seconds to reach the top. The next time you take the same stairs, it takes you 5 seconds to reach the top stair. During which of these trips up the stairs did you use more power to climb? Explain your answer in complete sentences with proper spelling, grammar, and other language mechanics.
Answers
Answer:
P₂ = 2 P₁
we conclude that in the second time the power used is double that in the first rise
Explanation:
In this exercise we are asked the power to climb the stairs, if we assume that we go up with constant speed, we use an energy equal to the potential energy due to the difference in height of the stairs, as this height is constant the potential energy does not change and therefore therefore the energy used by us does not change either.
Now we can analyze the required power,
P = W / t
From the analysis of the previous paragraph the work is equal to the energy used, according to the work energy theorem,
therefore the first time the power is
P₁ = E / 10
P₁ = 0.1 E
for the second time the power is
P₂ = E / 5
P₂ = 0.2 E
we see that the power in the second case is
P₂ = 2 P₁
Therefore, we conclude that in the second time the power used is double that in the first rise.
what is the pressure of the gas inside the apparatus on teh right if the outside pressure P atm = 750mmHg
Answers
Complete question:
Check file uploaded for diagram of the apparatus
Answer:
The pressure of the gas inside the apparatus is 1000 mmHg
Explanation:
Given;
outside pressure, Patm = 750 mmHg
change in height of the two columns, Δh = 25 cm = 250 mm
The pressure of the gas inside the apparatus is given by;
P(gas) = P(atm) + P(Hg)
where;
P(atm) is the outside pressure
P(Hg) is the gauge pressure = height difference of the two columns.
P(gas) = 750 mmHg + 250 mmHg
P(gas) = 1000 mmHg
Therefore, the pressure of the gas inside the apparatus is 1000 mmHg
The pressure of the gas inside the apparatus is 1000 mmHg.
Calculation of the pressure:Since
outside pressure, Patm = 750 mmHg
change in height of the two columns, Δh = 25 cm = 250 mm
so we know that
The pressure of the gas inside the apparatus is
P(gas) = P(atm) + P(Hg)
here;
P(atm) is the outside pressure
P(Hg) is the gauge pressure = height difference of the two columns.
So,
P(gas) = 750 mmHg + 250 mmHg
P(gas) = 1000 mmHg
Learn more about pressure here: https://brainly.com/question/24531151
What is the maximum distance allowed between the center of hole #2 and datum B as seen in the front view?
Answers
Answer:
4.003" (inches )
Explanation:
The maximum distance allowed between the center of hole #2 and datum B can be calculated by adding 4.000" + 0.003" ( perpendicularity of the of hole #2) as seen from the front view of the diagram .
Note :The hole 2 is sited below the workpiece when viewed from the front view while the Datum B is positioned on the left end of the workpiece also note that the diameter is
HELP PLEEAAAASSSEEEEEEE What is the definition of net force?
Answers
Answer:
the sum of all force being applied to an object.
Explanation:
A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stretched to a tension of 122 N. During what time interval will a transverse wave travel the entire length of the two wires
Answers
Answer:
The time taken is [tex]t = 0.356 \ s[/tex]
Explanation:
From the question we are told that
The length of steel the wire is [tex]l_1 = 31.0 \ m[/tex]
The length of the copper wire is [tex]l_2 = 17.0 \ m[/tex]
The diameter of the wire is [tex]d = 1.00 \ m = 1.0 *10^{-3} \ m[/tex]
The tension is [tex]T = 122 \ N[/tex]
The time taken by the transverse wave to travel the length of the two wire is mathematically represented as
[tex]t = t_s + t_c[/tex]
Where [tex]t_s[/tex] is the time taken to transverse the steel wire which is mathematically represented as
[tex]t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ][/tex]
here [tex]\rho_s[/tex] is the density of steel with a value [tex]\rho_s = 8920 \ kg/m^3[/tex]
So
[tex]t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ][/tex]
[tex]t_s = 0.235 \ s[/tex]
And
[tex]t_c[/tex] is the time taken to transverse the copper wire which is mathematically represented as
[tex]t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ][/tex]
here [tex]\rho_c[/tex] is the density of steel with a value [tex]\rho_s = 7860 \ kg/m^3[/tex]
So
[tex]t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ][/tex]
[tex]t_c =0.121[/tex]
So
[tex]t = t_c + t_s[/tex]
[tex]t = 0.121 + 0.235[/tex]
[tex]t = 0.356 \ s[/tex]
An airplane is in level flight over Antarctica, where the magnetic field of the earth is mostly directed upward away from the ground. As viewed by a passenger facing toward the front of the plane, is the left or the right wingtip at higher potential? Does your answer depend on the direction the plane is flying?
Answers
Answer:
It does not depend on direction of plane and the left windtips more potential
Explanation:
Because if the Fleming right hand rule is applied the the right han is pointed in the direction of flight, and the fingers are curled in the direction of the magnetic lines. Thus , the lines are vertical and so are pointing down, thus by the right hand rule, the electrons move to the left hand side of the plane, although the potentials are equal and opposite in direction
what is force??
...
Answers
Answer: force is a push or pull that results in the movement of an object ..
Explanation:
Hope this helps you!!!!
A 1500-kg car accelerates from 0 to 25 m/s in 7.0s with negligible friction and air resistance. What is the average power delivered by the engine? (1 hp 746 W)
Answers
Answer:
90 hp
Explanation:
Power = work / time
P = ½ (1500 kg) (25 m/s)² / 7.0 s
P = 67,000 W
P = 90 hp
A resistor is connected across an oscillating emf. The peak current through the resistor is 2.0 A. What is the peak current if:
a. The resistance R is doubled?
b. The peak emf εo is doubled?
c. The frequency ω is doubled?
Answers
Answer:
(a) When the resistance R is doubled, I = 1 A
(b) When the peak emf εo is doubled, I = 4 A
(c) When the frequency ω is doubled, I = 2 A
Explanation:
Given;
peak current through the resistor, I = 2.0 A
According to ohms law the peak current through the circuit is given by;
[tex]I = \frac{V}{R}[/tex]
(a) When the resistance R is doubled;
[tex]I = \frac{V_R}{R} \\\\I_1R_1 = I_2R_2\\\\I_2 = \frac{I_1R_1}{R_2} \\\\I_2 = \frac{2*R_1}{2R_1} \\\\I_2 = 1 \ A[/tex]
(b)When the peak emf εo is doubled
[tex]I = \frac{V}{R} = \frac{\epsilon_o}{R} \\\\R = \frac{\epsilon_ o}{I} \\\\\frac{\epsilon_ o_1}{I_1} = \frac{\epsilon_ o_2}{I_2} \\\\I_2 = \frac{\epsilon_ o_2 *I_1}{\epsilon _o_1} \\\\I_2 = \frac{2 \epsilon_ o_1 *2}{\epsilon _o_1} \\\\I_2 = 4 \ A[/tex]
(c) When the frequency ω is doubled
Peak current through resistor is independent of frequency
I₂ = 2.0 A