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Alexus [3.1K]
3 years ago
14

What is the momentum of a 10 kg object traveling at 4 m/s?

Physics
1 answer:
brilliants [131]3 years ago
4 0
Answer:40kg m/s
explanation: momentum =mass x velocity
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Define mass .Also write the table showing the relation of 'kg' with it's sub- multiples and multiples.​
lys-0071 [83]

Answer:

the quantity of matter contained in a body is called mass .si unit of mass is kilogram.weight is effected by gravity but mass is not effected by gravity.

3 0
2 years ago
Read 2 more answers
Please, pretty please help me!
Andrews [41]

Answer:

umm

Explanation:

1223565 578633 =334675

8 0
3 years ago
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How would you describe brass since it ia used in weapons, pipes, intruments and ect? A compound, Alloy, Element l, Or molecule?
Alik [6]
<span>Brass is an <u>alloy</u>. An alloy is a mixture of elements to form a unique material. Brass is a mixture of copper and zinc and the percentage of each element depends on the desired material. It has a higher malleability than bronze or zinc. Meaning that it can be bend easily into it desired form.</span>
5 0
3 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

(c) L =  114.28 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)

<u>L =  128.75 m</u>

<u></u>

(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>L =  182.56 m</u>

<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>L =  114.28 m</u>

<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)

<u>Mass of Gold = 7.68 kg = 7680 gram</u>

<u></u>

(e)

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

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

<u>Cost of Gold Wire = $ 307040</u>

7 0
4 years ago
How much energy is needed to change 100 g of 0o C ice to 0o C water? The latent heat of fusion for water L=335,000 J/kg. Hint: T
pychu [463]

Answer:

33.5 kJ

Explanation:

here there is no difference is made in the temperature. Only thing happens here is the conversion of the ice in to water of 0 degree. The heat energy taken from the outside is spent for this conversion.

we have ice 100g =0.1 kg

Appplying Q=mL

                  Q= 0.1 kg * 335 000 J kg^{-1}

                   Q = 33 500 J

                   Q = 33.5 kJ

3 0
4 years ago
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