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jasenka [17]
3 years ago
13

Convert the temperature of the air in a room air-conditioned to 20.0 degrees Celsius to equivalent temperatures on the Fahrenhei

t and Kelvin scales.
Physics
1 answer:
gizmo_the_mogwai [7]3 years ago
4 0

Answer:

                       Freezing Water         Boiling Water

Centigrade             0                               100

Fahrenheit             32                               212

Kelvin                    273                              373

One needs to know these values for an understanding of the temperature scale

  9/5 C + 32 = F           Check for boiling points

 5/9 (F - 32) = C

Example:

  F = 9/5 C + 32 = 9/5 * 20 + 32 = 68 F

And deg K = C + 273

So 20 deg C = 293 deg K

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HELP PLSSS I HAVE AN EXAM MONDAY AND I THINK THIS IS GONNA BE ON ITTTT
il63 [147K]
<h2>Answer:</h2>

(a) 3.2 x 10²s

(b) 0.9 m/s (S 13 E)

(c) 2.9 x 10²m

<h2>Explanation:</h2>

The sketch illustrating the scenario has been attached to this response.

As shown;

The fish swims due east with a velocity V_{x} = 0.2m/s

The river current has a velocity V_{y} due South = 0.9m/s

The resultant of the velocity is V

The width of the river is x = 64m

(a) To calculate how long it took the fish to get across the river, we know that velocity is the rate of change in distance, therefore we can use the relation;

V = \frac{d}{t}      -------------(i)

Where;

V = velocity of the fish = V_{x} = 0.2m/s

d = distance from the start to the end = width of the river = x = 64m

t = time taken to move for that distance

Make t subject of the formula in equation (i);

t = \frac{d}{V}

Substitute the values of d and V into the equation;

t = \frac{64m}{0.2m/s}

t = 320 s

t = 3.20 x 10²s

Therefore, the time taken for the fish to get across the river is 3.20 x 10²s

(b) The resulting vector of the fish is V whose magnitude is the algebraic sum of vectors  V_{x} and  V_{y}, and direction is given by θ. i.e

<em>The magnitude of the resulting vector is;</em>

|V| = \sqrt{(V_x)^2 + (V_y)^2}

|V| = \sqrt{(0.2)^2 + (0.9)^2}

|V| = \sqrt{(0.04) + (0.81)}

|V| = \sqrt{(0.85)}

|V| = 0.92m/s

|V| ≅ 0.9m/s

<em>The direction of the resulting vector θ and is given by;</em>

tan θ = \frac{V_y}{V_x}

tan θ = \frac{0.9}{0.2}

tan θ = 4.5

θ = tan⁻¹ ( 4.5)

θ = 77.47° South of East.

θ  ≈ 77.5° South of East.

Subtracting θ = 77.5° from 90° gives its value East of South

i.e

90 - 77.5 = 12.5° East of South

<em>This can also be written as S12.5°E</em>

<em>Approximating to the nearest whole number gives </em>S 13 E

Therefore, the resulting velocity of the fish is 0.9m/s in the direction S13°E

(c) When the fish arrives on the opposite bank, its distance from being at the point directly across from where it started is the product of the velocity of the river current and the time taken by the fish to get across the river. This point is equivalent to k as shown in the diagram.

Therefore;

distance = velocity of river current x time taken

distance = 0.9m/s x 3.20 x 10²s

distance = 2.88 x 10²m

distance ≅ 2.9 x 10²m

<em>Notice that the velocity of the river current is used since that's the velocity of the fish on the y-axis.</em>

<em />

<em />

7 0
3 years ago
In nuclear fission, a nucleus splits roughly in half. (a) What is the potential 2.00 × 10−14 m from a fragment that has 46 proto
Valentin [98]

Answer:

electric potential is 3.31 × 10^{6} V

potential energy is 152 MeV

Explanation:

given data

fragment charge  Q = 46 protons = 46 × 1.6 × 10^{-19} C

to find out

electric potential  and potential energy

solution

we know here distance from fragment d = 2 × 10^{-14} m

and constant for electric force k that is 9 × 10^{9} N-m²/C²

so that we can find electric potential = kQ/d

electric potential = 9 × 10^{9}[/tex ×46 × 1.6 × [tex]10^{-19} / ( 2 × 10^{-14} )

electric potential = 3.31 × 10^{6} V

and

we know relation between electric potential and potential

that is  V = U/q

so U will be = qV

now put all value

we get potential energy U

potential energy = 46 × 3.31 × 10^{6}

potential energy = 1.52 × 10^{8} eV

so potential energy = 152 MeV

4 0
3 years ago
What is the experimental group(s)?
Vinil7 [7]
<h3>I think it B The group(s) that gets the special treatment.</h3><h3 /><h3>I hope this is correct.</h3><h3 /><h3 /><h3 />
5 0
3 years ago
A man pushes a lawn mower on a level lawn with a force of 195 N. If 37% of this force is directed downward, how much work is don
densk [106]

Answer:

W = 1032.6 J

Explanation:

Net force by which man push the lawn mower is given as

F = 195 N

now it is given that 37% of this force is vertically downwards

so we will have

F_y = 0.37 \times 195

F_y = 72.15 N

now we also know that

F_x^2 + F_y^2 = F^2

here we have

F_x^2 + 72.15^2 = 195^2

F_x = 181.2 N

Now work done by this force to move the lawn mower is given as

W = F_x . d

W = (181.2)(5.7 m)

W = 1032.6 J

7 0
3 years ago
I don't know how to figure it out
drek231 [11]
Nice couch lol
and aluminum I think
4 0
3 years ago
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