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4 years ago

Which part of an atom has most of its mass? A. electrons B. neutrons C. nucleus D. protons

2 answers:

6 0

The nucleus has most of the atomic mass in an atom. The nucleus is made up of protons and neutrons.

pychu [463]4 years ago

6 0

Answer:

C. nucleus

Explanation:

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What can subatomic particles be broken down into?

AfilCa [17]

Subatomic particles cannot be broken down any further

5 0

3 years ago

A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder

QveST [7]

Answer:

$\Delta \theta = 47.57^{\circ} C$

Explanation:

given,

moles of air compressed, n = 1.70 mol

initial temperature, T₁ = 390 K

Power supply by the compressor, P = 7.5 kW

Heat removed = 1.3 kW

Angular frequency of the compressor, f = 110 rpm = 110/60 = 1.833 rps.

Time of compression = time of the hay revolution

= $\dfrac{1}{2}\ T$

= $\dfrac{1}{2}\times \dfrac{1}{f}$

= $\dfrac{1}{2}\times \dfrac{1}{1.833}$

=0.273 s

Using first law of thermodynamics

U = Q - W

now,

$\dfrac{\Delta U}{\Delta t} = \dfrac{\Delta Q}{\Delta t}- \dfrac{\Delta W}{\Delta t}$

Power supplied $\dfrac{\Delta W}{\Delta t}$ = 7.5 kW

heat removed $\dfrac{\Delta Q}{\Delta t}$ = 1.3 kW

now,

$\dfrac{\Delta U}{\Delta t} = 7.5 -1.3$

$\dfrac{\Delta U}{\Delta t} = 6.2 kW$

we know,

$\dfrac{\Delta U}{\Delta t}=\dfrac{nC_v\Delta \theta}{\Delta t}$

C_v for air = 5 cal/° mol

= 5 x 4.186 J/mol°C = 20.93 J/mol°C

now,

$\Delta \theta = \dfrac{\Delta U}{\Deta t}\times \dfrac{\Delta t}{n C_v}$

$\Delta \theta = 6.2\times 10^3 \times \dfrac{0.273}{1.7\times 20.93}$

$\Delta \theta = 47.57^{\circ} C$

the temperature change per compression stroke is equal to 47.57°C.

4 0

3 years ago

A particle travels along the curve from A to B in 5 s. It takes 8 s for it to go from B to C and then 10 s to go from C to A. De

Elza [17]

Answer:

5.1 m/s

Explanation:

The figure is missing: find it in attachment.

In order to find the average speed, we have to calculate the length of the total path, and divide it by the total time elapsed.

The curve from A to B is a quarter of a circle with a radius of r = 20 m, so its length is:

$AB=\frac{\pi r}{2}=\frac{\pi (20)}{2}=31.4 m$

The path BC is the hypothenuse of a right triangle with sides equal to 20 m and 30 m, so its length is

$BC=\sqrt{30^2+20^2}=36 m$

Finally, the length of the path AC is the sum of the side of 30 m and the radius of the curve, so

$AC=30 + 20 = 50 m$

So the total distance covered is

$d=AB+BC+AC=31.4+36+50=117.4 m$

The total time elapsed is

$t=5 s + 8 s + 10 s =23 s$

So, the average speed is

$v=\frac{d}{t}=\frac{117.4}{23}=5.1 m/s$

6 0

3 years ago

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

m₁ = 0.973kg

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

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

If 1.0kg copper block is used, T of the mixture will be

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

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

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

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

In a game of ice hockey, you use a hockey stick to hit a puck of mass 0.16 kg that slides on essentially frictionless ice. Durin

Grace [21]

Answer:

12N

Explanation:

given- mass, acelation

Fnet=ma= .16kg*75m/s2

Fnet=12 N only force no friction given.

8 0

3 years ago