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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.

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3 years ago

You place a cup of 210 degrees F coffee on a table in a room that is 68 degrees F, and 10 minutes later, it is 200 degrees F. Ap

Inga [223]

The answer is 35 minutes

The Newton's law of cooling is:
T(x) = Ta + (To - Ta)e⁻ⁿˣ

T(x) - the temperature of the coffee at time x
Ta - the ambient temperature
To - the initial temperature
n - constant

step 1. Calculate constant k:

We have:
T(x) = 200°F
x = 10 min
Ta = 68°F
To = 210°F
n = ?

T(x) = Ta + (To - Ta)e⁻ⁿˣ
200 = 68 + (210 - 68)e⁻ⁿ*¹⁰
200 = 68 + 142 * e⁻¹⁰ⁿ
200 - 68 = 142 * e⁻¹⁰ⁿ
132 = 142 * e⁻¹⁰ⁿ
e⁻¹⁰ⁿ = 132/142
e⁻¹⁰ⁿ = 0.93

Logarithm both sides with natural logarithm:
ln(e⁻¹⁰ⁿ) = ln(0.93)
-10n * ln(e) = -0.07
-10n * 1 = - 0.07
-10n = -0.07
n = -0.07 / - 10
n = 0.007

Step 2. Calculate time x when T(x) = 180°F:
We have:
T(x) = 180°F
x = ?
Ta = 68°F
To = 210°F
n = 0.007

T(x) = Ta + (To - Ta)e⁻ⁿˣ
180 = 68 + (210 - 68)e⁻⁰.⁰⁰⁷*ˣ
180 - 68 = 142 * e⁻⁰.⁰⁰⁷*ˣ
112 = 142 * e⁻⁰.⁰⁰⁷⁾*ˣ
e⁻⁰.⁰⁰⁷*ˣ = 112/142
e⁻⁰.⁰⁰⁷*ˣ = 0.79

Logarithm both sides with natural logarithm:
ln(e⁻⁰.⁰⁰⁷*ˣ) = ln(0.79)
-0.007x * ln(e) = -0.24
-0.007x * 1 = -0.24
-0.007x = -0.24
x = -0.24 / -0.007
x ≈ 35

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3 years ago

Read 2 more answers

A conducting sphere with a radius of 0.25 m has a total charge of 5.90 mC. A particle with a charge of −1.70 mC is initially 0.3

notsponge [240]

Explanation:

The given data is as follows.

$r_{1}$ = 0.25 m, q = 5.90 mC = $5.90 \times 10^{-3} C$

$r_{2}$ = 0.35 m, q = 1.70 mC = $1.70 \times 10^{-3} C$

(a) Now, we will calculate the electric potential as follows.

V = $k \frac{q}{r}$

First, we will calculate the initial and final electric potential as follows.

$V_{i} = 9 \times 10^{9} \times \frac{5.90 \times 10^{-3}}{0.25 m}$

= $212.4 \times 10^{6}$

or, = $2.124 \times 10^{8}$

$V_{f} = 9 \times 10^{9} \times \frac{1.70 \times 10^{-3}}{0.35 m}$

= $43.71 \times 10^{6}$

or, = $4.371 \times 10^{8}$

Hence, the value of change in electric potential is as follows.

$\Delta V = V_{f} - V_{i}$

= $4.371 \times 10^{8} - 2.124 \times 10^{8}$

= $2.247 \times 10^{8}$ V

Therefore, the difference in electric potential energy is $2.247 \times 10^{8}$ V.

(b) Now, we will calculate the potential energy as follows.

P.E = qV

= $-1.70 \times 10^{-3}C \times 2.247 \times 10^{8}$ V

= $-3.8199 \times 10^{5}$

Therefore, the change in the system's electric potential energy is $-3.8199 \times 10^{5}$ .

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3 years ago

if the force moving an object points at least partially in the opposite direction of the objects motion, the work is considered

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If the force moving an object points at least partially in the opposite direction of the objects motion, the work is considered to be negative.

Explanation:

Work done is the measure of displacements caused on an object by the amount of force acting on it. The work done can be positive or negative. If the direction of force acting on an object is along the direction of motion of the object then the work done is positive.

This is because, in this case the external force is helping the object to continue its state of motion. So, if the force acting on the object is in completely opposite or partially opposite direction to the direction of motion of the object, it will stop or slow down the motion of the object.

This kind of force will have negative direction to the motion of the object and hence the work done will also be negative.

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