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Lostsunrise [7]
3 years ago
8

Assuming two hypothetical maps that each cover a standard 8.5 by 11-inch sheet of paper, the larger-scale map would cover a larg

er geographic area than the smaller-scale map.
a) true
b) false
Physics
1 answer:
sukhopar [10]3 years ago
8 0

Answer:

<em>b) false</em>

Explanation:

The scale of a map is the ratio of a distance on the map to the corresponding distance on the ground. Scaling allow us to capture a large geographical area on a reduced platform while still retaining the relative sizes and positioning of places on the map to their real life sizes and positioning. If both maps cover a standard 8.5 by 11-inch sheet of paper, then the map with the smaller ratio will have the bigger geographical area.

To understand better, let us assume two geographical areas A and B. A is bigger than B. If we were to put them both on the same area of map paper, then we'll have to scale up the smaller geographical area B so as to fit into the map paper. This means that the geographical area with the smaller area B will have the larger scale on the map.

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A feather is dropped onto the surface of the moon. How far will the feather have fallen if it reaches the surface in 9.00 second
Natali5045456 [20]

The feather's vertical position y is determined by

y=\dfrac12g_{\text{moon}}t^2

We take the feather's starting position to be the origin, and the downward direction to be positive. Then

y=\dfrac12\left(1.63\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(9.00\,\mathrm s\right)^2=66.0\,\mathrm m

so the answer is D.

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3 years ago
I really need help asap.
Semmy [17]
The answer is answer chioce B.

8 0
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if a sea star egg is taken from the ocean and put into Elk Lake would you expect the egg to shrink swell are stay the same ?
mafiozo [28]
It would swell

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4 years ago
A car rolls along a surface with a coefficient of friction of .795 I f the car has a mass of 905 kg, what force is needed to acc
Mumz [18]
102.5 km/hr = 28.5 m/s
a = (28.5 - 0) / 10
a = 2.85 m/s²

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ma = F - μmg
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F = 905(2.85 + 0.795 x 9.81)
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4 0
3 years ago
There are two important isotopes of uranium, 235U
nydimaria [60]

Answer:

a) (vᵣₘₛ₁/vᵣₘₛ₂) = 1.00429

where vᵣₘₛ₁ represents the vᵣₘₛ for 235-UF6 and vᵣₘₛ₂ represents the vᵣₘₛ for 238-UF6.

b) T = 767.34 K

c) The answers point to a difficult seperation technique, as the two compounds of the different isotopes have very close rms speeds and to create a difference of only 1 m/s In their rms speeds would require a high temperature of up to 767.34 K.

Explanation:

The vᵣₘₛ for an atom or molecule is given by

vᵣₘₛ = √(3RT/M)

where R = molar gas constant = J/mol.K

T = absolute temperature in Kelvin

M = Molar mass of the molecules.

₁₂

Let the vᵣₘₛ of 235-UF6 be vᵣₘₛ₁

And its molar mass = M₁ = 349.0 g/mol

vᵣₘₛ₁ = √(3RT/M₁)

√(3RT) = vᵣₘₛ₁ × √M₁

For the 238-UF6

Let its vᵣₘₛ be vᵣₘₛ₂

And its molar mass = M₂ = 352.0 g/mol

√(3RT) = vᵣₘₛ₂ × √M₂

Since √(3RT) = √(3RT)

vᵣₘₛ₁ × √M₁ = vᵣₘₛ₂ × √M₂

(vᵣₘₛ₁/vᵣₘₛ₂) = (√M₂/√M₁) = [√(352)/√(349)]

(vᵣₘₛ₁/vᵣₘₛ₂) = 1.00429

b) Recall

vᵣₘₛ₁ = √(3RT/M₁)

vᵣₘₛ₂ = √(3RT/M₂)

(vᵣₘₛ₁ - vᵣₘₛ₂) = 1 m/s

[√(3RT/M₁)] - [√(3RT/M₂)] = 1

R = 8.314 J/mol.K, M₁ = 349.0 g/mol = 0.349 kg/mol, M₂ = 352.0 g/mol = 0.352 kg/mol, T = ?

√T [√(3 × 8.314/0.349) - √(3 × 8.314/0.352) = 1

√T (8.4538 - 8.4177) = 1

√T = 1/0.0361

√T = 27.7

T = 27.7²

T = 767.34 K

c) The answers point to a difficult seperation technique, as the two compounds of the different isotopes have very close rms speeds and to create a difference of only 1 m/s In their rms speeds would require a high temperature of up to 767.34 K.

5 0
3 years ago
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