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liq [111]
3 years ago
8

A data table shows the height of a person on his birthday each year for ten years. What is the the dependent variable? Provide e

vidence to support your claim, and explain your reasoning.
Physics
2 answers:
Anna35 [415]3 years ago
8 0

dep var = height

it depends on time ... year

iogann1982 [59]3 years ago
5 0

dep var = height

it depends on time ... year

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Accomplished silver workers in india can pound silver into incredibly thin sheets, as thin as 3.00 10-7 m (about one-hundredth o
postnew [5]
The density of silver is ρ = 10500 kg/m³ approximately.

Given:
m = 1.70 kg, the mass of silver
t = 3.0 x 10⁻⁷ m, the thickness of the sheet

Let A be the area.
Then, by definition,
m = (t*A)*ρ

Therefore
A = m/(t*ρ)
    = (1.7 kg)/ [(3.0 x 10⁻⁷ m)*(10500 kg/m³)]
    = 539.7 m²

Answer: 539.7 m²

8 0
3 years ago
A jetliner has a velocity of 95 m/s. What is the displacement of the jetliner at t=3.0 seconds?
charle [14.2K]
Distance = speed / time

speed = 95 m/s
time = 3 s

distance = 95 / 3 m

displacement = 95/3 m or 32 m (2 s.f.)
5 0
3 years ago
A playground merry-go-round has a mass of 50 kg and a diameter of 4.0 m. There are 4 children who want to ride on it. They have
mixer [17]

Answer:

B) I1 = 1680 kg.m^2          I2 = 1120 kg.m^2

C) V = 0.84m/s      T = 29.92s

D) ω2 = 0.315 rad/s

Explanation:

The moment of inertia when they are standing on the edge:

I1 = 1/2*M*R^2 + (m1+m2+m3+m4)*R^2   where M is the mass of the merry-go-round.

I1 = 1680 kg.m^2

The moment of inertia when they are standing half way to the center:

I2 = 1/2*M*R^2 + (m1+m2+m3+m4)*(R/2)^2

I2 = 1120 kg.m^2

The tangencial velocity is given by:

V = ω1*R = 0.84m/s

Period of rotation:

T = 2π / ω1 = 29.92s

Assuming that there is no friction and their parents are not pushing anymore, we can use conservation of the angular momentum to calculate the new angular velocity:

I1*ω1 = I2*ω2    Solving for ω2:

ω2 = I1*ω1 / I2 = 0.315 rad/s

5 0
3 years ago
Air enters a compressor with a pressure of 14.7 lbf/in.2 , a temperature of 70F, and a volumetric flow rate of 40 ft3 /s. Air ex
disa [49]

Answer:

210 hp

Explanation:

See attached picture.

3 0
3 years ago
How much heat (in kJ) is required to warm 13.0 g of ice, initially at -12.0 ∘C, to steam at 109.0 ∘C?
Sunny_sXe [5.5K]
<h2>Answer:</h2>

39.699 kJ

<h2>Explanation:</h2>

In this situation, there are a few transformations as follows;

(i) Heat required to warm the ice from -12°C to its melting point.

(ii) Heat required to melt the ice.

(iii) Heat required to boil the melted ice to boiling point (i.e to steam)

(iv) Heat required to vapourize the water

(v) Heat required to heat the steam from 100°C to 109.0°C

The sum of all the heat processes gives the heat required to warm the ice to steam;

<h3><em>Calculate each of these heat processes</em></h3>

<em>From (i);</em>

Let the heat required to warm the ice from -12.0°C to its melting point (0°C) be Q₁.

Q₁ = m x c x ΔT        -----------------------(i)

Where;

m = mass of ice = 13.0g

c = specific heat capacity of ice = 2.09 J/g°C

ΔT = final temperature - initial temperature = 0°C - (-12°C) = 12°C

Substitute these values into equation (i) as follows;

Q₁ = 13.0 x 2.09 x 12 = 326.04 J

<em>From (ii);</em>

Let the heat required to melt the ice be Q₂. This heat is called the heat of fusion and it is given by;

Q₂ = m x L        -----------------------(ii)

Where;

m = mass of ice = 13.0g

L = latent heat of fusion of ice = 333.6 J/g

Substitute these values into equation (ii) as follows;

Q₂ = 13.0 x 333.6

Q₂ = 4336.8 J

<em>From (iii);</em>

Let the heat required to boil the melted ice from 0°C to boiling point of 100°C be Q₃.

Q₃ =  m x c x ΔT        -----------------------(i)

Where;

m = mass of melted ice (water) which is still 13.0g

c = specific heat capacity of melted ice (water) = 4.2 J/g°C

ΔT = final temperature - initial temperature = 100°C - 0°C = 100°C

Substitute these values into equation (i) as follows;

Q₁ = 13.0 x 4.2 x 100 = 5460 J

<em>From (iv);</em>

Let the heat required to vaporize the water (melted ice) be Q₄. This heat is called the heat of vaporization and it is given by;

Q₄ = m x L        -----------------------(iv)

Where;

m = mass of ice = 13.0g

L = latent heat of vaporization of water = 2257 J/g

Substitute these values into equation (iv) as follows;

Q₄ = 13.0 x 2257

Q₄ = 29341 J

<em>From (v);</em>

Let the heat required to heat the steam from 100°C to 109°C be Q₅.

Q₅ =  m x c x ΔT        -----------------------(i)

Where;

m = mass of steam which is still 13.0g

c = specific heat capacity of steam = 2.01 J/g°C

ΔT = final temperature - initial temperature = 109.0°C - 100°C = 9°C

Substitute these values into equation (i) as follows;

Q₅ = 13.0 x 2.01 x 9 = 235.17J

<em>Finally:</em>

<em>Sum all the heat values together;</em>

Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Q = 326.04 + 4336.8 + 5460 + 29341 + 235.17

Q = 39699.01 J

Q = 39.699 kJ

Therefore, the amount of heat (in kJ) required is 39.699

7 0
2 years ago
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