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mel-nik [20]
3 years ago
15

Pls help quick too.................

Physics
1 answer:
statuscvo [17]3 years ago
4 0

Answer:

Pulley = Flag Pole

Screw = Side latch

Wedge = Scissors

Wheel and axle = Car

You might be interested in
A 0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of co
Genrish500 [490]

Answer: When 1.0kg of aluminium block is used, the final temperature of the mixture will be T = 36.2∘C

If 1.0kg copper block is used, T of the mixture will be = 17.4∘C

If 100g (0.1kg) of ice at 0∘C is used, T will be = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be= 147.1∘C

Explanation:

H = mcΘ

heat lost by block = heat gained by water

m₁c₁Θ₁ = m₂c₂Θ₂ where m₁ is mass of aluminium, m₂ is mass of water, c₁ is cAluminium, c₂ is cWater, Θ₁ is temperature change for aluminium, Θ₂ is temperature change for water.

0.5*900*(200-20) = m₁*4186*(20-0)

m₁ = 450*180/83270

<em>m₁ = 0.973kg</em>

<em>when 1.0kg of aluminium block is used, the final temperature of the mixture will be </em><em>T</em>

heat lost by block = heat gained by water

1.0*900*(200-T) = 0.973*4186*(T-0)

180000 - 900T = 4073T

4973T = 180000

T = 180000/4973 = 36.2∘C

<em>If 1.0kg copper block is used, T of the mixture will be</em>

heat lost by block = heat gained by water

1.0*387*(200-T) = 0.973*4186*(T-0)

77400 - 387T = 4073T

4460T = 77400

T = 77400/4460 = 17.4∘C

<em>If 100g (0.1kg) of ice at 0∘C is used, T will be</em>

<em>heat lost by block = heat gained by water + heat used in melting ice to form water at 0∘C</em>

heat used in melting 0.1kg of ice, H = ml, where l= 33600J/Kg

0.5*900*(200-T) = 0.1*4186*(T-0) + 0.1*33600J/Kg

90000 - 450T =  418.6T + 33600

418.6T + 450T = 90000 - 33600

868.6T = 56400

T = 56400/868.6 = 64.9∘C

If 25g (0.025Kg) of ice is used, T will be

0.5*900*(200-T) = 0.025*4186*(T-0) + 0.025*33600J/Kg

90000 - 450T =  104.65T + 8400

104.65T + 450T = 90000 - 8400

554.65T = 81600

T = 81600/554.65 = 147.1∘C

7 0
3 years ago
What is the density of 15 cm3 and 3 g?
lilavasa [31]

Answer:

0.2

Explanation:

3/15

6 0
2 years ago
Explain how energy balance sets planetary temperature? Imagine a planet colder than expected for energy balance and explain why
RUDIKE [14]

The planetary temperature energy balance is obtained by radiating back the absorbed radiation energy from outer-space, by the planet and thus acquiring thermal equilibrium.

What is the process of attaining thermal equilibrium by Earth?

The Stefan-Boltzmann law states that the more the temperature a planet has, the more it will radiate out to reach thermal equilibrium.

We know that outer space contains large masses of radiative energy freely distributed in its vast expanse. A small fraction of this energy is absorbed by the Earth through the atmosphere, surface land, clouds etc.

Now, radiative balance is achieved when a planet's surface continuously warms up until it reaches its peak at which point the same amount of absorbed energy can then be radiated back to space. The relative amount of energy radiated back by a planet is dependent upon the size of the planet.

A colder planet relatively absorbs lower amount of radiation energy from space. In some time, as the planet heats up enough, the energy is radiated back to the space attaining thermal equilibrium.

Learn more about Stefan-Boltzmann law here:

<u>brainly.com/question/14919749</u>

#SPJ4

6 0
1 year ago
The information on a can of soda indicates that the can contains 355 mL. The mass of a full can of soda is 0.369 kg, while an em
Vilka [71]

Answer:

\rho=995.50\ kg.m^{-3}

\bar w=9765.887\ N.m^{-3}

s=0.9955

Explanation:

Given:

  • volume of liquid content in the can, v_l=0.355\ L=3.55\times 10^{-4}\ L
  • mass of filled can, m_f=0.369\ kg
  • weight of empty can, w_c=0.153\ N

<u>So, mass of the empty can:</u>

m_c=\frac{w_c}{g}

m_c=\frac{0.153}{9.81}

m_c=0.015596\ kg

<u>Hence the mass of liquid(soda):</u>

m_l=m_f-m_c

m_l=0.369-0.015596

m_l=0.3534\ kg

<u>Therefore the density of liquid soda:</u>

\rho=\frac{m_l}{v_l} (as density is given as mass per unit volume of the substance)

\rho=\frac{0.3534}{3.55\times 10^{-4}}

\rho=995.50\ kg.m^{-3}

<u>Specific weight of the liquid soda:</u>

\bar w=\frac{m_l.g}{v_l}=\rho.g

\bar w=995.5\times 9.81

\bar w=9765.887\ N.m^{-3}

Specific gravity is the density of the substance to the density of water:

s=\frac{\rho}{\rho_w}

where:

\rho_w= density of water

s=\frac{995.5}{1000}

s=0.9955

3 0
3 years ago
Read 2 more answers
A rightward force of 12.0 N is applied to a 2.0-kg object to accelerate it across a horizontal
ad-work [718]

Answer:

below

Explanation:

Net accelerating force becomes  12-8 = 4 N

F = ma

4 = 2 * a

a = 2 m/s^2

8 0
1 year ago
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