Answer:
Explanation:
1.) Ratio of the speed of light in a vacuum to the speed of light in a medium is called refractive index. It is also known as index of refraction. It is a measure of light ray as they travel from one medium to another medium. We can determine the nature of a pair of media through their refractive indices.
2) Rule for how light is refracted at the boundary between two materials is known as the Snell's law. Snell's law states that the ratio of the angle of incidence to the angle of refraction is a constant for a given pair of media. The angle incidence is greater than the angle of refraction if the light ray travels from less dense medium to a denser medium and lesser than the angle of refraction if the light ray travels from denser medium to a less dense medium.
3) Process that occurs when the angle of incidence is greater than the critical angle is known as Total internal reflection. Total internal reflection only occurs when light rays travels from a denser medium to a less dense medium.
4.) Path of a particle of light is known as Ray. A ray is a particle of light that emerge from the sun or a reflecting (luminous) body. They are in form of straight lines striking the surface of a body.
5.) Dispersion is defined as the separation of white light into into component color. This is accompanied with the use of a triangular glass prism. The white light is an example of electromagnetic spectrum. The component colors are known as ROYGBIV (Red, Orange, Yellow, Green, Blue, Indigo and violet.). This combination of colors are called rainbow.
The separation of light into different frequencies to produce a rainbow is therefore known as DISPERSION
Please help! Average speed. Show work!
Answer:
3.78 m/s
Explanation:
Recall that the formula for average speed is given by
Speed = Distance ÷ Time taken
Where,
Speed = we are asked to find this
Distance = given as 340m
Time taken = 1.5 min = 1.5 x 60 = 90 seconds
Substituting the values into the equation:
Speed = Distance ÷ Time taken
= 340 meters ÷ 90 seconds
= 3.777777 m/s
= 3.78 m/s (round to nearest hundredth)
Answer:
3.78 m/sOption D is the right option.
solution,
Distance travelled=340 metre
Time taken= 1.5 minutes ( 1 minute=60 seconds)
=1.5* 60
=90 seconds
Now,
Average speed:
[tex] \ \frac{distance \: travelled \: }{time \: taken} \\ = \frac{340 \: m}{90 \: s} \\ = 3.78 \: metre \: per \: second[/tex]
Hope this helps...
Good luck on your assignment..
Calculate the angular momentum of a solid uniform sphere with a radius of 0.150 m and a mass of 13.0 kg if it is rotating at 5.70 rad/s about an axis through its center.
Answer:
The angular momentum of the solid sphere is 0.667 kgm²/s
Explanation:
Given;
radius of the solid sphere, r = 0.15 m
mass of the sphere, m = 13 kg
angular speed of the sphere, ω = 5.70 rad/s
The angular momentum of the solid sphere is given;
L = Iω
Where;
I is the moment of inertia of the solid sphere
ω is the angular speed of the solid sphere
The moment of inertia of solid sphere is given by;
I = ²/₅mr²
I = ²/₅ x (13 x 0.15²)
I = 0.117 kg.m²
The angular momentum of the solid sphere is calculated as;
L = Iω
L = 0.117 x 5.7
L = 0.667 kgm²/s
Therefore, the angular momentum of the solid sphere is 0.667 kgm²/s
PLS HELP ILL MARK U BRAINLIEST I DONT HAVE MUCH TIME!!
A football player of mass 103 kg running with a velocity of 2.0 m/s [E] collides head-
on with a 110 kg player on the opposing team travelling with a velocity of 3.2 m/s
[W]. Immediately after the collision the two players move in the same direction.
Calculate the final velocity of the two players.
Answer:
The final velocity of the two players is 0.69 m/s in the direction of the opposing player.
Explanation:
Since the players are moving in opposite directions, from the principle of conservation of linear momentum;
[tex]m_{1} u_{1}[/tex] - [tex]m_{2}u_{2}[/tex] = [tex](m_{1} + m_{2} )[/tex] v
Where: [tex]m_{1}[/tex] is the mass of the first player, [tex]u_{1}[/tex] is the initial velocity of the first player, [tex]m_{2}[/tex] is the mass of the second player, [tex]u_{2}[/tex] is the initial velocity of the second player and v is the final common velocity of the two players after collision.
[tex]m_{1}[/tex] = 103 kg, [tex]u_{1}[/tex] = 2.0 m/s, [tex]m_{2}[/tex] = 110 kg, [tex]u_{2}[/tex] = 3.2 m/s. Thus;
103 × 2.0 - 110 × 3.2 = (103 + 110)v
206 - 352 = 213 v
-146 = 213 v
v = [tex]\frac{-146}{213}[/tex]
v = -0.69 m/s
The final velocity of the two players is 0.69 m/s in the direction of the opposing player.
Carbon dioxide initially at 50 kPa, 400 K, undergoes a process in a closed system until its pressure and temperature are 2 MPa and 800 K, respectively. Assuming an ideal gas behaviour, find the entropy change of the carbon dioxide by assuming that the specific heats are constant. For the gas, take Cp = 0.846 kJ/kg.K and R = 0.1889 kJ/kg.K
Answer:
[tex]\Delta S=-0.11\frac{kJ}{kg*K}[/tex]
Explanation:
Hello,
In this case, we can compute the entropy change by using the following equation containing both pressure and temperature:
[tex]\Delta S=Cp\ ln(\frac{T_2}{T_1} )-R\ ln(\frac{p_2}{p_1} )[/tex]
Thus, we use the given data to obtain (2 MPa = 2000 kPa):
[tex]\Delta S=0.846\frac{kJ}{kg*K} \ ln(\frac{800K}{400K} )-0.1889\frac{kJ}{kg*K} \ ln(\frac{2000kPa}{50kPa} )\\\\\Delta S=-0.11\frac{kJ}{kg*K}[/tex]
Best regards.
your washer has a power of 350 watts and your dryer has a power of 1800 watts how much energy do you use to clean a load of clothes in 1 hour of washing and 1 hour of drying?
A. 1.29 x 10^3 J
B. 2.58 x 10^3 J
C. 1.55 x 10^7 J
D. 7.74 x 10^6 J
Answer:
7.74 x 10⁶ Joules
Explanation:
recall that "Watts" is the SI unit used for "energy per unit time"
Hence "Watts" may also be expressed as Joules / Second (or J/s)
We are given that the washer is rated at 350W (i.e. 350 Joules / s) and the dryer is rated at 1800W (i.e. 1800 Joules / s).
We are also given that the appliances are each run for 1 hour
1 hour = 60 min = (60 x 60) seconds = 3600 seconds
Hence the total energy used,
= Energy used by Washer in 1 hour + Energy used by dryer in 1 hour
= (350 J/s x 3600 s) + (1800 J/s x 3600 s)
= 3600 ( 350 + 1800)
= 3600 (2150)
= 7,740,000 Joules
= 7.74 x 10⁶ Joules
(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Suppose a spring has a natural length of 20 cm. If a 25-N force is required to keep it stretched to a length of 30 cm, how much work is required to stretch it from 20 cm to 25 cm?
(b) Find the area of the region enclosed by one loop of the curve r=2sin(5θ).
Answer:
a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve [tex]r(\theta) = 2\cdot \sin 5\theta[/tex] is [tex]4\pi[/tex].
Explanation:
a) The work, measured in joules, is a physical variable represented by the following integral:
[tex]W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx[/tex]
Where
[tex]x_{o}[/tex], [tex]x_{f}[/tex] - Initial and final position, respectively, measured in meters.
[tex]F(x)[/tex] - Force as a function of position, measured in newtons.
Given that [tex]F = k\cdot x[/tex] and the fact that [tex]F = 25\,N[/tex] when [tex]x = 0.3\,m - 0.2\,m[/tex], the spring constant ([tex]k[/tex]), measured in newtons per meter, is:
[tex]k = \frac{F}{x}[/tex]
[tex]k = \frac{25\,N}{0.3\,m-0.2\,m}[/tex]
[tex]k = 250\,\frac{N}{m}[/tex]
Now, the work function is obtained:
[tex]W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx[/tex]
[tex]W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}][/tex]
[tex]W = 0.313\,J[/tex]
The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.
b) Let be [tex]r(\theta) = 2\cdot \sin 5\theta[/tex]. The area of the region enclosed by one loop of the curve is given by the following integral:
[tex]A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta[/tex]
[tex]A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta[/tex]
By using trigonometrical identities, the integral is further simplified:
[tex]A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta[/tex]
[tex]A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta[/tex]
[tex]A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta[/tex]
[tex]A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)[/tex]
[tex]A = 4\pi[/tex]
The area of the region enclosed by one loop of the curve [tex]r(\theta) = 2\cdot \sin 5\theta[/tex] is [tex]4\pi[/tex].
Does there appear to be a simple mathematical relationship between the acceleration of an object (with fixed mass and negligible friction) and the force applied to the object (measured by the force probe mounted on the object)? Describe the mathematical relationship in words.
Answer:
the net force applied to an object is directly proportional to the acceleration undergone by that object
Explanation:
This verbal statement can be expressed in equation form as follows:
a = Fnet / m
1. A ski-plane with a total mass of 1200 kg lands towards the west on a frozen lake at 30.0
m/s. The coefficient of kinetic friction between the skis and the ice is 0.200. How far does
the plane slide before coming to a stop?
Answer:
d = 229.5 m
Explanation:
It is given that,
Total mass of a ski-plane is 1200 kg
It lands towards the west on a frozen lake at 30.0 m/s.
The coefficient of kinetic friction between the skis and the ice is 0.200.
We need to find the distance covered by the plane before coming to rest. In this case,
[tex]\mu mg=ma\\\\a=\mu g\\\\a=0.2\times 9.8\\\\a=1.96\ m/s^2[/tex]
It is decelerating, a = -1.96 m/s²
Now using the third equation of motion to find the distance covered by the plane such that :
[tex]v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}\\\\d=\dfrac{-(30)^2}{2\times -1.96}\\\\d=229.59\ m[/tex]
So, the plane slide a distance of 229.5 m.
Changing the camber of a wing by designing positive curvature in or lowering trailing edge flaps allows:______.
A. Higher Maximum Coefficients of Lift
B. Maximum Coefficient of Lift at lower Angle of Attack
C. Lower landing/approach angles of attack
D. A and C
E. All of the above
Answer:
E. All of the above
Explanation:
The wings that contain curvature is known as "camber," which in essence is half of a venturi, generating a greater-pressure area at the bottom of the wing, and a lesser-pressure area at the top. Creating an extra lift, the camber (curvature) of the wing is increased by extending (in an arc) the leading edge, typically by forcing or hinging the leading edge out on tracks.
The additional camber provides them with the additional lift needed for safe operation and control at slower aircraft speeds, such as when departing or landing.
By allowing wings to operate at a greater angle. A high lift coefficient is established and used as an angle of element for both attack and speed, when an airplane can travel extra steadily or take off and land in a smaller time with slats
Therefore the correct option is E.
A 1750kg bumpercar moving at 1.50m/s to the right collides elastically with a 1450kg car going to the left at 1.10m/s. The 1750kg car speed afterwards is 0.250m/s to the right. What is the speed of the 1450kg car after the collision?
The momentum in an elastic collision is conserved. Thus, the velocity of the 1450 kg car after collision will be 2.60 m/s.
What is momentum ?Momentum of a moving object is the product of its velocity and mass. For an elastic collision, the momentum and kinetic energy is conserved. Hence, the energy lost by an object is gained by another object.
Let u and V be the initial and final velocities and m1 and m2 be the colliding masses. Then,
m1u1 + m2u2 = m1v1 + m2v2.
The initial momentum in this collision is :
(1750 kg ×1.50 m/s) + (1450 kg × 1.10 m/s) = 4220 kg m/s
Final momentum = (1750 kg × 0.250 m/s) + (1450 kg × v2) = 4220 kg m/s
v2 = 2.60 m/s
Therefore, the final speed of the 1450 kg car is 2.60 m/s
Find more on elastic collision:
https://brainly.com/question/2356330
#SPJ5
Which of the following gives the magnitude of the average velocity (over the entire run) of an athlete running on a circular track with a circumference of 0.5 km, if that athlete runs a total length of 1.0 km in a time interval of 4 minutes?
a. O m/s
b. 2 m/s
c. 4.2 m/s
d. 16.8 m/s
Answer:
c. 4.2 m/s
Explanation:
The definition of the average velocity, measured in meters per second, is given by the following expression:
[tex]\bar v = \frac{x_{f}-x_{o}}{t_{f}-t_{o}}[/tex]
Where:
[tex]x_{o}[/tex], [tex]x_{f}[/tex] - Initial and final positions, measured in meters.
[tex]t_{o}[/tex], [tex]t_{f}[/tex] - Initial and final instants, measured in seconds.
Positions and instants must be written in meters and seconds, respectively:
[tex]x_{o} = 0\,m[/tex], [tex]x_{f} = 1000\,m[/tex].
[tex]t_{o} = 0\,s[/tex], [tex]t_{f} = 240\,s[/tex].
Finally, the average velocity of the athlete that runs a total length of 1.0 kilometer in a time interval of 4 minutes is:
[tex]\bar v = \frac{1000\,m-0\,m}{240\,s-0\,s}[/tex]
[tex]\bar v = 4.167\,\frac{m}{s}[/tex]
Hence, the best option is C.
What is the work done in stretching a spring by a distance of 0.5 m if the restoring force is 24N?
Answer:
3Nm
Explanation:
work = 0.5 x 12 x 0.5 = 3
The work done in stretching the spring by a distance of 0.5 m, with a restoring force of 24 N, is 6 joules.
To calculate the work done in stretching a spring, we can use the formula for work done by a spring:
Work = (1/2) * k *[tex]x^2[/tex]
where:
k = spring constant
x = distance the spring is stretched
Given that the restoring force (F) acting on the spring is 24 N, and the distance the spring is stretched (x) is 0.5 m, we can find the spring constant (k) using Hooke's law:
F = k * x
k = F / x
k = 24 N / 0.5 m
k = 48 N/m
Now, we can calculate the work:
Work = (1/2) * 48 N/m * [tex](0.5 m)^2[/tex]
Work = (1/2) * 48 N/m * [tex]0.25 m^2[/tex]
Work = 6 joules
Therefore, the work done in stretching the spring by a distance of 0.5 m, with a restoring force of 24 N, is 6 joules.
To know more about work done, here
brainly.com/question/2750803
#SPJ2
A crate of mass 9.2 kg is pulled up a rough incline with an initial speed of 1.58 m/s. The pulling force is 110 N parallel to the incline, which makes an angle of 20.2° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.10 m.A) How much work is done by the gravitational force on thecrate?
B) Determine the increase in internal energy of the crate-inclinesystem owing to friction.
C) How much work is done by the 100N force on the crate?
D) What is the change in kinetic energy of the crate?
E) What is the speed of the crate after being pulled 5.00m?
Given that,
Mass = 9.2 kg
Force = 110 N
Angle = 20.2°
Distance = 5.10 m
Speed = 1.58 m/s
(A). We need to calculate the work done by the gravitational force
Using formula of work done
[tex]W_{g}=mgd\sin\theta[/tex]
Where, w = work
m = mass
g = acceleration due to gravity
d = distance
Put the value into the formula
[tex]W_{g}=9.2\times(-9.8)\times5.10\sin20.2[/tex]
[tex]W_{g}=-158.8\ J[/tex]
(B). We need to calculate the increase in internal energy of the crate-incline system owing to friction
Using formula of potential energy
[tex]\Delta U=-W[/tex]
Put the value into the formula
[tex]\Delta U=-(-158.8)\ J[/tex]
[tex]\Delta U=158.8\ J[/tex]
(C). We need to calculate the work done by 100 N force on the crate
Using formula of work done
[tex]W=F\times d[/tex]
Put the value into the formula
[tex]W=100\times5.10[/tex]
[tex]W=510\ J[/tex]
We need to calculate the work done by frictional force
Using formula of work done
[tex]W=-f\times d[/tex]
[tex]W=-\mu mg\cos\theta\times d[/tex]
Put the value into the formula
[tex]W=-0.4\times9.2\times9.8\cos20.2\times5.10[/tex]
[tex]W=-172.5\ J[/tex]
We need to calculate the change in kinetic energy of the crate
Using formula for change in kinetic energy
[tex]\Delta k=W_{g}+W_{f}+W_{F}[/tex]
Put the value into the formula
[tex]\Delta k=-158.8-172.5+510[/tex]
[tex]\Delta k=178.7\ J[/tex]
(E). We need to calculate the speed of the crate after being pulled 5.00m
Using formula of change in kinetic energy
[tex]\Delta k=\dfrac{1}{2}m(v_{2}^2-v_{1}^{2})[/tex]
[tex]v_{2}^2=\dfrac{2\times\Delta k}{m}+v_{1}^2[/tex]
Put the value into the formula
[tex]v_{2}^2=\dfrac{2\times178.7}{9.2}+1.58[/tex]
[tex]v_{2}=\sqrt{\dfrac{2\times178.7}{9.2}+1.58}[/tex]
[tex]v_{2}=6.35\ m/s[/tex]
Hence, (A). The work done by the gravitational force is -158.8 J.
(B). The increase in internal energy of the crate-incline system owing to friction is 158.8 J.
(C). The work done by 100 N force on the crate is 510 J.
(D). The change in kinetic energy of the crate is 178.7 J.
(E). The speed of the crate after being pulled 5.00m is 6.35 m/s
A stretch spring has an elastic potential energy of 35 J when it is stretched 0.54m. What is the spring constant of the spring?
Answer:
240.1 N/m
Explanation:
Applying the formula for the potential energy stored in a stretched spring,
E = 1/2ke².................... Equation 1
Where E = potential energy, k = spring constant, e = extension.
make k the subject of the equation,
k = 2E/e².................. Equation 2
Given: E = 35 J, e = 0.54 m
Substitute into equation 2
k = 2(35)/0.54²
k = 70/0.2916
k = 240.1 N/m.
Hence the spring constant of the spring is 240.1 N/m
Answer:
240 N/m
Explanation:
edge2o2o
Q 6.30: What is the underlying physical reason for the difference between the static and kinetic coefficients of friction of ordinary surfaces
Answer:
the coefficient of static friction is larger than kinetic coefficients of friction
Explanation:
The coefficient of static friction is usually larger than the kinetic coefficients of friction because an object at rest has increasingly settled agreements with the surface it's resting on at the molecular level, so it takes more force to break these agreement.
Until this force is been overcome, kinetic coefficient of friction is not going to surface.
Note: coefficient of static friction is the friction between two bodies when the bodies aren't moving. Meanwhile, kinetic coefficient is the ratio of frictional force of a moving body to the normal reaction.
[tex]F_{s}[/tex] ≤μ[tex]_{s}[/tex]N(coefficient of static friction)
where [tex]F_{s}[/tex] is the static friction, μ[tex]_{s}[/tex] is the coefficient of static friction and N is the normal reaction
μ = [tex]\frac{F}{N}[/tex](kinetic coefficient of friction)
attached is diagram illustrating the explanation
Legacy issues $570,000 of 8.5%, four-year bonds dated January 1, 2019, that pay interest semiannually on June 30 and December 31. They are issued at $508,050 when the market rate is 12%.
1. Determine the total bond interest expense to be recognized.
Total bond interest expense over life of bonds:
Amount repaid:
8 payments of $24,225 $193,800
Par value at maturity 570,000
Total repaid 763,800
Less amount borrowed 645 669
Total bond interest expense $118.131
2. Prepare a straight-line amortization table for the bonds' first two years.
Semiannual Period End Unamortized Discount Carrying Value
01/01/2019
06/30/2019
12/31/2019
06/30/2020
12/31/2020
3. Record the interest payment and amortization on June 30. Note:
Date General Journal Debit Credit
June 30
4. Record the interest payment and amortization on December 31.
Date General Journal Debit Credit
December 31
Answer:
1) Determine the total bond interest expense to be recognized.
Total bond interest expense over life of bonds:
Amount repaid:
8 payments of $24,225: $193,800
Par value at maturity: $570,000
Total repaid: $763800 (193,800 + 570,000)
Less amount borrowed: $508050
Total bond interest expense: $255750 (763800 - 508,050)
2)Prepare a straight-line amortization table for the bonds' first two years.
Semiannual Interest Period End; Unamortized Discount; Carrying Value
01/01/2019 61,950 508,050
06/30/2019 54,206 515,794
12/31/2019 46,462 523,538
06/30/2020 38,718 531,282
12/31/2020 30,974 539,026
3) Record the interest payment and amortization on June 30:
June 30 Bond interest expense, dr 31969
Discount on bonds payable, Cr (61950/8) 7743.75
Cash, Cr ( 570000*8.5%/2) 24225
4) Record the interest payment and amortization on December 31:
Dec 31 Bond interest expense, Dr 31969
Discount on bonds payable, Cr 7744
Cash, Cr 24225
Use Kepler's third law to determine how many days it takes a spacecraft to travel in an elliptical orbit from a point 6 590 km from the Earth's center to the Moon, 385 000 km from the Earth's center.
Answer:
1.363×10^15 seconds
Explanation:
The spaceship travels an elliptical orbit from a point of 6590km from the earth center to the moon and 38500km from the earth center.
To calculate the time taken from Kepler's third Law :
T^2 = ( 4π^2/GMe ) r^3
Where Me is the mass of the earth
r is the average distance travel
G is the universal gravitational constant. = 6.67×10-11 m3 kg-1 s-2
π = 3.14
Me = mass of earth = 5.972×10^24kg
r =( r minimum + r maximum)/2 ......1
rmin = 6590km
rmax = 385000km
From equation 1
r = (6590+385000)/2
r = 391590/2
r = 195795km
From T^2 = ( 4π^2/GMe ) r^3
T^2 = (4 × 3.14^2/ 6.67×10-11 × 5.972×10^24) × 195795^3
= ( 4×9.8596/ 3.983×10^14 ) × 7.5059×10^15
= 39.4384/ 3.983×10^14 ) × 7.5059×10^15
= (9.901×10^14) × 7.5059×10^15
T^2 = 7.4321× 10^30
T =√7.4321× 10^30
T = 2.726×10^15 seconds
The time for one way trip from Earth to the moon is :
∆T = T/2
= 2.726×10^15 /2
= 1.363×10^15 secs
An electron has an initial velocity of (17.1 + 12.7) km/s, and a constant acceleration of (1.60 × 1012 m/s2) in the positive x direction in a region in which uniform electric and magnetic fields are present. If = (529 µT) find the electric field .
Answer:
Explanation:
Since B is perpendicular, it does no work on the electron but instead deflects it in a circular path.
q = 1.6 x 10-19 C
v = (17.1j + 12.7k) km/s = square root(17.1² + 12.7²) = 2.13 x 10⁴ m/s
the force acting on electron is
F= qvBsinΦ
F= (1.6 x 10⁻¹⁹C)(2.13.x 10⁴ m/s)(526 x 10⁻⁶ T)(sin90º)
F = 1.793x 10⁻¹⁸ N
The net force acting on electron is
F = e ( E+ ( vXB)
= ( - 1.6 × 10⁻¹⁹) ( E + ( 17.1 × 10³j + 12.7 × 10³ k)X( 529 × 10⁻⁶ ) (i)
= ( -1.6 × 10⁻¹⁹ ) ( E- 6.7k + 9.0j)
a= F/m
1.60 × 10¹² i = ( -1.6 × 10⁻¹⁹ ) ( E- 6.9 k + 7.56 j)/9.11 × 10⁻³¹
9.11 i = - ( E- 6.7 k + 9.0 j)
E = -9.11i + 6.7k - 9.0j
In a uniform electric field, the magnitude of torque is given by:-
Answer:
Electric dipole
Explanation:
the axis of a dipole makes an angle with the electric field. depending on the direction (clockwise/anticlockwise) we can get the torque (positive/negative).
hope this helps :D
A parallel-plate capacitor with circular plates of radius R is being discharged. The displacement current through a central circular area, parallel to the plates and with radius R/2, is 9.2 A. What is the discharging current?
Answer:
The discharging current is [tex]I_d = 36.8 \ A[/tex]
Explanation:
From the question we are told that
The radius of each circular plates is R
The displacement current is [tex]I = 9.2 \ A[/tex]
The radius of the central circular area is [tex]\frac{R}{2}[/tex]
The discharging current is mathematically represented as
[tex]I_d = \frac{A}{k} * I[/tex]
where A is the area of each plate which is mathematically represented as
[tex]A = \pi R ^2[/tex]
and k is central circular area which is mathematically represented as
[tex]k = \pi [\frac{R}{2} ]^2[/tex]
So
[tex]I_d = \frac{\pi R^2 }{\pi * [ \frac{R}{2}]^2 } * I[/tex]
[tex]I_d = \frac{\pi R^2 }{\pi * \frac{R^2}{4} } * I[/tex]
[tex]I_d = 4 * I[/tex]
substituting values
[tex]I_d = 4 * 9.2[/tex]
[tex]I_d = 36.8 \ A[/tex]
A coil has resistance of 20 W and inductance of 0.35 H. Compute its reactance and its impedance to an alternating current of 25 cycles/s.
Answer:
Reactance of the coil is 55 WImpedance of the coil is 59 WExplanation:
Given;
Resistance of the coil, R = 20 W
Inductance of the coil, L = 0.35 H
Frequency of the alternating current, F = 25 cycle/s
Reactance of the coil is calculated as;
[tex]X_L=[/tex] 2πFL
Substitute in the given values and calculate the reactance [tex](X_L)[/tex]
[tex]X_L =[/tex] 2π(25)(0.35)
[tex]X_L[/tex] = 55 W
Impedance of the coil is calculated as;
[tex]Z = \sqrt{R^2 + X_L^2} \\\\Z = \sqrt{20^2 + 55^2} \\\\Z = 59 \ W[/tex]
Therefore, the reactance of the coil is 55 W and Impedance of the coil is 59 W
A basketball rolls across a classroom floor without slipping, with its center of mass moving at a certain speed. A block of ice of the same mass is set sliding across the floor with the same speed along a parallel line. (i) Which object has more kinetic energy
Answer:
The two objects encounter a ramp sloping upward.
Explanation:
The basketball will travel farther up theramp
what is mean by the terminal velocity
Terminal Velocity is the constant speed that a falling thing reaches when the resistence of a medium prevents the thing to reach any further speed.
Best of Luck!
Given small samples of three liquids, you are asked to determine their refractive indexes. However, you do not have enough of each liquid to measure the angle of refraction for light retracting from air into the liquid. Instead, for each liquid, you take a rectangular block of glass (n= 1.52) and Place a drop of the liquid on the top surface f the block. you shine a laser beam with wavelength 638 nm in vacuum at one Side of the block and measure the largest angle of incidence for which there is total internal reflection at the interface between the glass and the liquid. Your results are given in the table.
Liquid A B C
θ 52.0 44.3 36.3
Required:
a. What is the refractive index of liquid A at this wavelength?
b. What is the refractive index of liquid B at this wavelength?
c. What is the refractive index of liquid C at this wavelength?
Answer:
A — 1.198B — 1.062C — 0.900Explanation:
The index of refraction of the liquid can be computed from ...
[tex]n_i\sin{(\theta_t)}=n_t[/tex]
where ni is the index of refraction of the glass block (1.52) and θt is the angle at which there is total internal refraction. nt is the index of refraction of the liquid.
For the given incidence angles, the computed indices of refraction are ...
A: n = 1.52sin(52.0°) = 1.198
B: n = 1.52sin(44.3°) = 1.062
C: n = 1.52sin(36.3°) = 0.900
what are the rays that come of the sun called? A. Ultraviolent rays B. Gamma rays C. soundwaves D. sonic rays
Answer:
The answer is ultraviolet rays...
Explanation:
...because the ozone layer protects us from the UV rays of the sun.
A 100-m long transmission cable is suspended between two towers. If the mass density is 2.01 kg/m and the tension in the cable is 3.00 x 104 N, what is the speed of transverse waves on the cable
Answer:
The speed is [tex]v =122.2 \ m/s[/tex]
Explanation:
From the question we are told that
The length of the wire is [tex]L = 100 \ m[/tex]
The mass density is [tex]\mu = 2.01 \ kg/m[/tex]
The tension is [tex]T = 3.00 *10^{4} \ N[/tex]
Generally the speed of the transverse cable is mathematically represented as
[tex]v = \sqrt{\frac{T}{\mu} }[/tex]
substituting values
[tex]v = \sqrt{\frac{3.0 *10^{4}}{2.01} }[/tex]
[tex]v =122.2 \ m/s[/tex]
Which statement describes one feature of a mineral's definite chemical composition?
It always occurs in pure form.
It always contains certain elements.
It cannot form from living or once-living materials.
It cannot contain atoms from more than one element.
N
Answer:
It always contains certain elements
Explanation:
Minerals can be defined as natural inorganic substances which possess an orderly internal structural arrangement as well as a particular, well known chemical composition, crystal structures and physical properties. Minerals include; quartz, dolomite, basalt, etc. Minerals may occur in isolation or in rock formations.
Minerals contain specific, well known chemical elements in certain ratios that can only vary within narrow limits. This is what we mean by a mineral's definite chemical composition. The structure of these minerals are all well known as well as their atom to atom connectivity.
The statement describes one feature of a mineral's definite chemical composition - It always contains certain elements.
A mineral is a naturally occurring chemical compound, usually of a crystalline form.
A mineral has one specific chemical composition.chemical composition that varies within a specific limited range and the atoms that make up the mineral must occur in specific ratiosthe proportions of the different elements and groups of elements in the mineral.Thus, The statement describes one feature of a mineral's definite chemical composition - It always contains certain elements.
Learn more:
https://brainly.com/question/690965
The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, Qin/eo, with Qin/e, where ε is the permittivity of the material. (Technically, Eo is called the vacuum permittivity.) Suppose that a 70 nC point charge is surrounded by a thin, 32-cm-diameter spherical rubber shell and that the electric field strength inside the rubber shell is 2500 N/C.
What is the permittivity of rubber?
Answer:
The permittivity of rubber is [tex]\epsilon = 8.703 *10^{-11}[/tex]
Explanation:
From the question we are told that
The magnitude of the point charge is [tex]q_1 = 70 \ nC = 70 *10^{-9} \ C[/tex]
The diameter of the rubber shell is [tex]d = 32 \ cm = 0.32 \ m[/tex]
The Electric field inside the rubber shell is [tex]E = 2500 \ N/ C[/tex]
The radius of the rubber is mathematically evaluated as
[tex]r = \frac{d}{2} = \frac{0.32}{2} = 0.16 \ m[/tex]
Generally the electric field for a point is in an insulator(rubber) is mathematically represented as
[tex]E = \frac{Q}{ \epsilon } * \frac{1}{4 * \pi r^2}[/tex]
Where [tex]\epsilon[/tex] is the permittivity of rubber
=> [tex]E * \epsilon * 4 * \pi * r^2 = Q[/tex]
=> [tex]\epsilon = \frac{Q}{E * 4 * \pi * r^2}[/tex]
substituting values
[tex]\epsilon = \frac{70 *10^{-9}}{2500 * 4 * 3.142 * (0.16)^2}[/tex]
[tex]\epsilon = 8.703 *10^{-11}[/tex]
Oceanographers often express the density of sea water in units of kilograms per cubic meter. If the density of sea water is 1.025 g/cm3 at 15ºC, what is its density in kilograms per cubic meter?
Answer: The density in kilograms per cubic meter is 1025
Explanation:
Density is defined as mass contained per unit volume.
Given : Density of sea water = [tex]1.025g/cm^3[/tex]
Conversion : [tex]1.025g/cm^3=?kg/m^3[/tex]
As 1 g = 0.001 kg
Thus 1.025 g =[tex]\frac{0.001}{1}\times 1.025=0.001025kg[/tex]
Also [tex]1cm^3=10^{-6}m^3[/tex]
Thus [tex]1.025g/cm^3=\frac{0.001025}{10^{-6}kg/m^3}=1025kg/m^3[/tex]
Thus density in kilograms per cubic meter is 1025
The image to the left shows Earth’s major plates. A geologist is studying a plate boundary indicated by the arrow. Fill in the blank to complete the statement about the plate boundary. This boundary is between the African Plate and the Plate.
Answer:
The Eurasian Plate
Explanation:
The Eurasian plate is one of the most extended on Earth, crossing all of Asia and Europe. The Eurasian plate is between the North American and the African Plates on the north and west sides. The Eurasian plate crushed up above the Indian plate. The Tibetan plateau and the Himalayan mountain range formed due to the crush between the Eurasian Plate and Indian Plate, which started 50 million years ago.
Answer:
The Eurasian plate.
Explanation:
I took an assinment on Edge 2020.