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3 years ago

I NEED HELP ASAP!!!!!!!!!!!! b) Contour lines represent areas of _________elevation

Physics

1 answer:

jeka943 years ago

6 0

Answer:

Contour line, a line on a map representing an imaginary line on the land surface, all points of which are at the same elevation above a datum plane, usually mean sea level. The diagram illustrates how contour lines show relief by joining points of equal elevation. Encyclopædia Britannica, Inc. Video Player is loading.

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The contribution of Tycho Brahe was primarily his

mihalych1998 [28]

Tycho Brahe ( 1546 - 1601 ) was a Danish astronomer known for his accurate astronomical and planetary observations. Tycho tried to produce a model with the best of both Ptolemy ( earth-centered solar system ) and Copernicus ( sun-centered solar system ).
Answer: B ) observation.

3 0

3 years ago

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 57.3 c

Mashcka [7]

Answer with Explanation:

We are given that

Diameter of pipe, $d_1=0.573 m$

$v_1=13.5 m/s$

$v_2=5.83 m/s$

Volume flow rate of the petroleum along the pipe= $Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})$

$Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s$

By equation of continuity

$A_1v_1=A_2v_2$

$\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2$

$d^2_2=\frac{v_1}{v_2}d^2_1$

$d_2=\sqrt{\frac{v_1}{v_2}}d_1$

$d_2=0.573\sqrt{\frac{13.5}{5.83}}$

$d_2=0.87 m$

$d_2=0.87\time 100=87 cm$

1 m=100 cm

4 0

3 years ago

The latent heat of fusion for Aluminium is 3.97 x 105. How much energy would be required to melt 0.75 kg of it?

RoseWind [281]

Answer:

$E = 2.9775\times10^5 J$

Explanation:

Given: The latent heat of fusion for Aluminum is $L = 3.97\times10^5 J/Kg$

mass to be malted m = 0.75 Kg

Energy require to melt E = mL

$E = 3.97\times10^5\times0.75 = 2.9775\times10^5 J$

Therefore, energy required to melt 0.75 Kg aluminum

$E = 2.9775\times10^5 J$

5 0

3 years ago

a 150 N force is used to pull a wooden box across a wooden surface at a constant velocity. what is the mass of the box?

IRINA_888 [86]

Answer:

The mass of the box:

m = 60 kg

Explanation:

Given:

F = 150 N

g = 10 m/s²

_________

m - ?

Coefficient of friction wood on wood:

μ = 0.25

Friction force:

F₁ = μ*m*g

Newton's Third Law:

F = F₁

F = μ*m*g

The mass of the box:

m = F / ( μ*g) = 150 / (0.25*10) = 60 kg

7 0

1 year ago

Hey guys, i'm at 10 grade at school i have a physics question about stress tension and young modulus ? Please this is my questio

makvit [3.9K]

Use the eq. of Young modulus Y=(F/A)/(∆l/lo)
dimana ∆l is the elongation of wire, lo is its initial length.
So ∆l = (F/A)lo/Y.
∆l = (1000N/(6.5 × 10^-7 m^2))×(2.5m)/(2.0 × 10^-11 N/m^2)
Use calculator to finish it.

5 0

4 years ago