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BigorU [14]
3 years ago
13

3 kg of wet clothes are hung on the middle of a clothesline with posts 6 ft apart. The clothesline sags down by 3 feet. What is

the total tension upon the clothesline?

Physics
2 answers:
vredina [299]3 years ago
5 0

Answer:

15√2 N.

Explanation:

Given that the mass of the wet clothes is, m=3kg

Here in the figure ABC the weight is hung from the middle then the sides AC and BC are equal will make their adjacent angle equal and the other angle which is ∠ACB will become 90°.

Now according the figure the tension is apply in two direction than by free body diagram we can solve newton equations.

2Tcos\theta=mg\\T=\frac{mg}{2cos\theta}\\ T=\frac{3\times10}{2cos(45^{\circ})}\\ T=15\sqrt{2}N

Therefore, the total tension upon the clothesline is, T=15\sqrt{2}N

miskamm [114]3 years ago
4 0

Answer:

Tension in string equals 14.715 Newtons

Explanation:

The situation is represented in the figure attached below:

For equilibrium of the clothes along y- axis we have

\sum F_{v}=0\\\\\Rightarrow 2Tcos(\theta )=W\\\\\therefore T=\frac{W}{2cos(\theta )}

Applying values we get

\sum F_{v}=0\\\\\Rightarrow 2Tcos(\theta )=W\\\\\therefore T=\frac{3\times 9.81}{2\times 1}=14.715N(\because cos(\theta )=\frac{3}{3}=1)

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A girl playing tug-of-war with her dog pulls the dog a distance of 8.0m by exerting a force at an angle of 18° with the horizont
AnnZ [28]

Answer:

25 N

Explanation:

Work is a product of force and perpendicular distance moved.

W=Fd where F is force exerted and d is perpendicular distance.

However, for this case, the distance is inclined hence resolving it to perpendicular so that it be along x-axis we have distance as dcos\theta

Therefore, W=Fdcos\theta

Making F the subject of the formula then

F=\frac {W}{dcos\theta} where \theta is the angle of inclination. Substituting 190 J for W then 18 degrees for \theta and 8 m for d then

F=\frac {190}{8cos18^{\circ}}\approx 25N

3 0
3 years ago
A 20 μF capacitor initially charged to 30 μC is discharged through a 1.5 kΩ resistor. Part A How long does it take to reduce the
Natasha_Volkova [10]

Answer:

it will take 36.12 ms to reduce the capacitor's charge to 10 μC

Explanation:

Qi= C×V

then:

Vi = Q/C = 30μ/20μ = 1.5 volts

and:

Vf = Q/C = 10μ/20μ = 0.5 volts

then:

v = v₀e^(–t/τ)  

v₀ is the initial voltage on the cap  

v is the voltage after time t  

R is resistance in ohms,  

C is capacitance in farads  

t is time in seconds  

RC = τ = time constant  

τ = 20µ x 1.5k = 30 ms  

v = v₀e^(t/τ)  

0.5 = 1.5e^(t/30ms)  

e^(t/30ms) = 10/3  

t/30ms = 1.20397

t = (30ms)(1.20397) = 36.12 ms

Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.

7 0
3 years ago
Menciona generaciones de particulas y explique cada una
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ENGLISH:

It distinguishes between the fermions, which are particles of matter, and the bosons, which carry forces. The matter particles include six quarks and six leptons. The six quarks are called the up, down, charm, strange, top and bottom quark. ... All of these matter particles fall into three “generations.”

SPANISH:

Distingue entre los fermiones, que son partículas de materia, y los bosones, que transportan fuerzas. Las partículas de materia incluyen seis quarks y seis leptones. Los seis quarks se denominan quark arriba, abajo, encanto, extraño, superior e inferior. ... Todas estas partículas de materia se dividen en tres "generaciones".

3 0
3 years ago
A supersaturated solution is one which A. has less solute dissolved than the solution should hold at that temperature. B. has mo
Anna35 [415]

Answer:

Option C is correct

Explanation:

A supersaturated solution is one that has more solute dissolved than the solution should hold at that temperature.

Examples include carbonated water, sugar syrup, honey.

A solution of a chemical compound can be dissolved in heated water to prepare a supersaturated solution. A solution becomes supersaturated as its temperature is changed.

3 0
3 years ago
Amplitude modulation is used in _____.
spayn [35]

Amplitude modulation is a modulation technique used in electronic communication, most commonly for transmitting information via a radio carrier wave. In amplitude modulation, the signal strength of the carrier wave is varied in proportion to that of the message signal being transmitted. The message signal is, for example, a function of the sound to be reproduced by a loudspeaker, or the light intensity of pixels of a television screen. This technique contrasts with frequency modulation, in which the frequency of the carrier signal is varied, and phase modulation, in which its phase is varied.

AM was the earliest modulation method used to transmit voice by radio. It was developed during the first quarter of the 20th century beginning with Landell de Moura and Reginald Fessenden's radiotelephone experiments. It remains in use today in many forms of communication; for example it is used in portable two-way radios, VHF aircraft radio, citizens band radio, and in computer modems in the form of QAM. AM is often used to refer to mediumwave AM radio broadcasting.

6 0
3 years ago
Read 2 more answers
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