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3 years ago

A student sects a leaf of length 7.2 cm to draw. Her drawing is 28.8 cm in length. What is the magnification of the drawing?

Physics

1 answer:

Alina [70]3 years ago

7 0

Answer:

A) x4

Explanation:

Magnification is equal to image size divided by the actual size, or M = I/A.

The image size is the student's drawing, which is 28.8 cm, and the actual size is 7.2 cm. Divide them, and cancel out the units, and you should get:

28.8 cm/7.2 cm = 4

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When you measure the boiling point of mercury, you are investigating a ___. a.chemical change b.chemical property c.physical cha

Amanda [17]

D. physical property

the bonds between molecules of mercury are breaking so it's physical and it's not changing the chemical composition of the substance

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3 years ago

For the following elementary reaction 2br• -> br2-. The rate of consumption of the reaction and the rate of formation of prod

Scorpion4ik [409]

Answer: $-\frac{1}{2}\times \frac{d[Br^.]}{dt}=+\frac{d[Br_2]}{dt}$

Explanation:

Rate of a reaction is defined as the rate of change of concentration per unit time.

Thus for reaction:

$2Br^.\rightarrow Br_2$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

$Rate=-\frac{d[Br^.]}{2dt}$

or $Rate=+\frac{d[Br_2]}{dt}$

Thus $-\frac{d[Br^.]}{2dt}=+\frac{d[Br_2]}{dt}$

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3 years ago

A reasonable estimate of the moment of inertia of an ice skater spinning with her arms at her sides can be made by modeling most

Oxana [17]

Answer:

A) $I_{total}$ = 1.44 kg m², B) moment of inertia must increase

Explanation:

The moment of inertia is defined by

I = ∫ r² dm

For figures with symmetry it is tabulated, in the case of a cylinder the moment of inertia with respect to a vertical axis is

I = ½ m R²

A very useful theorem is the parallel axis theorem that states that the moment of inertia with respect to another axis parallel to the center of mass is

I = $I_{cm}$ + m D²

Let's apply these equations to our case

The moment of inertia is a scalar quantity, so we can add the moment of inertia of the body and both arms

$I_{total}$ = $I_{body}$ + 2 $I_{arm}$

$I_{body}$ = ½ M R²

The total mass is 64 kg, 1/8 corresponds to the arms and the rest to the body

M = 7/8 m total

M = 7/8 64

M = 56 kg

The mass of the arms is

m’= 1/8 m total

m’= 1/8 64

m’= 8 kg

As it has two arms the mass of each arm is half

m = ½ m ’

m = 4 kg

The arms are very thin, we will approximate them as a particle

$I_{arm}$ = M D²

Let's write the equation

$I_{total}$ = ½ M R² + 2 (m D²)

Let's calculate

$I_{total}$ = ½ 56 0.20² + 2 4 0.20²

$I_{total}$ = 1.12 + 0.32

$I_{total}$ = 1.44 kg m²

b) if you separate the arms from the body, the distance D increases quadratically, so the moment of inertia must increase

6 0

3 years ago

How much energy (in kilojoules) is required to convert 200 mL of diethyl ether at its boiling point from liquid to vapor if its

LiRa [457]

Answer:

55.96kJ

Explanation:

Energy = mass of diethyl ether × enthalpy of vaporization of diethyl ether

Volume (v) = 200mL, density (d) = 0.7138g/mL

Mass = d × v = 0.7138 × 200 = 142.76g

Enthalpy of vaporization of diethyl ether = 29kJ/mol

MW of diethyl ether (C2H5)2O = 74g/mol

Enthalpy in kJ/g = 29kJ/mol ÷ 74g/mol = 0.392kJ/g

Energy = 142.76g × 0.392kJ/g = 55.96kJ

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Luden [163]

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3 years ago