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agasfer [191]
2 years ago
14

Good morning! Can someone please answer this, ill give you brainliest and you will earn 50 points.

Physics
2 answers:
german2 years ago
5 0

Here are the answers

  • Temperature and energy:-Ocean and surface temperature
  • Atmospheric composition:- Carbon dioxide and other greenhouse gases
  • Ocean and water-Ocean acidification and sea level.
  • Cryosphere :-Sea ice and glaciers
Stels [109]2 years ago
4 0

Temperature and energy ➡️ <em>Surface</em><em> </em><em>temperature</em><em> </em><em>and</em><em> </em><em>ocean</em><em> </em><em>temperature</em><em> </em>

Atmospheric composition ➡️ <em>Carbon-dioxide</em><em> </em><em>and</em><em> </em><em>other</em><em> </em><em>greenhouse</em><em> </em><em>gases</em>

Ocean and water ➡️ <em>Ocean</em><em> </em><em>acidification</em><em> </em><em>and</em><em> </em><em>sea</em><em> </em><em>level</em>

Cryosphere<em> </em>➡️ <em>Sea</em><em> </em><em>ice</em><em> </em><em>and</em><em> </em><em>galciers</em>

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A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i
Mariana [72]

Answer:

451.13 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

c = 1500/[0.133(45-20)]

c = 1500/(0.133×25)

c = 1500/3.325

c = 451.13 J/kg.°C

4 0
3 years ago
Explain how heat transfer is taking place in the form of conduction, convection &amp; radiation.
PIT_PIT [208]

Answer:

Heat can travel from one place to another in three ways: Conduction, Convection and Radiation. ... Thermal energy is transferred from hot places to cold places by convection. Convection occurs when warmer areas of a liquid or gas rise to cooler areas in the liquid or gas.

3 0
3 years ago
An aircraft maintenance technician walks past a tall hangar door that acts like a single slit for sound entering the hangar. Out
Sphinxa [80]

Answer:

The first minimum would be observed at 41.57°

Explanation:

v = 340m/s = speed of sound

f = 610Hz

d = 0.840m

λ = ?

Mλ = wsinθ

m = mth order minima

λ = wavelength incident on the single slit

θ = angular position of the mth minima

But, λ = v / f

λ = 340 / 610 = 0.557m

θ = sin⁻(mλ/d)

θ = sin⁻ [(1 * 0.557) / 0.840]

θ = sin⁻ 0.6635

θ = 41.57°

The first minimum would be observed at 41.57°

4 0
3 years ago
A 150kg person stands on a compression spring with spring constant 10, 000 N/m and norminal.Length of 0.50.What is the togal len
Eva8 [605]

Answer:

Total length of spring 0.647 m

Explanation:

We have given mass of the person m = 150 kg

Acceleration due to gravity g=9.8m/sec^2

Spring constant k = 10000 N/m

Nominal length of spring = 0.50

According to hook's law

mg=kx

150\times 9.8=10000\times x

x = 0.147 m

So total length of spring = 0.50+0.147 = 0.647 m

4 0
3 years ago
An auditorium measures 40.0 m 3 20.0 m 3 12.0 m. The density of air is 1.20 kg/m3. What are (a) the volume of the room in cubic
Burka [1]

We use the formula,

m = V\rho

Here, m is the mass, V is the volume and  \rho density

Also

V = l w h

Here l is length, w is width and h is height.

(a) The volume of the room,

V=40.0 \ m \times 3 20.0 \ m \times 3 12.0\ m = 3.99 \times 10^{6} m^3

The volume of the room in cubic feet,

V = 3.99 \times 10^{6} m^3 \times(\frac{3.281 \ ft}{m} )^3 = 4.3 \times 10^7 \ ft^3

(b) Now the mass of the air in room,

m= (3.99 \times 10^{6} \ m^3) (1.20 \ kg/m^3) = 4.8 \times 10^6 kg.

Therefore, the weight of the air in room,

W = mg= 4.8 \times 10^6 kg \times 9.8 m/s^2 = 46.9 \times 10^6 \ N \\\\ W = 4.69 \times 10^7 \ N.

The weight of air in the room in pounds,

W = 4.69 \times 10^7 \ N ( \frac{1 \ lb}{4.448 \ N} ) = 1.1 \times 10^7 \ lb


3 0
3 years ago
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