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3 years ago

A boiler is being used to heat water. The graph shows the temperature of

Physics

1 answer:

Vinvika [58]3 years ago

4 0

Answer:

a. Liquid state

b. At point R. The physical state of water at its boiling point temperature of 100 degree Celsius will be both liquid state as well as gaseous state.

c. at 110 degrees it is in gaseous state

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Now assume that Eq. 6-14 gives the magnitude of the air drag force on the typical 20 kg stone, which presents to the wind a vert

chubhunter [2.5K]

Answer:

362.41 km/h

Explanation:

F = Force

m = Mass = 84 kg

g = Acceleration due to gravity = 9.81 m/s²

C = Drag coefficient = 0.8

ρ = Density of air = 1.21 kg/m³

A = Surface area = 0.04 m²

v = Terminal velocity

F = ma

$F=\frac{1}{2}\rho CAv^2\\\Rightarrow mg=\frac{1}{2}\rho CAv^2\\\Rightarrow v=\sqrt{2\frac{mg}{\rho CA}}\\\Rightarrow v=\sqrt{2\frac{20\times 9.81}{1.21\times 0.8\times 0.04}}\\\Rightarrow v=100.66924\ m/s$

Converting to km/h

$100.66924\times 3.6=362.41\ km/h$

The terminal velocity of the stone is 362.41 km/h

5 0

3 years ago

Read 2 more answers

100 POINTS FOR CORRECT ANSWER/EXPLANATION

Shalnov [3]

Answer:

6 N

Explanation:

Let's start with the small block m on top. There are four forces:

Weight force mg pulling down, normal force N₁ pushing up, tension force T pulling right, and friction force N₁μ pushing left (opposing the direction of motion).

Now let's look at the large block M on bottom. There are seven forces:

Normal force N₁ pushing down (opposite and equal from block m),

Friction force N₁μ pushing right (opposite and equal from block m),

Weight force Mg pulling down,

Tension force T pulling right,

Applied force F pulling left,

Normal force N₂ pushing up,

and friction force N₂μ pushing right (opposing the direction of motion).

So you've correctly identified the free body diagrams.

Now apply Newton's second law. Sum of forces in the y direction for block m:

∑F = ma

N₁ − mg = 0

N₁ = mg

Sum of forces in the x direction:

∑F = ma

T − N₁μ = 0

T = N₁μ

T = mgμ

Sum of forces in the y direction for block M:

∑F = ma

-N₁ − Mg + N₂ = 0

N₂ = N₁ + Mg

N₂ = mg + Mg

Sum of forces in the x direction:

∑F = ma

N₁μ + T − F + N₂μ = 0

F = N₁μ + T + N₂μ

F = mgμ + mgμ + (mg + Mg)μ

F = gμ(3m + M)

Since M = 2m:

F = 5gμm

Plug in values:

F = 5 (10 m/s²) (0.400) (0.300 kg)

F = 6 N

8 0

3 years ago

Read 2 more answers

Imagine that the earth and the-moon have positive charges of the same magnitud. How big isäºthe charge necesary to produce an el

lions [1.4K]

Answer:

5.7 x 10^12 C

Explanation:

Let the charge on earth and moon is q.

mass of earth, Me = 5.972 x 10^24 kg

mass of moon, Mm = 7.35 x 10^22 kg

Let d be the distance between earth and moon.

the gravitational force between them is

$F_{g}=G\frac{M_{e} \times M_{m}}{d^{2}}$

The electrostatic force between them is

$F_{e}=\frac{Kq^{2}}{d^{2}}$

According to the question

1 % of Fg = Fe

$0.01 \times 6.67\times10^{-11}\frac{5.97 \times 10^{24}\times7.35 \times 10^{22}}{d^{2}}=9 \times 10^{9}\frac{q^{2}}{d^{2}}$

$2.927 \times 10^{35}=9 \times10^{9}q^{2}$

$3.25 \times 10^{25}=q^{2}$

q = 5.7 x 10^12 C

Thus, the charge on earth and the moon is 5.7 x 10^12 C.

6 0

3 years ago

Solve the problem.

gulaghasi [49]

As the shock waves travel in concentric outward circles from the epicenter, and the diameter is measured 120 miles,
area of a circle =πr*r

d=120
r=120/2r=6060*60=36003600*π=11309.734
11309.734 square miles

5 0

3 years ago

A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w

trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

$KE = PE$