All categories

3 years ago

In an experiment the mass of a calorimeter is 36.35 g . Express in micrometer ,millimetre and kg.

Physics

1 answer:

Andru [333]3 years ago

4 0

Answer:

1. 36.35 g = 36.35E15 micrometer.

II. 36.35 g = 363.5 millimetre.

III. 36.35 g = 0.03635 kilogram.

Explanation:

Given the following data;

Mass of calorimeter = 36.35 grams

To convert the mass in grams (g) to;

I. Micrometer

Conversion:

1 g = 1 exp 15 um

36.35 g = X um

Cross-multiplying, we have;

X = 36.35 * 1 exp 15 = 36.35 exp 15 um

36.35 g = 36.35E15 micrometer

II. Millimetre

Conversion:

1 g = 1 milliliter

36.35 g = X milliliter

Cross-multiplying, we have;

X = 36.35 * 1 = 36.35 milliliter

Next, we would convert milliliter to millimetre;

1 milliliter = 10 millimetre

36.35 milliliter = X millimetre

Cross-multiplying, we have;

X = 36.35 * 10 = 363.5 millimetre

36.35 g = 363.5 millimetre

III. Kilogram

Conversion:

1000 grams = 1 kilogram

36.35 g = X kilogram

Cross-multiplying, we have;

X * 1000 = 36.35 * 1

Dividing both sides by 1000, we have;

X = 36.35/1000 = 0.03635 kilogram

36.35 g = 0.03635 kilogram

Note:

g is the symbol for grams.
Exp (E) means exponential = 10
um is the symbol for micrometer.

You might be interested in

complete the following equation that represents the complete combustion of hexane in a car engine.C6H14 + O2

mina [271]

The balanced combustion reaction for butane is;

2C4H10 + 13O2 → 8CO2 + 10H2O

Molar mass of water = 18.02 g/mol

Molar mass of oxygen = 32 g/mol

2.46g H2O(1 mol H2O/18.02g)(13 mol O2/10 mol H2O)(32g O2/1 mol O2) = 5.68g O2

5 0

3 years ago

At a certain place, Earth's magnetic field has magnitude B =0.703 gauss and is inclined downward at an angle of 75.4° to the hor

Irina18 [472]

Answer:

The charge flows in coulombs is

$dq=1.843x10^{-5}C$

Explanation:

The current magnitude of current is given by the resistance and the induced Emf as:

$I=N*\frac{dF}{Rdt}$

$\frac{dq}{dt}=\frac{dF}{Rdt}=dq=N*\frac{dF}{R}$

$dq=\frac{N*\beta*A*(Cos(\alpha_f)-Cos(\alpha_i}{R}$

$N=1300$ , $\beta=0.703$ , $A=\pi*r^2=\pi*0.10^2=0.01\pi m^2$ , $R=99.4+202=301.4$ Ω

$\alpha_f=14.6$ , $\alpha_i=165.4$

Replacing :

$dq=\frac{1300*0.703x10^{-4}*0.01\pi*(0.9667-(-0.9667))}{202+99.4}$

$dq=1.843x10^{-5}C$

5 0

3 years ago

A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along

storchak [24]

Answer:

$v = \frac{kQ}{a}$

Explanation:

We define the linear density of charge as:

$\lambda = \frac{Q}{L}$

Where L is the rod's length, in this case the semicircle's length L = πr

The potential created at the center by an differential element of charge is:

$dv = \frac{kdq}{r}$

where k is the coulomb's constant

r is the distance from dq to center of the circle

Thus.

$v = \int_{}^{}\frac{kdq}{a}$

$v = \frac{k}{a}\int_{}^{}dq$

$v = \frac{kQ}{a}$ Potential at the center of the semicircle

4 0

2 years ago

If the Earth was the size of a basketball, about 24 centimeters in diameter, about how thick is the Earth's crust?

zalisa [80]

C)yea there you go hope that helped

4 0

3 years ago

Which energy sources can transfer thermal energy to a heat pump? Check all that apply.

choli [55]

Answer:1.Air, 2.the ground,3.water

Explanation:I just got it right:)

3 0

3 years ago