Answer:
Explanation:
Using the lens formula
1//f = 1/u+1/v
f is the focal length of the lens
u is the object distance
v is the image distance
For convex lens
The focal length of a convex lens is positive and the image distance can either be negative or positive.
Given f = 20cm and u = 10cm
1/v = 1/f - 1/u
1/v = 1/20-1/10
1/v = (1-2)/20
1/V = -1/20
v = -20/1
v = -20 cm
Since the image distance is negative, this shows that the nature of the image formed by the convex lens is a virtual image
For concave lens
The focal length of a concave lens is negative and the image distance is negative.
Given f = -20cm and u = 10cm
1/v = 1/f - 1/u
1/v = -1/20-1/10
1/v = (-1-2)/20
1/V = -3/20
v = -20/3
v = -6.67 cm
Since the image distance is negative, this shows that the nature of the image formed by the concave lens is a virtual image
Light of wavelength 500 nm falls on two slits spaced 0.2 mm apart. If the spacing between the first and third dark fringes is to be 4.0 mm, what is the distance from the slits to a screen?
Answer:
L = 0.8 m
Explanation:
Since, the distance between first and third dark fringes is 4 mm. Therefore, the fringe spacing between consecutive dark fringes will be:
Δx = 4 mm/2 = 2 mm = 2 x 10⁻³ m
but,
Δx = λL/d
λ = wavelength of the light = 500 nm = 5 x 10⁻⁷ m
d = slit spacing = 0.2 mm = 0.2 x 10⁻³ m
L = Distance between slits and screen = ?
Therefore, using the values, we get:
2 x 10⁻³ m = (5 x 10⁻⁷ m)(L)/(0.2 x 10⁻³)
L = (2 x 10⁻³ m)(0.2 x 10⁻³ m)/(5 x 10⁻⁷ m)
L = 0.8 m
A resistor made of Nichrome wire is used in an application where its resistance cannot change more than 1.35% from its value at 20.0°C. Over what temperature range can it be used (in °C)?
Answer:
Pls seeattached file
Explanation:
A resistor made of Ni chrome wire is used in an application where its resistance cannot be more than 1.35 % so its temperature range will be from 33.75 to -33.75 °C.
What is Resistance?Electrical resistance, or resistance to electricity, is a force that opposes the flow of current. Ohms are used to expressing resistance values.
When there is an electron difference between two terminals, electricity will flow from high to low. In opposition to that flow is resistance. As resistance rises, the current declines. On the other side, when the resistance falls, the current rises.
According to the question,
R = R₀ (1 + α ΔT)
(1 + 0.0135)R₀ = R₀(1 + α ΔT)
ΔT = (1 + 0.0135) / α
= 0.0135 / 0.0004
= 33.75 °C.
ΔT = [(1 - 0.0135) -1]/0.004
= -33.75 °C
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g In the atmosphere, the shortest wavelength electromagnetic waves are called A. infrared waves. B. ultraviolet waves. C. X-rays. D. gamma rays. E.
Answer:gamma ray
Explanation:
Two protons, A and B, are next to an infinite plane of positive charge. Proton B is twice as far from the plane as proton A. Which proton has the larg
Answer:
They both have the same acceleration
A car travels at 45 km/h. If the driver breaks 0.65 seconds after seeing the traffic light turn yellow, how far will the car continue to travel before it begins to slow?
Answer:
8.1 m
Explanation:
Convert km/h to m/s.
45 km/h × (1000 m/km) × (1 h / 3600 s) = 12.5 m/s
Distance = speed × time
d = (12.5 m/s) (0.65 s)
d = 8.125 m
What is the difference between matter and energy
Answer:
Everything in the Universe is made up of matter and energy. Matter is anything that has mass and occupies space. ... Energy is the ability to cause change or do work. Some forms of energy include light, heat, chemical, nuclear, electrical energy and mechanical energy.
Explanation:
The linear density rho in a rod 3 m long is 8/ x + 1 kg/m, where x is measured in meters from one end of the rod. Find the average density rhoave of the rod.
Answer:
The average density of the rod is 1.605 kg/m.
Explanation:
The average density of the rod is given by:
[tex] \rho = \frac{m}{l} [/tex]
To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, as follows:
[tex] \int_{0}^{3} \frac{8}{3(x + 1)}dx [/tex]
[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{(x + 1)}dx [/tex] (1)
Using u = x+1 → du = dx → u₁= x₁+1 = 0+1 = 1 and u₂ = x₂+1 = 3+1 = 4
By entering the values above into (1), we have:
[tex] \rho = \frac{8}{3} \int_{0}^{3} \frac{1}{u}du [/tex]
[tex]\rho = \frac{8}{3}*log(u)|_{1}^{4} = \frac{8}{3}[log(4) - log(1)] = 1.605 kg/m[/tex]
Therefore, the average density of the rod is 1.605 kg/m.
I hope it helps you!
The average density of the rod is [tex]1.605 \;\rm kg/m^{3}[/tex].
Given data:
The length of rod is, L = 3 m.
The linear density of rod is, [tex]\rho=\dfrac{8}{x+1} \;\rm kg/m[/tex].
To find the average density we need to integrate the linear density from x₁ = 0 to x₂ = 3, The expression for the average density is given as,
[tex]\rho' = \int\limits^3_0 { \rho} \, dx\\\\\\\rho' = \int\limits^3_0 { \dfrac{m}{L}} \, dx\\\\\\\rho' = \int\limits^3_0 {\dfrac{8}{3(x+1)}} \, dx[/tex]............................................................(1)
Using u = x+1
du = dx
u₁= x₁+1 = 0+1 = 1
and
u₂ = x₂+1 = 3+1 = 4
By entering the values above into (1), we have:
[tex]\rho' =\dfrac{8}{3} \int\limits^3_0 {\dfrac{1}{u}} \, du\\\\\\\rho' =\dfrac{8}{3} \times [log(u)]^{4}_{1}\\\\\\\rho' =\dfrac{8}{3} \times [log(4)-log(1)]\\\\\\\rho' =1.605 \;\rm kg/m^{3}[/tex]
Thus, we can conclude that the average density of the rod is [tex]1.605 \;\rm kg/m^{3}[/tex].
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wo 10-cm-diameter charged rings face each other, 25.0 cm apart. Both rings are charged to + 20.0 nC . What is the electric field strength
Complete question:
Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:
a) the midpoint between the two rings?
b) the center of the left ring?
Answer:
a) the electric field strength at the midpoint between the two rings is 0
b) the electric field strength at the center of the left ring is 2712.44 N/C
Explanation:
Given;
distance between the two rings, d = 25 cm = 0.25 m
diameter of each ring, d = 10 cm = 0.1 m
radius of each ring, r = [tex]\frac{0.1}{2} = 0.05 \ m[/tex]
the charge on each ring, q = 20 nC
Electric field strength for a ring with radius r and distance x from the center of the ring is given as;
[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}}[/tex]
The electric field strength at the midpoint;
the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m
[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9}*0.125*20*10^{-9}}{(0.125^2 + 0.05^2)^{3/2}} \\\\E = 9210.5 \ N/C[/tex]
[tex]E_{left} = 9210.5 \ N/C[/tex]
The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;
[tex]E_{right} = -9210.5 \ N/C[/tex]
The electric field strength at the midpoint;
[tex]E_{mid} = E_{left} + E_{right}\\\\E_{mid} = 9210.5 \ N/C - 9210.5 \ N/C\\\\E_{mid} = 0[/tex]
(b)
The distance from the right ring to center of the left ring, x = 0.25 m.
[tex]E = \frac{KxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9} *0.25*20*10^{-9}}{(0.25^2 + 0.05^2)^{3/2}} \\\\E = 2712.44 \ N/C[/tex]
Matter's resistance to a change in motion is called _____ and is directly proportional to the mass of an object
Answer:
Matter's resistance to a change in motion is called INERTIA and is directly proportional to the mass of an object.
Explanation:
Two identical planets orbit a star in concentric circular orbits in the star's equatorial plane. Of the two, the planet that is farther from the star must have
Answer:
The planet that is farther from the star must have a time period greater.
Explanation:
We can determine the ratio of the period's planet with the radius of the circular orbit in the star's equatorial plane:
[tex] T = 2\pi*\sqrt{\frac{r^{3}}{GM}} [/tex] (1)
Where:
r: is the radius of the circular orbit of the planet and the star
T: is the period
G: is the gravitational constant
M: is the mass of the planet
From equation (1) we have:
[tex] T = 2\pi*\sqrt{\frac{r^{3}}{GM}} = k*r^{3/2} [/tex] (2)
Where k is a constant
From equation (2) we have that of the two planets, the planet that is farther from the star must have a time period greater.
I hope it helps you!
A/An ____________________ is a small, flexible tube with a light and lens on the end that is used for examination. Question 96 options:
Answer:
"Endoscope" is the correct answer.
Explanation:
A surgical tool sometimes used visually to view the internal of either a body cavity or maybe even an empty organ like the lung, bladder, as well as stomach. There seems to be a solid or elastic tube filled with optics, a source of fiber-optic light, and sometimes even a sample, epidurals, suction tool, and perhaps other equipment for sample analysis or recovery.What is the direction of the net gravitational force on the mass at the origin due to the other two masses?
Answer:
genus yds it's the
Explanation:
xmgxfjxfjxgdfjusufzjyhmfndVFHggssjtjhryfjftjsrhrythhrsrhrhsfhsgdagdah vhj
This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A. A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
What happens to end A of the rod when the ball approaches it closely this first time?a. It is strongly repelled.b. It is strongly attracted.c. It is weakly attracted.d. It is weakly repelled.e. It is neither attracted nor repelled.
Answer:
e. It is neither attracted nor repelled.
Explanation:
Electrostatic attraction or repulsion occurs between two or more charged particles or conductors. In this case, if the negatively charged ball is brought close to the neutral end A of the rod, there would be no attraction or repulsion between the rod end A and the negatively charged ball. This is because a charged particle or conductor has no attraction or repulsion to a neutral particle or conductor.
Three resistors, each having a resistance, R, are connected in parallel to a 1.50 V battery. If the resistors dissipate a total power of 3.00 W, what is the value of R
Answer:
The value of resistance of each resistor, R is 2.25 Ω
Explanation:
Given;
voltage across the three resistor, V = 1.5 V
power dissipated by the resistors, P = 3.00 W
the resistance of each resistor, = R
The effective resistance of the three resistors is given by;
R(effective) = R/3
Apply ohms law to determine the current delivered by the source;
V = IR
I = V/R
I = 3V/R
Also, power is calculated as;
P = IV
P = (3V/R) x V
P = 3V²/R
R = 3V² / P
R = (3 x 1.5²) / 3
R = 2.25 Ω
Therefore, the value of resistance of each resistor, R is 2.25 Ω
When a mercury thermometer is heated, the mercury expands and rises in the thin tube of glass. What does this indicate about the relative rates of expansion for mercury and glass
Answer:
This means that mercury has a higher or faster expansion rate than glass
Explanation:
This is because When a container expands, the reservoir in the glass expands at the same rate as the glass. Thus, if there is something in a glass and both expand at the same rate, they have no change - but if the contents expand faster, they will fill the container to a higher level, and if the contents expand slower, they will fill the container to a lower level (relative to the new size of the container).
If the
refractive index of benzere is 2.419,
what is the speed of light in benzene?
Answer:
[tex]v=1.24\times 10^8\ m/s[/tex]
Explanation:
Given that,
The refractive index of benzene is 2.419
We need to find the speed of light in benzene. The ratio of speed of light in vacuum to the speed of light in the medium equals the refractive index. So,
[tex]n=\dfrac{c}{v}\\\\v=\dfrac{c}{n}\\\\v=\dfrac{3\times 10^8}{2.419}\\\\v=1.24\times 10^8\ m/s[/tex]
So, the speed of light in bezene is [tex]1.24\times 10^8\ m/s[/tex].
3. What are the first steps that you should take if you are unable to get onto the Internet? (1 point)
O Check your router connections then restart your router.
O Plug the CPU to a power source and reboot the computer.
O Adjust the display properties and check the resolution.
Use the Control Panel to adjust the router settings.
Answer:
Check your router connections then restart your router.
Explanation:
Answer:
Check your router connections then restart your router.
Explanation:
Most internet access comes from routers so the problem is most likely the router.
A plastic dowel has a Young's Modulus of 1.50 ✕ 1010 N/m2. Assume the dowel will break if more than 1.50 ✕ 108 N/m2 is exerted.
(a) What is the maximum force (in kN) that can be applied to the dowel assuming a diameter of 2.40 cm?
______Kn
(b) If a force of this magnitude is applied compressively, by how much (in mm) does the 26.0 cm long dowel shorten? (Enter the magnitude.)
mm
Answer:
a
[tex]F = 67867.2 \ N[/tex]
b
[tex]\Delta L = 2.6 \ mm[/tex]
Explanation:
From the question we are told that
The Young modulus is [tex]Y = 1.50 *10^{10} \ N/m^2[/tex]
The stress is [tex]\sigma = 1.50 *10^{8} \ N/m^2[/tex]
The diameter is [tex]d = 2.40 \ cm = 0.024 \ m[/tex]
The radius is mathematically represented as
[tex]r =\frac{d}{2} = \frac{0.024}{2} = 0.012 \ m[/tex]
The cross-sectional area is mathematically evaluated as
[tex]A = \pi r^2[/tex]
[tex]A = 3.142 * (0.012)^2[/tex]
[tex]A = 0.000452\ m^2[/tex]
Generally the stress is mathematically represented as
[tex]\sigma = \frac{F}{A}[/tex]
=> [tex]F = \sigma * A[/tex]
=> [tex]F = 1.50 *10^{8} * 0.000452[/tex]
=> [tex]F = 67867.2 \ N[/tex]
Considering part b
The length is given as [tex]L = 26.0 \ cm = 0.26 \ m[/tex]
Generally Young modulus is mathematically represented as
[tex]E = \frac{ \sigma}{ strain }[/tex]
Here strain is mathematically represented as
[tex]strain = \frac{ \Delta L }{L}[/tex]
So
[tex]E = \frac{ \sigma}{\frac{\Delta L }{L} }[/tex]
[tex]E = \frac{\sigma }{1} * \frac{ L}{\Delta L }[/tex]
=> [tex]\Delta L = \frac{\sigma * L }{E}[/tex]
substituting values
[tex]\Delta L = \frac{ 1.50*10^{8} * 0.26 }{ 1.50 *10^{10 }}[/tex]
[tex]\Delta L = 0.0026[/tex]
Converting to mm
[tex]\Delta L = 0.0026 *1000[/tex]
[tex]\Delta L = 2.6 \ mm[/tex]
Water flows at speed v in a pipe of radius R. At what speed does the water flow through a constriction in which the radius of the pipe is R/3
Answer:
v₂ = 9 v
Explanation:
For this exercise in fluid mechanics, let's use the continuity equation
v₁ A₁ = v₂ A₂
where v is the velocity of the fluid, A the area of the pipe and the subscripts correspond to two places of interest.
The area of a circle is
A = π R²
let's use the subscript 1 for the starting point and the subscript 2 for the part with the constraint
In this case v₁ = v and the area is
A₁ = π R²
in the second point
A₂= π (R / 3)²
we substitute in the continuity equation
v π R² = v₂ π R² / 9
v = v₂ / 9
v₂ = 9 v
If Superman really had x-ray vision at 0.12 nm wavelength and a 4.1 mm pupil diameter, at what maximum altitude could he distinguish villains from heroes, assuming that he needs to resolve points separated by 5.4 cm to do this?
Answer:
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
Explanation:
Given:
wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m
Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m
Separation distance (D) = 5.4 cm = 0.054 m
Find:
Maximum altitude to see(L)
Computation:
Resolving power = 1.22(λ / d)
D / L = 1.22(λ / d)
0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]
0.054 / L = 1.22 [0.03 × 10⁻⁶]
L = 0.054 / 1.22 [0.03 × 10⁻⁶]
L = 0.054 / [0.0366 × 10⁻⁶]
L = 1.47 × 10⁶
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
A 750 gram grinding wheel 25.0 cm in diameter is in the shape of a uniform solid disk. (we can ignore the small hole at the center). when it is in use, it turns at a consant 220 rpm about an axle perpendicular to its face through its center. When the power switch is turned off, you observe that the wheel stops in 45.0 s with constant angular acceleration due to friction at the axle. What torque does friction exert while this wheel is slowing down?
Answer:
Torque = 0.012 N.m
Explanation:
We are given;
Mass of wheel;m = 750 g = 0.75 kg
Radius of wheel;r = 25 cm = 0.25 m
Final angular velocity; ω_f = 0
Initial angular velocity; ω_i = 220 rpm
Time taken;t = 45 seconds
Converting 220 rpm to rad/s we have;
220 × 2π/60 = 22π/3 rad/s
Equation of rotational motion is;
ω_f = ω_i + αt
Where α is angular acceleration
Making α the subject, we have;
α = (ω_f - ω_i)/t
α = (0 - 22π/3)/45
α = -0.512 rad/s²
The formula for the Moment of inertia is given as;
I = ½mr²
I = (1/2) × 0.75 × 0.25²
I = 0.0234375 kg.m²
Formula for torque is;
Torque = Iα
For α, we will take the absolute value as the negative sign denotes decrease in acceleration.
Thus;
Torque = 0.0234375 × 0.512
Torque = 0.012 N.m
What is the magnitude of the applied electric field inside an aluminum wire of radius 1.4 mm that carries a 4.5-A current
Answer:
Explanation:
From the question we are told that
The radius is [tex]r = 1.4 \ mm = 1.4 *10^{-3} \ m[/tex]
The current is [tex]I = 4.5 \ A[/tex]
Generally the electric field is mathematically represented as
[tex]E = \frac{J}{\sigma }[/tex]
Where [tex]\sigma[/tex] is the conductivity of aluminum with value [tex]\sigma = 3.5 *10^{7} \ s/m[/tex]
J is the current density which mathematically represented as
[tex]J = \frac{I}{A}[/tex]
Here A is the cross-sectional area which is mathematically represented as
[tex]A = \pi r^2[/tex]
[tex]A = 3.142 * (1.4*10^{-3})^2[/tex]
[tex]A = 6.158*10^{-6} \ m^2[/tex]
So
[tex]J = \frac{ 4.5 }{6.158*10^{-6}}[/tex]
[tex]J = 730757 A/m^2[/tex]
So
[tex]E = \frac{ 730757}{3.5*10^{7} }[/tex]
[tex]E = 0.021 \ N/C[/tex]
1. (I) If the magnetic field in a traveling EM wave has a peak magnitude of 17.5 nT at a given point, what is the peak magnitude of the electric field
Answer:
The electric field is [tex]E = 5.25 V/m[/tex]
Explanation:
From the question we are told that
The peak magnitude of the magnetic field is [tex]B = 17.5 nT = 17.5 *10^{-9}\ T[/tex]
Generally the peak magnitude of the electric field is mathematically represented as
[tex]E = c * B[/tex]
Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
So
[tex]E = 3.0 *10^{8} * 17.5 *10^{-9}[/tex]
[tex]E = 5.25 V/m[/tex]
The peak magnitude of the electric field will be "5.25 V/m".
Magnetic fieldAccording to the question,
Magnetic field's peak magnitude, B = 17.5 nT or,
= 17.5 × 10⁻⁹ T
Speed of light, c = 3.0 × 10⁸ m/s
We know the relation,
→ E = c × B
By substituting the values, we get
= 3.0 × 10⁸ × 17.5 × 10⁻⁹
= 5.25 V/m
Thus the above approach is appropriate.
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An unstable particle at rest spontaneously breaks into two fragments of unequal mass. The mass of the first fragment is 3.00 10-28 kg, and that of the other is 1.86 10-27 kg. If the lighter fragment has a speed of 0.844c after the breakup, what is the speed of the heavier fragment
Answer: Speed = [tex]3.10^{-31}[/tex] m/s
Explanation: Like in classical physics, when external net force is zero, relativistic momentum is conserved, i.e.:
[tex]p_{f} = p_{i}[/tex]
Relativistic momentum is calculated as:
p = [tex]\frac{mu}{\sqrt{1-\frac{u^{2}}{c^{2}} } }[/tex]
where:
m is rest mass
u is velocity relative to an observer
c is light speed, which is constant (c=[tex]3.10^{8}[/tex]m/s)
Initial momentum is zero, then:
[tex]p_{f}[/tex] = 0
[tex]p_{1}-p_{2}[/tex] = 0
[tex]p_{1} = p_{2}[/tex]
To find speed of the heavier fragment:
[tex]\frac{mu_{1}}{\sqrt{1-\frac{u^{2}_{1}}{c^{2}} } }=\frac{mu_{2}}{\sqrt{1-\frac{u^{2}_{2}}{c^{2}} } }[/tex]
[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=\frac{3.10^{-28}.0.844.3.10^{8}}{\sqrt{1-\frac{(0.844c)^{2}}{c^{2}} } }[/tex]
[tex]\frac{1.86.10^{-27}u_{1}}{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }=1.42.10^{-19}[/tex]
[tex]1.86.10^{-27}u_{1} = 1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } }[/tex]
[tex](1.86.10^{-27}u_{1})^{2} = (1.42.10^{-19}.{\sqrt{1-\frac{u^{2}_{1}}{(3.10^{8})^{2}} } })^{2}[/tex]
[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38}.(1-\frac{u_{1}^{2}}{9.10^{16}} )[/tex]
[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -[2.02.10^{-38}(\frac{u_{1}^{2}}{9.10^{16}} )][/tex]
[tex]3.46.10^{-54}.u_{1}^{2} = 2.02.10^{-38} -2.24.10^{-23}.u^{2}_{1}[/tex]
[tex]3.46.10^{-54}.u_{1}^{2}+2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]
[tex]2.24.10^{-23}.u^{2}_{1} = 2.02.10^{-38}[/tex]
[tex]u^{2}_{1} = \frac{2.02.10^{-38}}{2.24.10^{-23}}[/tex]
[tex]u_{1} = \sqrt{9.02.10^{-62}}[/tex]
[tex]u_{1} = 3.10^{-31}[/tex]
The speed of the heavier fragment is [tex]u_{1} = 3.10^{-31}[/tex]m/s.
Two separate disks are connected by a belt traveling at 5m/s. Disk 1 has a mass of 10kg and radius of 35cm. Disk 2 has a mass of 3kg and radius of 7cm.
a. What is the angular velocity of disk 1?
b. What is the angular velocity of disk 2?
c. What is the moment of inertia for the two disk system?
Explanation:
Given that,
Linear speed of both disks is 5 m/s
Mass of disk 1 is 10 kg
Radius of disk 1 is 35 cm or 0.35 m
Mass of disk 2 is 3 kg
Radius of disk 2 is 7 cm or 0.07 m
(a) The angular velocity of disk 1 is :
[tex]v=r_1\omega_1\\\\\omega_1=\dfrac{v}{r_1}\\\\\omega_1=\dfrac{5}{0.35}\\\\\omega_1=14.28\ rad/s[/tex]
(b) The angular velocity of disk 2 is :
[tex]v=r_2\omega_2\\\\\omega_2=\dfrac{v}{r_2}\\\\\omega_2=\dfrac{5}{0.07}\\\\\omega_2=71.42\ rad/s[/tex]
(c) The moment of inertia for the two disk system is given by :
[tex]I=I_1+I_2\\\\I=\dfrac{1}{2}m_1r_1^2+\dfrac{1}{2}m_2r_2^2\\\\I=\dfrac{1}{2}(m_1r_1^2+m_2r_2^2)\\\\I=\dfrac{1}{2}\times (10\times (0.35)^2+3\times (0.07)^2)\\\\I=0.619\ kg-m^2[/tex]
Hence, this is the required solution.
a transformer changes 95 v acorss the primary to 875 V acorss the secondary. If the primmary coil has 450 turns how many turns does the seconday have g
Answer:
The number of turns in the secondary coil is 4145 turns
Explanation:
Given;
the induced emf on the primary coil, [tex]E_p[/tex] = 95 V
the induced emf on the secondary coil, [tex]E_s[/tex] = 875 V
the number of turns in the primary coil, [tex]N_p[/tex] = 450 turns
the number of turns in the secondary coil, [tex]N_s[/tex] = ?
The number of turns in the secondary coil is calculated as;
[tex]\frac{N_p}{N_s} = \frac{E_p}{E_s}[/tex]
[tex]N_s = \frac{N_pE_s}{E_p} \\\\N_s = \frac{450*875}{95} \\\\N_s = 4145 \ turns[/tex]
Therefore, the number of turns in the secondary coil is 4145 turns.
A high school physics student claims her muscle car can achieve a constant acceleration of 10 ft/s/s. Her friend develops an accelerometer to confirm the feat. The accelerometer consists of a 1 ft long rod (mass=4 kg) with one end attached to the ceiling of the car, but free to rotate. During acceleration, the rod rotates. What will be the angle of rotation of the rod during this acceleration? Assume the road is flat and straight.
Answer: Ф = 17.2657 ≈ 17°
Explanation:
we simply apply ET =0 about the ending of the rod
so In.g.L/2sinФ - In.a.L/2cosФ = 0
g.sinФ - a.cosФ = 0
g.sinФ = a.cosФ
∴ tanФ = a/g
Ф = tan⁻¹ a / g
Ф = tan⁻¹ ( 10 / 32.17405)
Ф = tan⁻¹ 0.31080948777
Ф = 17.2657 ≈ 17°
Therefore the angle of rotation of the rod during this acceleration is 17.2657 ≈ 17°
In the direction perpendicular to the drift velocity, there is a magnetic force on the electrons that must be cancelled out by an electric force. What is the magnitude of the electric field that produces this force
Answer:
E = VdB
Explanation:
This is because canceling the electric and magnetic force means
q.vd. B= we
E= Vd. B
A Galilean telescope adjusted for a relaxed eye is 36.2 cm long. If the objective lens has a focal length of 39.5 cm , what is the magnification
Answer:
The magnification is [tex]m = 12[/tex]
Explanation:
From the question we are told that
The object distance is [tex]u = 36.2 \ cm[/tex]
The focal length is [tex]v = 39.5 \ cm[/tex]
From the lens equation we have that
[tex]\frac{1}{f} = \frac{1}{u} + \frac{1}{v}[/tex]
=> [tex]\frac{1}{v} = \frac{1}{f} - \frac{1}{u}[/tex]
substituting values
[tex]\frac{1}{v} = \frac{1}{39.5} - \frac{1}{36.2}[/tex]
[tex]\frac{1}{v} = -0.0023[/tex]
=> [tex]v = \frac{1}{0.0023}[/tex]
=> [tex]v =-433.3 \ cm[/tex]
The magnification is mathematically represented as
[tex]m =- \frac{v}{u}[/tex]
substituting values
[tex]m =- \frac{-433.3}{36.2}[/tex]
[tex]m = 12[/tex]
The sun generates both mechanical and electromagnetic waves. Which statement about those waves is true?
OA. The mechanical waves reach Earth, while the electromagnetic waves do not.
OB. The electromagnetic waves reach Earth, while the mechanical waves do not.
OC. Both the mechanical waves and the electromagnetic waves reach Earth.
OD. Neither the mechanical waves nor the electromagnetic waves reach Earth.
Answer: The correct answer for this question is letter (B) The electromagnetic waves reach Earth, while the mechanical waves do not. The sun generates both mechanical and electromagnetic waves. Space, between the sun and the earth is a nearly vacuum. So mechanical wave can not spread out in the vacuum.
Hope this helps!
Answer:
The electromagnetic waves reach Earth, while the mechanical waves do not
