A 100 mm cube of a substance weighs 8 n. what is density and relative density
1. A 100 mm cube of a substance weighs 8 n. what is density and relative density
Jawaban:
800 kg/m³
Penjelasan:
Rumus dari massa jenis adalah:
ρ = m / V
Diket:
Panjang = 100mm = 0,1 m
Volume kubus = r³ = 0.001 m³
berat = 8 newton,
massa = berat/percepatan gravitasi
= 8/10 = 0,8 kg
massa jenis = 0,8 / 0.001
= 800 kg/m³
Jadikan jawaban terbaik yaa.
2. 1. A cube has length of 7 cm a width of 8 cm and a height of 2 cm. What is its volume? 2. The density if aluminum is 2.7 g/ml. If the aluminum has a volume of 20 ml, what is the mass of the aluminum? 3. A marble has a volume of 20 ml. It also has a mass of 10g. Calculate the density of the marble. Tolong sertakan cara mengerjakannya
1. its volume = 7 × 8 × 2
= 112 cm
2.mass = the density × volume
= 2,7 g / ml × 20 ml
= 54 gram
3.the density= mass / volume
= 10 / 20
= 0,5 g / ml
semoga bermanfaat
3. the density of ball weighing 0,5 kg and diameter 10 cm is....
We know if radii = 1/2 dia
r = 10/2
r = 5 cm
The ball volume formula is 4/3 phi r r r
= 4 3,14 5 5 5 / 3
= 523,59877 cm3, make it simple
= 523,6 cm3
Then I change 5 kg to 5000 gram (its equal, of course).
Density or Rho in greek (similar with letter p, lets use p) formula is mass/volume
p = 5000 g / 523,6 cm3
p = 9,54927 gram / cm3 (make it easy as pie)
p = 9,55 gram / cm3
4. 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
5. Two moles of aluminum consist of...... atoms (L = 6.02 x 10^23)
x = n x L
x = 2 x 6,02 x 10^23
x = 12,02 x 10^23
x = 1,202 x 10^24
6. Diketahui luas sebuah bola logam 2,0096 m2 , massanya 2000 kg a. Berapa densitynya ? b. Berapa Relative Density (RD) nya ?
Jawaban:
a. Density = Massa / Luas = 2000 kg / 2,0096 m2 = 993,9 kg/m3
b. Relative Density (RD) = Density benda / Density air = 993,9 kg/m3 / 1000 kg/m3 = 0,9939
7. unit terbesar dan terkecil density( kg/m3)
Jawaban:
The density (lebih tepatnya, kepadatan massa volumetrik , juga dikenal sebagai massa jenis ), suatu zat adalah yang massa per satuan volume yang . Simbol yang paling sering digunakan untuk kepadatan adalah ρ (huruf Yunani kecil rho ), meskipun huruf Latin D juga dapat digunakan. Secara matematis, massa jenis didefinisikan sebagai massa dibagi volume: (1)
maaf kalau salah
8. A metal hemispherical bowl has an external diameter of 50.8 cm and thickness of 2.54cm.i) given that the empty bowl weights 97.9 kg, find the density, in kg/m cube, of the metal which the bowl is made of.ii) if the bowl is completely filled with a liquid of density 3 1.75kg/m cube, find the mass of liquid in grams.
Jawaban:
A metal hemispherical bowl has an external diameter of 50.8 cm and thickness of 2.54cm.
i) given that the empty bowl weights 97.9 kg, find the density, in kg/m cube, of the metal which the bowl is made of.
ii) if the bowl is completely filled with a liquid of density 3 1.75kg/m cube, find the mass of liquid in grams.
9. Berapa fraksi aluminum yang tercelup dalam air raksa ketika batang aluminum mengapung di atas air raksa? Massa jenis aluminium adalah 2700 kg/m
Dik:
[tex] rho_{air raksa} = 13600[/tex] kg/m³
[tex] rho_{al} = 2700[/tex] kg/m³
[tex] W = F_{A} \\ m_{al}.g = rho_{air raksa}.g. V_{tercelup} \\ rho_{al}.V_{total}.g = rho_{air raksa}.g. V_{tercelup} \\ rho_{al}.V_{total} = rho_{air raksa}. V_{tercelup} \\ V_{tercelup} = \frac{rho_{al}}{rho_{air raksa}}.V_{total} \\ V_{tercelup} = \frac{2700}{13600}.V_{total} \\ V_{tercelup} = 0,1985 V_{total}[/tex]
10. a stone of density 3500 kg/m3 if The Rock Fragments Has Volumes Of 700 cm3 Then A Mass Of Stone Is.... A. 20 Kg B. 200 Kg C. 2.450 G D. 2,450 KG
Jawaban:
sebuah batu massa jenisnya 3500 kg/m3 jika pecahan batu tersebut volumenya 700 cm3 maka massa batu tersebut adalah....
A.20 Kg
B.200 Kg
C. 2.450 G
D. 2.450 KG
Penjelasan:
wee tukang translate
11. A bar of gold has a mass of 3,8 kg. If the density of gold is 19 g/cm3, what is the volume of the gold?
Penjelasan:
so, the bar of gold has a mass : 3,8 kg ⇒ 3800 g
and the density of gold is : 19 g/cm³
this is the formula : Volume = mass / density
V = 3800 g / 19 g/cm³
V = 200 cm³
the answer is : V = 200 cm³
12. A metal has a density of 750 kg / m3. If the mass is 30 kg. Calculate the volume of the metal
Penjelasan:
HgvjbfvkfwfjudnjgbjhvbjutaadbkkJawaban:
0,04 m³
Penjelasan:
Diket :
Ρ= 750 kg / m3
m = 30 kg.
Dit :
v
Jawab:
p = m/v
750 = 30/v
750v = 30
v = 0,04 m³
Semoga membantu : )
13. There is 20 kg of aluminium with the temperature is 100°C. Then the aluminium temperature increase to 110°C. What is the quantity of thermal energy needed to raise the temperature of the aluminum if specific heat capacity of aluminum is 900 J/kg°C? tolong sertakan caranya jg
Jawaban:
m: 20 kg
initial temperature: 100⁰C
final temperature: 110⁰C
heat capacity: 900 J/kg ⁰C
quantity of thermal energy?
Q=m.c. T
=20.900.(110-100)
=18.000(10)
=180.000 Joule
14. a block has density of 2500 kg/m^3 and when in air it weight 25 newton. determine the weight of the block in water
Jawaban:
15 Newton
Penjelasan:
[tex]w = m × g \\
25 = m × 10 \\
m = 2.5 \: kg \\
ρ = \frac{m}{v} \\ 2500 = \frac{2.5}{v} \\ v = {10}^{ - 3} \: {m}^{3}[/tex]
[tex]Fa = \: ρf \times g \times v \: \\ Fa = \: {10}^{3} \times 10 \times {10}^{ - 3} = 10 \: N[/tex]
[tex]w'=w - Fa =25 - 10 = 15 \: N[/tex]
Fa = Archimedes Force
ρf = fluid density (water ρf = 1 000 kg/m³)
w' = weight of the block in water
15. 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
16. debatang aluminum memiliki kalor Jenis 900j/kg°c, jika massa aluminum tersebut 3 Kg tentukan kalor yang diperlukan untuk menaikan suhu 40° menjadi 80°
Jawaban:
252.000 joule
Penjelasan:
Berapa besar kalor yang diperlukan untuk menaikan suhu sebatang alumunium yang massanya 7 kg dan 10°C menjadi 50°C, jika kalor jenis alumunium 900 J/kg°C?
Diketahui:
Massa aluminium (m) = 7 kg
Kalor jenis aluminium (c) = 900 J/kg°C
ΔT (perubahan suhu) = T2 - T1
= 50°C - 10°C
= 40°C
Ditanya: Q (kalor)
Penyelesaian soal:
Q = m x c x ΔT
Q = 7 kg x 900 J/kg°C x 40°C
Q = 252.000 Joule
Berarti, aluminium tersebut membutuhkan kalor sebesar 252.000 Joule
JADIKAN JAWABAN TERBAIK MU YAA
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17. 6.5 an aluminum bar 125 mm (5.0 in.) long and having a square cross section 16.5 mm (0.65 in.) on an edge is pulled in tension with a load of 66,700 n (15,000 lb ), and experiences an elongation of 0.43 mm ( in.). as- suming that the deformation is entirely elas- tic, calculate the modulus of elasticity of the aluminum.
Panjang, L = 125 mm
Luas permukaan, A = 16,5 mm²
Gaya, F = 66.700 N
Pertambahan panjang, ∆L = 0,43 mm
Modulus Elastisitas
M = F.L/A.∆L
M = (66.700 x 125)/(16,5 x 0,43)
M = 8.337.500/7,095
M = 1,12 x 10⁶ N/mm²
M = 1,12 x 10¹² N/m
18. a tin have density 1.850 kg/m³ and volume 8 m³. mass of a tin is ...
mass of a tin is
1.850 : 8=
19. Aluminum bermassa 2 kg dengan suhu 20°C dipanaskan hingga suhu 120°C. Jika kalor jenis aluminum 900 j/kg°C, maka jumlah kalor yg di perlukan adalah..
semoga benar dan menjadi jawaban terbaik.semoga tdk salah
20. Berapakah kalor yang diperlukan untuk melebur 3 kg aluminum jika kalor lebur alumunium , 403.000 j/kg
Diketahui:
m = 3 kg
L = 403.000 J/kg
Ditanya:
Q = ...?
Jawab:
Q = m · L
= 3 × 403.000
= 1.209.000Joule= 1,209MegaJoule
Jadi, kalor yang diperlukan untuk meleburkan aluminium tersebut adalah 1,209MegaJoule.
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