Answer:
object distance p = 13.33 cm
Explanation:
For this problem of finding the image of an object we must use the constructor equation
1 / f = 1 / p + 1 / q
where f is the focal length, p and q are the distances to the object and the image, respectively.
In this case they indicate the focal length f = 10 cm, since the person has hyperopia, the image must be formed q = 40 cm, let's find where the object is (p)
1 / p = 1 / f - 1 / q
1 / p = 1/10 - 1/40
1 / p = 0.075
p = 13.33 cm
A plastic rod that has been charged to − 15 nC touches a metal sphere. Afterward, the rod's charge is − 5.0 nC.
1) What kind of charged particle was transferred between the rod and the sphere, and in which direction?
A) electrons transferred from rod to sphere.
B) electrons transferred from sphere to rod.
C) protons transferred from rod to sphere.
D) protons transferred from sphere to rod.
2) How many charged particles were transferred?
Answer:
B) electrons transferred from sphere to rod.
(2) 1.248 x 10¹¹ electrons were transferred
Explanation:
Given;
initial charge on the plastic rod, q₁ = 15nC
final charge on the plastic rod, q₂ = - 5nC
let the charge acquired by the plastic rod = q
q + 15nC = -5nC
q = -5nC - 15nC
q = -20 nC
Thus, the plastic rod acquired excess negative charge from the metal sphere.
Hence, electrons transferred from sphere to rod
B) electrons transferred from sphere to rod.
2) How many charged particles were transferred?
1.602 x 10⁻¹⁹ C = 1 electron
20 x 10⁻⁹ C = ?
= 1.248 x 10¹¹ electrons
Thus,1.248 x 10¹¹ electrons were transferred
1. Electrons transferred from sphere to rod.
Option B is correct.
2. There are [tex]6.24*10^{10}[/tex] electrons transferred from sphere to rod.
Given that initial charge on the plastic rod, q₁ = 15nC
final charge on the plastic rod, q₂ = - 5nC
let us consider that the charge absorbed by the plastic rod is q
[tex]q - 15nC = -5nC\\q = -5nC +15nC\\q = -10 nC[/tex]
Thus, the plastic rod acquired excess negative charge from the metal sphere.
Therefore, electrons transferred from sphere to rod
The charge on one electron is, 1.602 x 10⁻¹⁹ C .
Number of electrons, [tex]n=\frac{10*10^{-9} }{1.602*10^{-19} }= 6.24*10^{10}[/tex]
Thus,[tex]6.24*10^{10}[/tex] electrons were transferred.
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You want to create a spotlight that will shine a bright beam of light with all of the light rays parallel to each other. You have a large concave spherical mirror and a small lightbulb. Where should you place the lightbulb?
a. at the point, because all rays bouncing off the mirror will be parallel.
b. at the focal point of the mirror
c. at the radius of curvature of the mirror
d. none of the above, you cant make parallel rays wilth a concave mirror
Answer:
Explanation:
Concave mirrors is otherwise known as converging mirrors: These are mirrors that are caved inwards (reflecting surface is on the outside curved part). It is called a converging mirror due to the fact that light converges to a point when it strikes and reflects from the surface of the mirror. This type of mirror is used to focus light; parallel rays that are directed towards it will be concentrated to a point.
For a concave mirror to reflect light with properties that are the same as a spotlight (directed light rays parallel to each other), one has to consider its property to gather light to a point after reflecting. Meaning that, we can achieve the spotlight by locatng the point where the rays will be parallel, this point is called the focal point.
Therefore, the light bulb should be placed at the focal point of the mirror.
: A spaceship is traveling at the speed 2t 2 1 km/s (t is time in seconds). It is pointing directly away from earth and at time t 0 it is 1000 kilometers from earth. How far from earth is it at one minute from time t 0
Answer:
145060km
Explanation: Given that
speed = dx/dt = 2t^2 +1
integrate
x = 2/3t^3 + t + c (c is constant, x is in km, t is in second)
given that at t=0, x = 1000
so 1000 = 2/3 X (0)^3 + 0 + c
or c = 1000
So x = 2/3t^3 + t + 1000
for t = 1 min = 60s
x = 2/3 X 60^3 + 60 + 1000
x = 2/3×216000+ 1060
x = 144000+1060
= 145060km
At one minute, it will be 145060km far from the earth
A rod of length L is hinged at one end. The moment of inertia as the rod rotates around that hinge is ML2/3. Suppose a 2.50 m rod with a mass of 3.00 kg is hinged at one end and is held in a horizontal position. The rod is released as the free end is allowed to fall. What is the angular acceleration as it is released?
Answer:
6 rad/s²
Explanation:
Sum the torques about the hinge.
∑τ = Iα
mg(L/2) = mL²/3 α
g/2 = L/3 α
α = 3g/(2L)
α = 3 (10 m/s²) / (2 × 2.50 m)
α = 6 rad/s²
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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A flat, circular loop has 18 turns. The radius of the loop is 15.0 cm and the current through the wire is 0.51 A. Determine the magnitude of the magnetic field at the center of the loop (in T).
Answer:
The magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.
Explanation:
Given;
number of turns of the flat circular loop, N = 18 turns
radius of the loop, R = 15.0 cm = 0.15 m
current through the wire, I = 0.51 A
The magnetic field through the center of the loop is given by;
[tex]B = \frac{N\mu_o I}{2R}[/tex]
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
[tex]B = \frac{N\mu_o I}{2R} \\\\B = \frac{18*4\pi*10^{-7} *0.51}{2*0.15} \\\\B = 3.846 *10^{-5} \ T[/tex]
Therefore, the magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻⁵ T.
A red laser beam goes from crown glass with refraction index n=1.3 to air (n=1) with an incident angle of 0.23 radians. What is the angle of refraction in degrees?
Answer:
θ = 10.28º
Explanation:
To find the angle of refraction use the equation of refraction
n₁ sin θ₁ = n₂ sin θ₂
where index 1 is for incident light and index 2 is for refracted light.
sin θ₂ = n₁ / n₂ sin θ
let's calculate
sin = 1 / 1.3 sin 0.23
sin = 0.175
θ= 0.17528 rad
let's reduce to degrees
θ = 0.17528 rad (180ª / pi rad)
θ = 10.28º
Rays of light coming from the sun (a very distant object) are near and parallel to the principal axis of a concave mirror. After reflecting from the mirror, where will the rays cross each other at a single point?
The rays __________
a. will not cross each other after reflecting from a concave mirror.
b. will cross at the center of curvature.
c. will cross at the point where the principal axis intersects the mirror.
d. will cross at the focal point. will cross at a point beyond the center of curvature.
A concave mirror is an example of curved mirrors. So that the appropriate answer to the given question is option D. The rays will cross at the focal point.
A concave mirror is a type of mirror in which its inner part is the reflecting surface, while its outer part is the back of the mirror. This mirror reflects all parallel rays close to the principal axis to a point of convergence. It can also be referred to as the converging mirror.
In this type of mirror, all rays of light parallel to the principal axis of the mirror after reflection will cross at the focal point.
Therefore, the required answer to the given question is option D. i.e The rays will cross at the focal point.
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A painter sets up a uniform plank so that he can paint a high wall. The plank is 2 m long and weighs 400 N. The two supports holding up the plank are placed 0.2 m from either end. Show that the upwards force on each of the planks is 200 N. Draw a sketch.
The upward force on each supporting plank is 200 N
The given parameters include;
weight of the plank, W₁ = 400 Nlength of the plank, l = 2 mupward force of each supporting plank, = W₂ and W₃To show that the upward force of each supporting plank is 200 N, make the following sketch.
W₂ W₃
↑ ↑
-----------------------------------------------------------------------
0.2m ↓ 0.2m
400 N
The two supporting planks keeps the 2m plank in equilibrium position. If the plank is in equilibrium position the sum of the upward forces equals sum of the downward force.W₂ + W₃ = 400 N
But the distance of each supporting plank from the end is equal, (0.2m).
Then, W₂ = W₃
2W₂ = 400 N
W₂ = 400N/2
W₂ = 200 N
W₃ = 200 N
Therefore, the upward force on each supporting plank that keeps the plank in equilibrium position is 200 N.
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A car starts from rest and accelerates with a constant acceleration of 5 m/s2 for 4 s. The car continues for 18 s at constant velocity. How far has the car traveled from its starting point
"When red light in vacuum is incident at the Brewster angle on a certain glass slab, the angle of refraction is"
Complete Question
When red light in vacuum is incident at the Brewster angle on a certain glass slab, the angle of refraction is [tex]36.0 ^o[/tex] . What are
(a) the index of refraction of the glass and
(b) the Brewster angle?
Answer:
a
[tex]n_r = 1.376[/tex]
b
[tex]i = 54^o[/tex]
Explanation:
From the question we are told that
The angle of refraction is [tex]r = 36.0 ^o[/tex]
Generally according Brewster law
[tex]i + r = 90[/tex]
Here [tex]i[/tex] is the angle of incidence which is also the Brewster angle
So
[tex]i + 36.0 = 90[/tex]
[tex]i = 54^o[/tex]
Now the refractive index is mathematically represented as
[tex]n_r = tan (i)[/tex]
substituting values
[tex]n_r = tan (54)[/tex]
[tex]n_r = 1.376[/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
In a double-slit experiment using light of wavelength 486 nm, the slit spacing is 0.600 mm and the screen is 2.00 m from the slits. Find the distance along the screen between adjacent bright fringes.
Answer:
The distance is [tex]y = 0.00162 \ m[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 486 \ nm = 486 *10^{-9} \ m[/tex]
The slit spacing is [tex]d = 0.600 \ mm = 0.60 *10^{-3} \ m[/tex]
The distance of the screen is [tex]D = 2.0 \ m[/tex]
Generally the distance along the screen between adjacent bright fringes is mathematically represented as
[tex]y = \frac{\lambda * D}{d}[/tex]
substituting values
[tex]y = \frac{ 486 *10^{-9} * 2}{0.6*10^{-3}}[/tex]
[tex]y = 0.00162 \ m[/tex]
At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.45 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.10 ✕ 10−5 T
Answer:
The speed of the proton is 4059.39 m/s
Explanation:
The centripetal force on the particle is given by;
[tex]F = \frac{mv^2}{r}[/tex]
The magnetic force on the particle is given by;
[tex]F = qvB[/tex]
The centripetal force on the particle must equal the magnetic force on the particle, for the particle to remain in the circular path.
[tex]\frac{mv^2}{r} = qvB\\\\r = \frac{mv^2}{qvB} \\\\r = \frac{mv}{qB}[/tex]
where;
r is the radius of the circular path moved by both electron and proton;
⇒For electron;
[tex]r = \frac{(9.1*10^{-31})(7.45*10^6)}{(1.602*10^{-19})(1.1*10^{-5})}\\\\r = 3.847 \ m[/tex]
⇒For proton
The speed of the proton is given by;
[tex]r = \frac{mv}{qB}\\\\mv = qBr\\\\v = \frac{qBr}{m} \\\\v = \frac{(1.602*10^{-19})(1.1*10^{-5})(3.847)}{1.67*10^{-27}} \\\\v = 4059.39 \ m/s[/tex]
Therefore, the speed of the proton is 4059.39 m/s
The microwaves in a microwave oven are produced in a special tube called a magnetron. The electrons orbit the magnetic field at 2.4 GHz, and as they do so they emit 2.4 GHz electromagnetic waves. What is the strength of the magnetic field?
Answer:
The magnetic field is 0.0857 T.
Explanation:
The electrons orbit the magnetic field with a centripetal force equal to
F = [tex]\frac{mv^{2} }{r}[/tex]
also, the force on an electron in a magnetic field is gotten as
F = Bqv
equating this two equations give
[tex]\frac{mv^{2} }{r}[/tex] = Bqv
mv/r = Bq
where m is the mass of the electron = 9.11 x 10^-31 kg
v is the the linear speed of the electron
B is the magnetic field on the electron
r is the radius of the orbital movement
q is the charge on an electron = 1.602 x 10^-19 C
but, the linear speed v = ωr
where ω is the angular speed of the electron
substituting into equation above, we have
mωr/r = Bq
which reduces to
mω = Bq
finally, w know that the angular speed is related to the frequency of the electron by
ω = 2πf
we then finally have
2mπf = Bq
where f is the frequency emitted by the electron = 2.4 GHz = 2.4 x 10^9 Hz
substituting values into the equation, we have
2 x 9.11 x 10^-31 x 3.142 x 2.4 x 10^9 = B x 1.602 x 10^-19
B = (1.3734 x 10^-20)/(1.602 x 10^-19) = 0.0857 T
= 85.7 mT
The difference between a DC and an AC generator is that
a. the DC generator has one unbroken slip ring.
b. the AC generator has one unbroken slip ring
c. the DC generator has one slip ring splitin two halves.
d. the AC generator has one slip ring split in two halves.
e The DC generator has twounbroken sip rings
Answer:
The AC generator has one unbroken slip ring
Explanation:
In physics, the application of electromagnetic induction can be seen in generators and dynamos. Electromagnetic induction is the process of generating electricity using magnets. It found applications in generators and the types of generator they found application is in AC and DC generator.
An AC generator is also called a Dynamo. A DC generator contains what is called a SPLIT RING fixed to the end of the coil which can be separated and coupled back according to the name "split". An AC generator also called a Dynamo makes use of a SLIP ring which cannot be divided into two. It comes as an entity. The presence of this rings is what differentiates a DC generator from an AC generator.
We can replace split rings with slip rings when converting a DC generator to an AC generator and vice versa.
It can therefore be concluded that the difference between a DC and an AC generator is that the AC generator has one unbroken slip ring.
Sonar is used to determine the speed of an object. A 38.0-kHz signal is sent out, and a 40.0-kHz signal is returned. If the speed of sound is 341 m/s, how fast is the object moving?
Answer:
The velocity is [tex]v = 8.743 \ m/s[/tex]
Explanation:
From the question we are told that
The frequency of the signal sent out is [tex]f_s = 38.0 \ kHz = 38.0 *10^{3} \ Hz[/tex]
The frequency of the signal received is [tex]f_r = 40.0 \ kHz = 40.0 *10^{3} \ Hz[/tex]
The speed of sound is [tex]v_s = 341 \ m/s[/tex]
Generally the frequency of the sound received is mathematically represented as
[tex]f_r = f_s [\frac{v_s + v}{v_s - v} ][/tex]
where v is the velocity of the object
=> [tex]40 *10^{3} = 38 *10^{3} * [\frac{341 + v}{341 - v} ][/tex]
=> [tex]1.05263 = \frac{341+v }{341-v}[/tex]
=> [tex]358.94 - 1.05263v = 341 + v[/tex]
=> [tex]17.947 = 2.05263 v[/tex]
=> [tex]v = 8.743 \ m/s[/tex]
A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision.
(a) Is she nearsighted or farsighted?
(b) What type of lens (converging or diverging) is needed to correct her vision?
(c) What focal length contact lens is needed, and what is its power in diopters?
Answer:
(a) nearsighted
(b) diverging
(c) the lens strength in diopters is 1.33 D, and considering the convention for divergent lenses normally prescribed as: -1 33 D
Explanation:
(a) The person is nearsighted because he/she cannot see objects at distances larger than 75 cm.
(b) the type of correcting lens has to be such that it counteracts the excessive converging power of the eye of the person, so the lens has to be diverging (which by the way carries by convention a negative focal length)
(c) the absolute value of the focal length (f) is given by the formula:
[tex]f=\frac{1}{d} =\frac{1}{0.75} = 1.33\,D[/tex]
So it would normally be written with a negative signs in front indicating a divergent lens.
Approximating the eye as a single thin lens 2.70 cmcm from the retina, find the focal length of the eye when it is focused on an object at a distance of 265 cmcm
Answer:
0.37 cm
Explanation:
The image is formed on the retina which is at a constant distance of 2.70 cm to the lens. Therefore, image distance = 2.70 cm.
The object is at a distant of 265 cm to the lens of the eye.
From lens formula,
[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]
where: f is the focal length, u is the object distance and v is the image distance.
Thus, u = 265.00 cm and v = 2.70 cm.
[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{265}[/tex] + [tex]\frac{27}{10}[/tex]
= [tex]\frac{10+7155}{2650}[/tex]
[tex]\frac{1}{f}[/tex] = [tex]\frac{7165}{2650}[/tex]
⇒ f = [tex]\frac{2650}{7165}[/tex]
= 0.37
The focal length of the eye is 0.37 cm.
You need to make a spring scale to measure the mass of objects hung from it. You want each 1.0 cm length along the scale to correspond to a mass difference of 0.10 kg. What should be the value of the spring constant?
Answer:
The spring constant should be:
[tex]k= 98\, \frac{N}{m}[/tex]
Explanation:
Use Hooke's law for this problem, knowing that the magnitude of the force (F) on the spring equals the stretching it experiences [tex]\Delta x[/tex] times the spring constant "k":
[tex]F=k\,\Delta x[/tex]
in our case, since the mass hanging is given in kg, we need to multiply it by "g" to get the force exerted:
Then if we add to the spring in its relaxed state, a mass of 0.10 kg, and we want for that a displacement of 1 cm (0.01 m), then the value of the spring constant should be:
[tex]k=\frac{F}{\Delta x} \\k=\frac{9.8\,(0.1)}{0.01} \, \frac{N}{m} \\k= 98\, \frac{N}{m}[/tex]
A fish in a flat-sided aquarium sees a can of fish food on the counter. To the fish's eye, the can looks to be 43 cmcm outside the aquarium.
Required:
What is the actual distance between the can and the aquarium?
Answer:
The actual distance is [tex]d_a = 0.3233\ m[/tex]
Explanation:
From the question we are told that
The distance of the can is d = 43 cm = 0.43 m
Generally the actual distance is mathematically represented as
[tex]d_a = [\frac{ n_a }{n_w} ]* d[/tex]
Where [tex]n_a , n_w[/tex] are the refractive index of air and water and their value is
[tex]n_a = 1 , \ \ \ n_w = 1.33[/tex]
So
[tex]d_a = [\frac{ 1 }{1.33} ]* 0.43[/tex]
[tex]d_a = 0.3233\ m[/tex]
A tank whose bottom is a mirror is filled with water to a depth of 19.6 cm. A small fish floats motionless a distance of 6.40 cm under the surface of the water.
A) What is the apparent depth of the fish when viewed at normal incidence?
B) What is the apparent depth of the image of the fish when viewed at normal incidence?
Answer:
A. 4.82 cm
B. 24.66 cm
Explanation:
The depth of water = 19.6 cm
Distance of fish = 6.40 cm
Index of refraction of water = 1.33
(A). Now use the below formula to compute the apparent depth.
[tex]d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\= \frac{1}{1.33} \times 6.40 \\= 4.82 cm.[/tex]
(B). the depth of the fish in the mirror.
[tex]d_{real} = 19.6 cm + (19.6 cm – 6.40 cm) = 32.8 cm[/tex]
Now find the depth of reflection of the fish in the bottom of the tank.
[tex]d_{app} = \frac{n_{air}}{n_{water}} \times d_{real} \\d_{app} = \frac{1}{1.33} \times 32.8 = 24.66\\[/tex]
A guitar string is 90 cm long and has a mass of 3.5g . The distance from the bridge to the support post is L=62cm, and the string is under a tension of 540N . What are the frequencies of the fundamental and first two overtones? Express your answers using two significant figures. Enter your answers in ascending order separated by commas.
Answer:
[tex]v_1 = 301 Hz[/tex]
[tex]v_2 = 601 \ \ Hz[/tex]
[tex]v_3 = 901 \ Hz[/tex]
Explanation:
From the question we are told that
The length of the string is [tex]l = 90 \ cm = 0.9 \ m[/tex]
The mass of the string is [tex]m_s = 3.5 \ g =0.0035 \ kg[/tex]
The distance from the bridge to the support post [tex]L = 62 \ c m = 0.62 \ m[/tex]
The tension is [tex]T = 540 \ N[/tex]
Generally the frequency is mathematically represented as
[tex]v = \frac{n}{2 * L } [\sqrt{ \frac{T}{\mu} } ][/tex]
Where n is and integer that defines that overtones
i.e n = 1 is for fundamental frequency
n = 2 first overtone
n =3 second overtone
Also [tex]\mu[/tex] is the linear density of the string which is mathematically represented as
[tex]\mu = \frac{m_s}{l}[/tex]
=> [tex]\mu = \frac{0.0035 }{ 0.9 }[/tex]
=> [tex]\mu = 0.003889 \ kg/m[/tex]
So for n = 1
[tex]v_1 = \frac{1}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]
[tex]v_1 = 301 \ Hz[/tex]
So for n = 2
[tex]v_2 = \frac{2}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]
[tex]v_2 = 601 \ Hz[/tex]
So for n = 3
[tex]v_3 = \frac{3}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]
[tex]v =901 \ Hz[/tex]
Two long straight wires are suspended vertically. The wires are connected in series, and a current from a battery is maintained in them. What happens to the wires? What happens if the battery is replaced by an a-c source?
Answer:
(i) When a battery is connected inseries to two long parallel wires, the currents in the two wires will be in opposite directions. Due to which a force of repulsion will be acting between them and they are moving further apart.
(ii) When a battery is connected in parallel to two long parallel wires, the currents in the two wires will be in same direction. Due to it, a force of attraction will be acting between them and they are coming closer to each other.
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Explanation:
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What is the current in milliamperes produced by the solar cells of a pocket calculator through which 4.2 C of charge passes in 2.7 h
Answer:
0.432mAExplanation:
Current produced by the solar cells of the pocket calculator is expressed using the formula I = Q/t where;
Q is the charge (in Columbs)
t is the time (in seconds)
Given parameters
Q = 4.2C
t = 2.7 hrs
t = 2.7*60*60
t = 9720 seconds
Required
Current produced by the solar cell I
Substituting the given values into the formula;
I = 4.2/9720
I = 0.000432A
I = 0.432mA
Hence, the current in milliamperes produced by the solar cells of a pocket calculator is 0.432mA
A spherical balloon has a radius of 6.95 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of air is 1.29 kg/m3. The skin and structure of the balloon has a mass of 950 kg. Neglect the buoyant force on the cargo volume itself. Determine the largest mass of cargo the balloon can lift. Express your answer to two significant figures and include the appropriate units.
volume of balloon
= 4/3 T R3
= 4/3 x 3.14 x 6.953
= 1405.47 m3
uplift force
= volume of balloon x density of air x 9.8
= = 1405.47 x 1.29 x 9.8
= 1813.05 x 9.8 N
weight of helium gas
= volume of balloon x density of helium x
9.8
= 1405.47 x .179 x 9.8
= 251.58 x 9.8 N
Weight of other mass = 930 x 9.8 N Total weight acting downwards
= 251.58 x 9.8 +930 x 9.8
= 1181.58 x 9.8 N
If W be extra weight the uplift can balance
1181.58 × 9.8 + W × 9.8 = 1813.05 * 9.8
1181.58+W=1813.05
W= 631.47 kg
A laser emits photons having an energy of 3.74 × 10–19 J. What color would be expected for the light emitted by this laser? (c = 3.00 × 108 m/s, h = 6.63 × 10–34 J ⋅ s)
Answer:
The wavelength of the emitted photons 532 nm, corresponds to a visible light having GREEN color.
Explanation:
Given;
energy of the emitted photons, E = 3.74 x 10⁻¹⁹ J
speed of light, c = 3 x 10⁸ m/s
Planck's constant, h = 6.63 x 10⁻³⁴ J.s
The wavelength of the emitted light will be calculated by applying energy of photons;
[tex]E = hf[/tex]
where;
E is the energy emitted light
h is Planck's constant
f is frequency of the emitted photon
But f = c / λ
where;
λ is the wavelength of the emitted photons
[tex]E = \frac{hc}{\lambda} \\\\\lambda = \frac{hc}{E} \\\\\lambda = \frac{6.63*10^{-34} *3*10^{8}}{3.74*10^{-19}} \\\\\lambda = 5.318 *10^{-7} \ m\\\\\lambda = 531.8 *10^{-9} \ m\\\\\lambda = 531.8 \ nm[/tex]
λ ≅ 532 nm
the wavelength of the emitted photons is 532 nm.
Therefore, the wavelength of the emitted photons 532 nm, corresponds to a visible light having GREEN color.
front wheel drive car starts from rest and accelerates to the right. Knowing that the tires do not slip on the road, what is the direction of the friction force the road applies to the rear tire
Answer:
The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.
Explanation:
This question suggests that the car is accelerating forward. Thus, the easiest way for us to know what friction is doing is for us to know what happens when we turn friction off.
Now, if there is no friction and the car is stopped, if we push down on the accelerator, it will make the front wheels to spin in a clockwise manner. This spin occurs on the frictionless surface with the rear wheels doing nothing while the car doesn't move.
Now, if we apply friction to just the front wheels, the car will accelerate forward while the back wheels would be dragging along the road and not be spinning. Thus, friction opposes the motion and as such, it must act im a direction opposite to where the car is going. This must be static friction.
The frictional force the road applies to the rear tire is static friction and it acts opposite to the direction in which the car is traveling.
You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and hang a 1.8 g mass from it. This stretches the spring to a length of 5.1 cm . You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.3 cm .
Requried:
What is the magnitude of the charge (in nC) on each bead?
Answer:
2.2nC
Explanation:
Call the amount by which the spring’s unstretched length L,
the amount it stretches while hanging x1
and the amount it stretches while on the table x2.
Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,
we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,
applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,
where ke is the Coulomb constant. Combining these,
we get q = √(mgx2(L+x2)²/x1ke =2.2nC
"Can we consider light wave as a single frequency wave? Either Yes or No, explain the reason of your answer. "
Answer:
Well, yes.
We can have an isolated light wave that is defined by only one frequency (and one wavelenght). But this is not a really common situation, most of the light that we can see in nature, is actually a composition of different waves with different frequencies.
Even if we have, for example, a red laser, the actual frequency of the light that comes from the laser may be in a range of frequencies, so the actual wave is a composition of different waves with really close frequencies.
An example of a light wave defined by only one frequency can be, for example, the photon that comes out of a change in energy of an electron.
Here we have a single photon, with a single frequency, that is modeled as a single frequency wave.