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?
1. 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
2. Today used to measure the weight of gemstones or the amount of gold per 24 parts of pure gold, ……… originally the weight of a seed of the carob tree.
Jawaban:
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3. calculate the volume of the prism
Jawaban:
artinya; hitunglah volume prisma
Penjelasan:
rumus untuk mencari volume prisma adalah:
V=luas alas × tinggi.V=1/2×(panjang×lebar×tinggi)Baca juga: Cara Menghitung Luas Permukaan Prisma.V=s³V=panjang×lebar×tinggi.V=(1/2×5×sisi×apotema)×tinggi prisma.Jawaban:
Prism volume formula/rumus volume Prisma
↓
Volume=Luas alas×tinggi
Large/luas=(2 x ((alas x tinggi) : 2)) + (keliling alas x tinggi prisma).
4. A 2.00-g sample of a large biomolecule was dissolved in 15.0 g carbon tetrachloride. The boiling point of this solution was determined to be 77.85 oC. Calculate the molar mass of the biomolecule. For carbon tetrachloride, the boiling-point constant is 5.03oC .kg/mol, and the boiling point of pure carbon tetrachloride is 76.50oC
we will use boiling point formula:
ΔT = i Kb m
when ΔT is the temperature change from the pure solvent's boiling point to the boiling point of the solution = 77.85 °C - 76.5 °C = 1.35
and Kb is the boiling point constant =5.03
and m = molality
i = vant's Hoff factor
so by substitution, we can get the molality:
1.35 = 1 * 5.03 * m
∴ m = 0.27
when molality = moles / mass Kg
0.27 = moles / 0.015Kg
∴ moles = 0.00405 moles
∴ The molar mass = mass / moles
= 2 g / 0.00405 moles
= 493.8 g /mol
5. A cylinder is of radius 10 cm and the heigth of cylinder is 21 cm calculate the volume of the cylinder
Jawab:
6594 cm^3
Penjelasan dengan langkah-langkah:
vol cylinder =
Jawaban:
6.600
Penjelasan dengan langkah-langkah:
Volume
phi × r² × height
22/7 × 10 × 10 × 21
= 6600
semoga bermanfaat
6. Today used to measure the weight of gemstones or the amount of gold per 24 parts of pure gold, ……… originally the weight of a seed of the carob tree.(a)was a carat(b)a carat was(c)which was a carat(d)that a carat was
Jawaban:
(a)was a carat
maap kalo salah
7. the volume of a cube is 343cm³.calculate the total surface area,in cm²,of the cube
[tex] \sqrt[3]{343} = 7[/tex]
[tex]6 \times {7}^{2} = [/tex]
[tex]294 \: cm^{2} [/tex]
8. The density of steel is 7850 kgm Calculate themass of a steel sphere of radius 0.15 m. (Firstcalculate the volume of the sphere using theformula V = $rt and then use the density equation.)
Penjelasan:
[tex]V sphere = \frac{4}{3} \pi {r}^{3} \\ = \frac{4}{3} \times 3.14 \times {(0.15)}^{3} = 0.01413 \: {m}^{3} [/tex]
[tex]ρ = \frac{m}{V} \\ 7850= \frac{m}{0.01413} \\ m = 7850 \times 0.01413 \\ = 110.9205 \: kg[/tex]
9. calculate the perimeter of a semicircle of diameter 8cm
Aku jawab pake bahasa Indonesia ya ^^
Jari jari = 8:2 = 4cm
Keliling lingkaran : 2xpi x r
—> 2 x 3,14 x 4cm
—>25,12cm
10. calculate the length of the simple pendulum having time period of 3 seconds (g=9.8)
Penyelesaian:
[tex] \displaystyle (T)^2 = \left(2\pi \sqrt{\frac{l}{g}}\right)^2 \\ T^2 = 4\pi^2 \frac{l}{g} \\ l = \frac{T^2\cdot g}{4\pi^2}[/tex]
Tinggal masukkan angka-angkanya
[tex] \displaystyle l \approx \frac{(3)^2\cdot (9.8)}{4(3.14)^2} \\ l \approx 2.2364 \; m [/tex]
Jawaban:
[tex] \displaystyle \boxed{\bold{l \approx 2.2364 \; m }}[/tex]
11. A cylinder has a height of 15cm and a volume of 500cm. Calculate the radius of the cylinder.
Jawaban:
10.6
Penjelasan dengan langkah-langkah:
diketahui:
t = 15 cm
v = 500 cm
ditanya:
r= ?...
rumus:
[tex] {r}^{2} = \frac{v}{\pi.t} [/tex]
[tex] = \frac{500}{3.14 \times 15} [/tex]
[tex] = \frac{500}{47.1} [/tex]
[tex] = 10.6[/tex]
sorry if this answer doesn't help
12. The density of carbon tethacholride is 1.59 g/ml. Calculate the volume of 26.4 grams of the liquid.tolong dijawab besok dikumpulkan
V= m/d
= 26,4/1,59
= 16,604 ml3 ( ml pangkat 3 )
13. Calculate the difference (selisih) between the volume of the following!
penyelesaian :
= p × l × t - p × l × t
= 3 × 3 × 6 - 2 × 3 × 2
= 54 - 12
= 42 cm³
Jawaban dan langkah langkah penyelesaian ada di dalam lampiran ya semoga membantu.
Jangan lupa apabila membantu jadikan jawaban tercerdas ya terimakasih ^_^
14. calculate the volume of the space above Please answer!!!!!
Penjelasan dengan langkah-langkah:
Volume bangun 1 : Kubus : S×S×S
=10×10×10
=100×10
=1.000 cm³
volume bangun 2 : Balok : p×l×t
=20×12×10
=240×10
=2.400
V.gabungan = 1.000 + 2.400 =3.400 Cm³
15. 2. A sample of strontium exists as a mixture of four isotopes. Information about three of these isotopes is given in the table.(15 points) Mass number 86 87 88 Abundance 9.86% 7.00% 82.58% a. Calculate the abundance of the fourth isotope. abundance % b. The relative atomic mass of this sample of strontium is 87.71. Calculate the mass number of the fourth isotope.
Jawaban:
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16. Calculate the KE of the liquid flowing in a pipe 5 cm inner diameter at the rate of 500 kg/min. The density of the liquid is 1.15 g/cm³.
Jawaban:
The kinetic energy (KE) of the liquid flowing in the pipe can be calculated as follows:
First, we need to determine the velocity of the liquid. This can be calculated using the equation:
v = Q / A
where v is the velocity (m/s), Q is the flow rate (m^3/s) and A is the cross-sectional area of the pipe (m^2).
To convert the flow rate from kg/min to m^3/s, we need to multiply the flow rate by the conversion factor of 1/60. Then, we need to convert the density from g/cm³ to kg/m³:
Q = (500 kg/min) * (1/60 s/min) * (1000 g/kg) = 50/6 m^3/s
ρ = 1.15 g/cm³ * (1000 cm³/m³) = 1150 kg/m³
Now, we can find the cross-sectional area of the pipe:
A = (π/4) * d^2
d = 5 cm
A = (π/4) * (5 cm)^2 = 19.63 cm^2
A = (19.63 cm^2) * (0.01 m/cm^2) = 0.1963 m^2
Using the velocity and the cross-sectional area, we can now find the volume flow rate of the liquid:
v = Q / A = 50/6 m^3/s / 0.1963 m^2 = 156.41 m/s
Finally, we can find the kinetic energy of the liquid:
KE = 0.5 * ρ * v^2 * A
KE = 0.5 * (1150 kg/m³) * (156.41 m/s)^2 * (0.1963 m^2) = 174444.44 J/s = 174.44 kJ/s
17. A glass house is in the shape as shown in the figure . calculate total area of glass needed and calculate the volume of the glass house
Jawab:
Total Luas seluruh bangunan = L prisma segitiga + L balok
= 248 m² + 292 m² = 540 m²
Total Volume seluruh bangunan = V prisma segitiga + V balok
= 60 m³ + 336 m³ = 396 m³
Penjelasan dengan langkah-langkah:
Bangunan ini adalah gabungan dari prisma segitiga dan balok.
Diketahui :
Prisma segitiga : alas = 5 m
tinggi alas = 6 m
tinggi prisma = 11 m - 7 m = 4 m
Ditanya : L = ?
V = ?
L = 2 X Luas alas + Keliling alas + tinggi
= 2 X (1/2 X 5X6) + (5+5+6) x 4
= 2 x (15 + 16 x 4)
= 2 x 124 = 248 m²
Volume = Luas alas X tinggi
= (1/2 X 5X6) X 4
= 60 m³
Sekarang kita hitung balok.
Diketahui : p = 8 m
l = 6 m
t = 7 m
Ditanya : L = ?
V = ?
L balok = 2 (pl + pt + lt)
= 2 (8X6 + 8X7 + 6X7)
= 2 (48+56+42)
= 2 x 146 = 292 m²
Volume balok = pxlxt
= 8 X 6 X 7
= 336 m³
Total Luas seluruh bangunan = L prisma segitiga + L balok
= 248 m² + 292 m² = 540 m²
Total Volume seluruh bangunan = V prisma segitiga + V balok
= 60 m³ + 336 m³ = 396 m³
18. a 63.5 sample of an unindentified metal absorb 355 J of heat when it's temperature changed by 4.56°C. calculate the specific heat capacity of the metal
Jawaban:
maaf kalo salah......
................
19. calculate the volume of pyramid
Unknown:
High = 6 cm
Side plinth = 5 cm
Asked:volume of pyramid...?
Answer:V = ⅓ × plinth area × high
V = ⅓ × (s × s) × high
V = ⅓ × (5 cm × 5 cm) × 6 cm
V = ⅓ × 25 cm × 6 cm
V = ⅓ × 150 cm³
V = 150 cm³/3
V = 50cm³
20. A cube with a volume of 27 cm³. Calculate the total surface area of the cube, in cm².
Jawaban:
54cm²
___________________________
To find the surface area of a cube, first find the side lengths
formula :
V = s³
or
[tex] \tt s = \sqrt[3]{V} [/tex]
[tex] \tt s = \sqrt[3]{27} [/tex]
[tex] \tt s = \sqrt[ \cancel3]{ {3}^{ \cancel3} } [/tex]
[tex] \tt s = 3cm[/tex]
___________________________
Surface area :
[tex] \tt S = 6 \times {s}^{2} [/tex]
[tex] \tt S = 6 \times {3}^{2} [/tex]
[tex] \tt S = 6 \times 9[/tex]
[tex] \boxed{\tt \red{S =54 {cm}^{2}} }[/tex]
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