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2 years ago

If you want to lift a 5-kg box to a height of 2 meters, how much work must be

Physics

1 answer:

BlackZzzverrR [31]2 years ago

5 0

Answer:100joules

Explanation:

Mass(m)=5kg

Acceleration due to gravity(g)=10m/s^2

Height(h)=2 meters

Work=m x g x h

Work=5 x 10 x 2

Work=100joules

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Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times g

ollegr [7]

Answer:

The beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

Explanation:

Given;

velocity of wave on the string with lower tension, v₁ = 35.2 m/s

the fundamental frequency of the string, F₁ = 258 Hz

velocity of wave on the string with greater tension;

$v_1 = \sqrt{\frac{T_1}{\mu }$

where;

v₁ is the velocity of wave on the string with lower tension

T₁ is tension on the string

μ is mass per unit length

$v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}$

Where;

T₁ lower tension

T₂ greater tension

v₁ velocity of wave in string with lower tension

v₂ velocity of wave in string with greater tension

From the given question;

T₂ = 1.1 T₁

$v_2^2 = \frac{T_2v_1^2}{T_1} \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s$

Fundamental frequency of wave on the string with greater tension;

$f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1} =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz$

Beat frequency = F₂ - F₁

= 270.6 - 258

= 12.6 Hz

Therefore, the beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

6 0

3 years ago

High-voltage power lines are a familiar sight throughout the country. The aluminum wire used for some of these lines has a cross

goblinko [34]

Answer:

Explanation:

For resistance of a wire , the formula is as follows

R = ρ L / S

where ρ is specific resistance , L is length and S is cross sectional area

Given L = 14 000 m ,

S = 4.8 x 10⁻⁴ m²

specific resistance of aluminum = 2.8 x 10⁻⁸ ohm-meter

Putting the values in the formula

R = 2.8 x 10⁻⁸ x 14 x 10³ / (4.8 x 10⁻⁴ )

R = 0.8167 ohm .

= .82 ohm .

5 0

3 years ago

Question 6

sleet_krkn [62]

Answer:

B: Process #1: Energy is decreasing Process#2: Energy is increasing

6 0

3 years ago

A stone is thrown vertically upward with a speed of 18.0 m/s. How fast is it moving when it reaches a height of 11.0m? How long

steposvetlana [31]

Given: v0= 18.0 m/s, y0=0m, yf=11m, g=-9.81 m/s^2

v0= initial velocity, vf= final velocity, y0= initial height, yf= final height, g= gravity, sqrt()= square root, ^2=squared

vf^2=v0^2 + (2)(g)(yf-y0)
vf^2=(18.0 m/s)^2+(2)(-9.81 m/s^2)(11 m-0m)
vf^2=18.0 m/s)^2 + (-19.62 m/s^2)(11 m)
vf^2=(324 m^2/s^2) - (215.82 m^2/s^2)
vf^2=108.18 m^2/s^2
vf=sqrt(108.18 m^2/s^2)
vf=10.4 m/s

8 0

3 years ago

Uranium-235 undergoes fission, forming krypton-92, barium-141, and 3

Ahat [919]

Answer:

Explanation:

Some of the mass

8 0

2 years ago

Read 2 more answers