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2 years ago

What is the mass of an 291pound persond

Physics

1 answer:

Luden [163]2 years ago

4 0

Well, it also depends on the height... but say if they were 5'3" and 291 pounds...
their BMI would be

51.55

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What are the fundamental units involved in pascal<br>

Llana [10]

Answer:

<h2>FUNDAMENTAL UNITS INVOLVED ARE : NEWTON AND SECOND .</h2>

<h2>FORMULA OF PRESSURE = </h2>

4 0

3 years ago

Read 2 more answers

A car travels at a speed of 40 m/s for 29.0 s;<br>what is the distance traveled by the car?

lianna [129]

Answer: 1160 m

Explanation:

Speed = distance / time. Plug in 40 m/s for speed and 29 s for time in order to get the distance, 1160 m.

8 0

1 year ago

Calculate the amount of heat (in kJ) required to convert 97.6 g of water to steam at 100° C. (The molar heat of vaporization of

Ratling [72]

Answer:

221.17 kJ

Explanation: Note the heat of vaporization is in kJ/mol,then to determine the number of moles of water: divide the mass by 18. Then multiply the number of moles by the molar heat of vaporization of water.

N = 97.6 ÷ 18

Q=molar heat *moles

Q = (40.79) * (97.6 ÷ 18)

This is approximately 221.17 kJ

4 0

2 years ago

You want to produce three 2.00-mm-diametercylindrical wires, each with a resistance of 1.00 Ω at room temperature. One wire is g

Vlada [557]

Answer:

(a) L = 128.75 m

(b) L = 182.56 m

(d) Mass of Gold = 7.68 kg = 7680 gram

(e) Cost of Gold Wire = $ 307040

Explanation:

The resistance of the wire is given as:

R = ρL/A

where,

R = Resistance

ρ = resistivity

L = Length

A = cross-sectional area

(a)

For Gold Wire:

ρ = 2.44 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.44 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.44 x 10⁻⁸ Ω.m)

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(b)

For Copper Wire:

ρ = 1.72 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (1.72 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(1.72 x 10⁻⁸ Ω.m)

<u></u>

(c)

For Aluminum Wire:

ρ = 2.75 x 10⁻⁸ Ω.m

A = πd²/4 = π(2 x 10⁻³ m)²/4 = 3.14 x 10⁻⁶ m²

R = 1 Ω

Therefore,

1 Ω = (2.75 x 10⁻⁸ Ω.m)L/(3.14 x 10⁻⁶ m²)

L = (1 Ω)(3.14 x 10⁻⁶ m²)/(2.75 x 10⁻⁸ Ω.m)

<u></u>

(d)

Density = Mass/Volume

Mass = (Density)(Volume)

Volume of Gold = AL = (3.14 x 10⁻⁶ m²)(128.75 m) = 4.04 x 10⁻⁴ m³

Therefore,

Mass of Gold = (1.9 x 10⁴ kg/m³)(4.04 x 10⁻⁴ m³)

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(e)

Cost of Gold Wire = (Unit Price of Gold)(Mass of Gold)

Cost of Gold Wire = ($ 40/gram)(7680 grams)

7 0

3 years ago

A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If

Salsk061 [2.6K]

Complete question:

A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?

Answer:

The current in the circuit 7 ms later is 0.2499 A

Explanation:

Given;

Ideal inductor, L = 45-mH

Resistor, R = 60-Ω

Ideal voltage supply, V = 15-V

Initial current at t = 0 seconds:

I₀ = V/R

I₀ = 15/60 = 0.25 A

Time constant, is given as:

T = L/R

T = (45 x 10⁻³) / (60)

T = 7.5 x 10⁻⁴ s

Change in current with respect to time, is given as;

$I(t) = I_o(1-e^{-\frac{t}{T}})$

Current in the circuit after 7 ms later:

t = 7 ms = 7 x 10⁻³ s

$I(t) = I_o(1-e^{-\frac{t}{T}})\\\\I =0.25(1-e^{-\frac{7*10^{-3}}{7.5*10^{-4}}})\\\\I = 0.25(0.9999)\\\\I = 0.2499 \ A$

Therefore, the current in the circuit 7 ms later is 0.2499 A

6 0

3 years ago