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A bag of cement of Weight 1000N hangs from ropes. Two of the ropes make angles of 1=60 and 2=30 with the horizontal.if the system is in equilibrium,find the tension T1,T2andT3 in the ropes
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n Section 12.3 it was mentioned that temperatures are often measured with electrical resistance thermometers made of platinum wire. Suppose that the resistance of a platinum resistance thermometer is 125 Ω when its temperature is 20.0°C. The wire is then immersed in boiling chlorine, and the resistance drops to 99.6 Ω. The temperature coefficient of resistivity of platinum is α = 3.72 × 10−3(C°)−1. What is the temperature of the boiling chlorine?
b) 6.67x10-19hz
c) 3x108hz
d) 1.5hz
Show calculation
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Answer:
1.5 x 10¹⁸hz
Explanation:
Given parameters:
Wavelength = 2 x 10⁻¹⁰m
Unknown:
Frequency = ?
Solution:
To find the frequency, use the expression below;
V = f x wavelength
V is the speed of light = 3 x 10⁸m/s
f is the frequency
Now;
Insert the parameters
3 x 10⁸ = 2 x 10⁻¹⁰ x frequency
Wavelength = = 1.5 x 10¹⁸hz
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Answer:
V1 = -3.260 m/s, V2 = 1.303 m/s
Explanation:
Let mass of the left glider m1 = 0.157 kg and velocity v1 = 0.850 m/s
mass of the right glider m2 = 0.306 Kg and v2 = -2.26 m/s (-ve sign mean it is opposite to direction of left glider)
To Find: Final Velocity of Left Glider is V1=? m/s and Velocity of right Glider is V2 =? m/s (After Collision)
from law of conservation of momentum and energy we deduce a formula:
V1 = (m1-m2) v1 /(m1+m2) + 2 m2 v2/(m1+m2)
V1 = (0.157 kg - 0.306 Kg) × 0.850 m/s / (0.157 kg + 0.306 Kg) + 2 ×0.306 kg × -2.26 m/s / (0.157 kg + 0.306 Kg)
V1 = -0.273 -2.987
V1 = -3.260 m/s
and V2 Formula
V2 = (m2-m1) v2/(m1+m2) + 2 m1 v1/(m1+m2)
V2 = (0.157 kg - 0.306 Kg) × -2.26 m/s / (0.157 kg + 0.306 Kg) + 2 ×0.157 kg × 0.850 m/s / (0.157 kg + 0.306 Kg)
V2 = 0.727 + 0.576
V2 = 1.303 m/s
-0.149, 0.463
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that counts.
However many turns the primary has, the secondary should have
about TEN TIMES that number. Then the transformer will multiply
the primary voltage by 10 ... 120 volts of AC at the primary will
become 1,200 volts of AC at the secondary.
Note that it HAS TObe AC. If the transformer is supplied with DC,
then 120 volts at the primary becomes zero volts at the secondary
and a big cloud of stinky smoke in the room.
Answers
Answer:
The wavelength of the wave is 1 m
Explanation:
Given;
mass of the string, m = 20 g = 0.02 kg
length of the string, L = 3.2 m
tension on the string, T = 2.5 N
the frequency of the wave, f = 20 Hz
The velocity of the wave is given by;
where;
μ is mass per unit length = 0.02 kg / 3.2 m
μ = 6.25 x 10⁻³ kg/m
The wavelength of the wave is given by;
λ = v / f
λ = (20 m/s )/ (20 Hz)
λ = 1 m
Therefore, the wavelength of the wave is 1 m
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Answer:
doubling the size of the tax more than doubles the deadweight loss while less than doubles the revenue generated
Explanation:
(a)
The quantity of rooms rented before tax, Q1 = 1000 rooms.
The quantity of rooms rented after the imposition of tax Q2 = 900 rooms.
Size of the tax = $10
Price paid by buyer = $108
Price received by seller = $98
Deadweight loss = 1/2 x (Q2 — Q1) x (size of the tax)
Deadweight loss = 1/2 x (1000 — 900) x ($10) = $500
Tax revenue generated = size of tax * (Q2) = $10 x (900) = $9000
b)
The quantity of rooms rented before tax, Q1 = 1000 rooms
The quantity of rooms rented after the imposition of tax, Q2 = 800 rooms Size of the tax = $20
Price paid by buyer = $116
Price received by seller = $96
Deadweight loss = 1/2 x (Q2 — Q1) x (size of the tax)
New Deadweight loss = 1/2 x (1000 — 800) x ($20) = $2000
Thus, dead weight loss quadruples post doubling the size of tax. New Tax revenue generated = size of tax x (Q2) = $20 x (800) = $16000 Thus, revenue generated less than doubles post doubling the size of tax.
Answers
Answer:
y = 80.2 mille
Explanation:
The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening
θ = 1.22 λ/ d
in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m
θ = 1.22 550 10⁻⁹ / 0.002
θ = 3.355 10⁻⁴ rad
Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi
tan θ = y / L
y = L tan θ
y = 2,389 10⁵ tan 3,355 10⁻⁴
y = 8.02 10¹ mi
y = 80.2 mille
This is the smallest size of an object seen directly by the eye
Final answer:
An individual with 20/20 vision can observe the moon from a maximum distance of around 6200 km or 3850 miles. Beyond this distance, it might be difficult to distinguish the moon from other celestial objects without using a telescope. The use of a telescope can expand this range significantly.
Explanation:
The detailed observation of a lunar eclipsed, when viewed without any form of optical aid like a telescope, is contingent on many factors, one of which is the human eye's angular resolution—the eye's ability to differentiate between two separate points of light. For an average human eye with 20/20 vision, the angular resolution is approximately 0.02 degrees.
To calculate the maximum distance at which the moon could be observed clearly with the eye, the formula for small angle approximation can be used, which in this context is: Distance = Size / Angle = (2159.14 miles) / (0.02 degrees in radians). This calculates to a distance of approximately 6200 km or 3850 miles.
Beyond this distance, distinguishing the moon from other celestial bodies might be challenging using just the eye. Utilizing a high-powered telescope would significantly extend this range by magnifying the image, allowing clearer detail over much greater distances.
Learn more about Observing Moon here:
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