A Copper Wire Has Diameter 0 5 Mm

A cube of copper has a mass of 600kg. copper has a density of 9.3g/cm³. Calculate the length of each side of the cube to the nearest mm.

Daftar Isi

1. A cube of copper has a mass of 600kg. copper has a density of 9.3g/cm³. Calculate the length of each side of the cube to the nearest mm.

This is how you solve it

volume = 600.000 / 9,3 (we make 600 kg to 600.000 g because the density use gram to)

Volume = 64.516 cm³ (approximately)

Lenght = 254 cm

Lenght = 254,0 mm

2. One of the element used to produce wire is copper. The chemical code for copper is...

Copper = tembaga = Cu(Cuprum)

3. Please Answer Correctly 1. If a 1.5 m length of resistance wire has a resistance of 1.9 Ω and the wire has a diameter of 0.9 mm, calculate the resistivity of the wire. 2. The length of a piece of wire is 4 m and the resistance is 1.6 Ω. If the electrical resistance of the wire is 8 x 10-7 Ωm, what is the cross-sectional area of the wire? 3. A 2m long piece of wire has a cross-section area of 0.15 mm2. What is the resistance of the wire if its resistivity is 2x10-8 Ωm? 4. A 2m long wire with a cross-sectional area of 1 mm2 has a resistance of 16 Ω. What is the resistance of the wire if the cross-sectional area is 2 mm2? 5. If the resistance of a wire is 16 Ω, the length of the wire is 4 m. What is the resistance of the wire if it is 1 m long? 6. The length of a wire is 1m. If the wire is now cut to 0.5 m, the resistance of the wire becomes .... (same/half times larger / twice larger) 7. If the 1.25 m length of resistance wire has a resistance of 3 Ω and the wire has a diameter of 0.5 m. The resistivity of the wire is...

Jawaban:

Tolong Jawab Dengan Benar

1. Jika kawat hambatan sepanjang 1,5 m memiliki hambatan 1,9 dan kawat memiliki diameter 0,9 mm, hitunglah hambatan kawat tersebut.

2. Panjang seutas kawat adalah 4 m dan hambatannya 1,6 . Jika hambatan listrik kawat adalah 8 x 10-7 m, berapa luas penampang kawat tersebut?

3. Sepotong kawat sepanjang 2m memiliki luas penampang 0,15 mm2. Berapakah hambatan kawat jika resistivitasnya 2x10-8 m?

4. Sebuah kawat sepanjang 2m dengan luas penampang 1 mm2 memiliki hambatan 16 . Berapakah hambatan kawat jika luas penampangnya 2 mm2?

5. Jika hambatan sebuah kawat adalah 16 , panjang kawat tersebut adalah 4 m. Berapakah hambatan kawat jika panjangnya 1 m?

6. Panjang seutas kawat adalah 1m. Jika kawat sekarang dipotong menjadi 0,5 m, hambatan kawat

menjadi .... (sama/setengah kali lebih besar / dua kali lebih besar)

7. Jika panjang kawat hambatan 1,25 m memiliki hambatan 3 dan kawat memiliki diameter 0,5 m. Resistivitas kawat tersebut adalah...

Penjelasan:

ini arti nya doang yaヾ(＾-＾)ノjawaban nya sendiri aja takut nya salah klo aku yg Jawabಥ‿ಥ makasih

4. Agung has a 5 meter long wire (kawat). the wire is used to make a cube frame with a length of 35 cm. the rest of wire is

[tex]the \: length \: of \: a \: cube \: frame \\ 12 \: edges \times side \: length \\ 12 \times 35 \: cm \\ 420 \: cm \\ \\ the \: rest \: of \: wire \\ 500 \: cm - 420 \: cm \\ 80 \: cm[/tex]

5. Zaim has a piece of wire lenght 125π cm. The wire is cut to form 10 circles. The diameter of the circles differ from each other, in ascending order, by 2 cm. Calculate the diameter of the smallest circle.

Jawab:

3.5 cm

Penjelasan dengan langkah-langkah:

Zaim's wire length = 125π

Divided into 10 circles with 2 cm differ.

Assume that the smallest one has x diameter. So, the largest one has x+18 diameter.

The diameter for each (in descending order) is x, x+2, x+4, x+6, x+8, x+10, x+12, x+14, x+16, x+18.

The sum of circumference of all circles must be equal to wire length.

Formula for each circle circumference is πd.

The sum of circumference = π * (sum of all diameters) = [tex]\pi(10x + 90)[/tex]

Wire length = 125π

So,

[tex]\pi(10x + 90) = 125\pi \\10x + 90 = 125\\10x = 35\\x = 3.5[/tex]

With the smallest circle diameter equal to x, then the smallest circle has 3.5 cm diameter

6. umur David masih 24× dari umur Copper, dan umur Field masih 48× dari Copper. jika umur David 10 tahun. maka, berapa umur Copper dan Field? ini hanya kuis, nanti aku jawabkan

Jawaban:

umur copper : 5 bulan

umur field : 20 tahun

Penjelasan:

David = 24x copper

David = 10 tahun = 120 bulan

10 tahun = 24 x C

C = 10 tahun : 24 = 120 bulan : 24 = 5 bulan

jadi umur copper adalah 5 bulan

umur field = 48 x copper = 48 x 5 bulan

=240bulan

jadi umur field = 240 bulan : 12 bulan = 20 tahun

semoga membantu

7. 9. Below is the object that can be attracted by magnets .... * O A. aluminium can O B. copper wire C. iron nail D. rubber eraser

Answer:

C. Iron nail

-AnswerbyInjuniee

Hopeithelpsandgoodluck!

MAKETHEBRAINLIESTANSWER!

8. Where should you attach a wire

artinya: kemana kamu harus memasang kawat ?
On the corner of the wall

9. mr wong cuts a coil of wire into two pieces in the ratio 5:1 the leghth of the coil of wire is 180 cm.how long is the shorter piece of wire

ratio = 5 : 1
sum ratio = 5+1 = 6
Long = 180 cm
shorter piece = 1/6* 180 cm
=180 cm / 6
= 30 cm

good luck , follow yeah ,thank

10. pengertian wire diameter

The wire diameter (d) in a specification is always mesured before weaving. It may be slightly altered during the weaving process. The thicker the wire diameter the stronger and more abrasion resistant is the wire cloth. The heavier the wire the smaller the open area.

11. A copper wire of cross-sectional area 2.0 mm2 carries a current of 10a. how many electrons pass through a given cross-section of the wire in one second ?

A copper wire of cross-sectional area 2.0 mm2 carries a current of 10a. how many electrons pass through a given cross-section of the wire in one second?

Sebuah kawat tembaga memiliki area potongan melintang 2.0 mm2 membawa arus 10a. Berapa jumlah elektron yang lewat melalui potongan melintang kabel tersebut dalam satu detik?

Jumlah elektron yang lewat melalui cross-section kabel tersebut adalah 6.25 x 10¹⁹ elektron/detik

Pembahasan:

Pertanyaan ini sedikit menjebak karena menyebut area potongan melintang yang dapat diabaikan.

Apa itu ampere?

Satu Ampere sama dengan satu Coulomb per detik

(1 A = 1 C).

Jadi,

10A = 10 Coulumb/detik

Apa itu coulomb?

Satu Coulomb (C) sama dengan kira-kira 6.241 x 10¹⁸ muatan dasar,

Atau 1 C = 6.241 x 10¹⁸ x Muatan Dasar

(Dimana Muatan Dasar memiliki konstan = 1.6 x 10⁻¹⁹ C)

= 6.241 x 10¹⁸ x 1.60217662 x 10⁻¹⁹ C

Jika,

1 C = 6.241 x 10¹⁸ elektron/detik

10 C = (10 C x 6.241 x 10¹⁸elektron/detik) / 1 C

= 6.241 x 10¹⁸ elektron/detik

Sekarang, muatan dasar selalu memiliki konstan 1.6 x 10⁻¹⁹ C, yang berarti anda akan mendapatkan 6.25 x 10¹⁸ elektron dalam satu Coulomb.

Jadi, jawabannya adalah:

10C x (6.25 x 10¹⁸ elektron/detik)/1 C = 6.25 x 10¹⁸elektron/detik.

Pelajari lebih lanjut:

Penjelasan tentang Arus Listrik https://brainly.co.id/tugas/8039693

Penjelasan tentang Tegangan listrik, Arus listrik, Hambatan listrik https://brainly.co.id/tugas/24426162

#BelajarBersamaBrainly #SPJ4

12. What is the resistance of a nichrome wire 1 mm in diameter and 1 m in length?

jawaban:

tentukan luas penampang kawat, A

A = 1/4 πd²

A = 1/4 × 3,14 × 10⁻⁸

A = 7,85 × 10⁻⁹ m²

hambatan pada kawat ditentukan dengan persamaan

R = ρ × L / A

R = 10⁻⁶ × 1 / (7,85 × 10⁻⁹)

R = 127,4 Ω

dengan menggunakan hukum ohm

V = I × R

I = V / R

I = 5 / 127,4

I = 0,039 A

I = 39 mA

jadi arus yang mengalir pada kawat tersebut adalah 39 mA

maap kalo salah

13. The weight of a piece of wire is directly proportional to its length. A piece of wire is 25 cm long and has a weight of 6 grams. What is the weight of another piece of the same wire which is 30 cm long.

step-by-step explanation :

According to the question

25 cm piece of wire has weight of 6 grams

So, 1 cm of wire will has a weight of 0.24 grams

( as 25cm weight = 6grams,

1cm = 6/25 grams = 0.24 grams)

As 1 cm = 0.24 grams so,

30 cm = 0.24 x 30 grams

= 7.2 grams

14. a long aluminum wire of diameter 3 mm is extruded at a temperature of 370°c. the wire is subjected to cross air flow at 30°c at a velocity of 6 m/s. determine the rate of heat transfer from the wire to the air per meter length when it is first exposed to the air.

answer:

Explanation:

Given:

Diameter of aluminum wire, D = 3mm

Temperature of aluminum wire, T_{s}=280^{o}CT

=280

Temperature of air, T_{\infinity}=20^{o}CT

\infinity

=20

Velocity of air flow V=5.5m/sV=5.5m/s

The film temperature is determined as:

\begin{gathered}T_{f}=\frac{T_{s}-T_{\infinity}}{2}\\\\=\frac{280-20}{2}\\\\=150^{o}C\end{gathered}

−T

\infinity

280−20

=150

from the table, properties of air at 1 atm pressure

At T_{f}=150^{o}CT

=150

Thermal conductivity, K = 0.03443 W/m^oCK=0.03443W/m

C ; kinematic viscosity v=2.860 \times 10^{-5} m^2/sv=2.860×10

−5

/s ; Prandtl number Pr=0.70275Pr=0.70275

The reynolds number for the flow is determined as:

\begin{gathered}Re=\frac{VD}{v}\\\\=\frac{5.5 \times(3\times10^{-3})}{2.86\times10^{-5}}\\\\=576.92\end{gathered}

Re=

2.86×10

−5

5.5×(3×10

−3

)

=576.92

sice the obtained reynolds number is less than 2\times10^52×10

, the flow is said to be laminar.

The nusselt number is determined from the relation given by:

Nu_{cyl}= 0.3 + \frac{0.62Re^{0.5}Pr^{\frac{1}{3}}}{[1+(\frac{0.4}{Pr})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{Re}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}Nu

cyl

=0.3+

[1+(

0.4

)

]

0.62Re

0.5

[1+(

282000

)

]

\begin{gathered}Nu_{cyl}= 0.3 + \frac{0.62(576.92)^{0.5}(0.70275)^{\frac{1}{3}}}{[1+(\frac{0.4}{(0.70275)})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{576.92}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}\\\\=12.11\end{gathered}

cyl

=0.3+

[1+(

(0.70275)

0.4

)

]

0.62(576.92)

0.5

(0.70275)

[1+(

282000

576.92

)

]

=12.11

The covective heat transfer coefficient is given by:

Nu_{cyl}=\frac{hD}{k}Nu

cyl

Rewrite and solve for hh

\begin{gathered}h=\frac{Nu_{cyl}\timesk}{D}\\\\=\frac{12.11\times0.03443}{3\times10^{-3}}\\\\=138.98 W/m^{2}.K\end{gathered}

cyl

\timesk

3×10

−3

12.11×0.03443

=138.98W/m

The rate of heat transfer from the wire to the air per meter length is determined from the equation is given by:

\begin{gathered}Q=hA_{s}(T_{s}-T{\infin})\\\\=h\times(\pi\timesDL)\times(T_{s}-T{\infinity})\\\\=138.92\times(\pi\times3\times10^{-3}\times1)\times(280-20)\\\\=340.42W/m\end{gathered}

Q=hA

−T∞)

=h×(π\timesDL)×(T

−T\infinity)

=138.92×(π×3×10

−3

×1)×(280−20)

=340.42W/m

The rate of heat transfer from the wire to the air per meter length is Q=340.42W/mQ=340.42W/m

Penjelasan:

15. A wire made of an unknown substance has a resistance of 125 mω. the wire has a length of 1.8 m and a cross-sectional area of 2.35×10 -5 m2 .what is the resistivity of the substance from which the wire is made? give your answer in scientific notation to one decimal place.

Jawaban:

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Penjelasan:

ghhvhhhhh hh

16. 5. (a) A copper ball has a mass of 1 kg. Calculate the radius of the ball, given that the density of copper is 8900 kg m^-3tolong ya kak makasih..

Jawaban:

3 x 10^-2 m

Penjelasan:

Massa = 1 kg

Massa Jenis = 8900 kg/m³

Massa Jenis = Massa / Volume

8900 kg/m³ = 1 kg/Volume

Volume = 1/8900 m³

V = 4/3 πr³

1/8900 = 4/3 x π x r³

r³ = 3/4π x 8900

r = 0,03 m = 3 x 10^-2 m

17. 0 c) 75 Fahrenheit = d) 60° Reamur renheit 3. A 50cm long copper rod has an initial temperature of 30°C. The coefficient of expansion of copper is 1.8 105FC. The length of copper at 50°C is ... cm. Plissa tolongg yang tauu jawab yaa

75 Fahrenheit = 24,1°C

60° Reamur = 56,7°C

A 50cm long copper rod has an initial temperature of 30°C. The coefficient of expansion of copper is 1.8*10^-5 FC. The length of copper at 50°C is 50 + (50 x 1.8 x 10^-5 x 50) = 50 + 0,9 = 50,9 cm.

18. John cuts a coil of wire into two pieces in the ratio of 5 : 1. The length of the coil of wire is 180 cm. How long is the long wire?

150 cm

Penjelasan dengan langkah-langkah:

the length of the long wire is

5/(1+5) x 180 cm = 5/6 x 180 cm

= 150 cm

19. Wich wire allows a stronger current to pass through the short wire or the long wire

Long wire have bigger resistance which means less current to go to, shorter wire have smaller resistance. Remember the ohms law I = V/R.
Bigger R means lesser current.

20. rido has wire 10 m long. the wire will be made cuboid skeleton with dimension 9cm×7cm×4cm. the number of cuboid skeletons can be made is

Keliling balok = 4 (9 + 7 + 4)
= 4 (20)
= 80 cm

10 m = 1000 cm

1000 / 80 = 12 balok dan sisa kawat 4 cm

Roving cuboid = 4 (9 + 7 + 4) = 4 (20) = 80 cm
10 m = 1000 cm
1000/80 = 12 cuboids and the rest of the wire 4 cm

smoga membantu.. tandai sbg jawaban terbaik ya

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