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Studentka2010 [4]
3 years ago
14

The piece of wood shown above has a mass of 20 grams. Calculate its volume and density. Then

Physics
1 answer:
kozerog [31]3 years ago
4 0

We have that The Density and Volume is given as

\rho=0.67\\\\\V=30cm^3

The factors that affect Density are Mass and Volume

From the question we are told that

The piece of wood shown above has a mass of 20 grams.

Wood parameters

2cm

5cm

3cm

Generally the equation for the Volume  is mathematically given as

V=L*B*H\\\\\V=2*5*3\\\\V=30cm^3

Generally the equation for the Density is mathematically given as

\rho=\frac{m}{v}\\\\\\rho=\frac{20}{30}\\\\\rho=0.67

Therefore the factors that affect Density are Mass and Volume

For more information on this visit

brainly.com/question/19694949?referrer=searchResults

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A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p
Semenov [28]

Answer:

f_n=3.75N

Explanation:

From the question we are told that:

Frictional force F=0.150N

Coefficient of kinetic friction \mu=0.04

Generally the equation for Normal for is mathematically given by

 f_n=\frac{F}{\mu}

Therefore

 f_n=\frac{0.150}{0.04}

 f_n=3.75N

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3 years ago
When using a spring scale, the measurements you obtain will be in ____.
SpyIntel [72]
The answer is letter c

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a typical cmall flashlight contains two batteries each having na emf of 2.0 v connected in series with a bulb havin ga resistanc
Helen [10]

Answer:

P = 0.25 W

Explanation:

Given that,

The emf of the battry, E = 2 V

The resistance of a bulb, R = 16 ohms

We need to find the power delivered to the bulb. We know that, the formula for the power delivered is given by :

P=\dfrac{V^2}{R}\\\\P=\dfrac{2^2}{16}\\\\=0.25\ W

So, 0.25 W power is delivered to the bulb.

5 0
3 years ago
What trend does the reactivity of nonmetals show in a periodic table? random changes without any trends on the periodic table ch
Naddik [55]

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A narrow beam of light from a laser travels through air (n = 1.00) and strikes the surface of the water (n = 1.33) in a lake at
Natalka [10]

Answer:

A) d = 11.8m

B) d = 4.293 m

Explanation:

A) We are told that the angle of incidence;θ_i = 70°.

Now, if refraction doesn't occur, the angle of the light continues to be 70° in the water relative to the normal. Thus;

tan 70° = d/4.3m

Where d is the distance from point B at which the laser beam would strike the lakebottom.

So,d = 4.3*tan70

d = 11.8m

B) Since the light is moving from air (n1=1.00) to water (n2=1.33), we can use Snell's law to find the angle of refraction(θ_r)

So,

n1*sinθ_i = n2*sinθ_r

Thus; sinθ_r = (n1*sinθ_i)/n2

sinθ_r = (1 * sin70)/1.33

sinθ_r = 0.7065

θ_r = sin^(-1)0.7065

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Thus; xonsidering refraction, distance from point B at which the laser beam strikes the lake-bottom is calculated from;

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d = 4.293 m

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