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Bogdan [553]
2 years ago
15

No interior de um calorímetro de capacidade térmica 6 cal/°C encontram-se 85 g de um líquido a 18°C. Um bloco de cobre de massa

120 g e calor específico 0,094 cal/g°C, aquecido a 100°C, é colocado dentro do calorímetro. O equilíbrio térmico se estabelece a 42°C. Determine o calor específico do líquido.

Physics
2 answers:
MrRissso [65]2 years ago
4 0

Answer:

Heat Exchange

Explanation:

<em>Within a 6 cal / ° C thermal capacity calorimeter is 85 g of a liquid at 18 ° C. A block of copper of mass 120 g and specific heat 0.094 cal / g ° C, heated to 100 ° C, is placed within the calorimeter. The thermal equilibrium is established at 42 ° C. Determine the specific heat of the liquid.</em>

I will add the answer as an image, as the system encounters Bug and warns that there are external links in the response.

Lyrx [107]2 years ago
3 0

a fórmula do calor sensível é Q = m.c.ΔT


m = massa

c = calor específico do corpo

ΔT = variação de temperatura (Tf - Ti)


quando ocorre o equilíbrio térmico, a soma dos calores trocados entre os corpos é 0:

QA + QB + QC = 0

sabendo que a capacidade térmica é o produto m.c e substituindo os demais valores fornecidos no enunciado:


QA = calor sensível do calorímetro = (6 cal/°C).(42-18)

QB = calor sensível do líquido =  (85g).c.(42-18)

QC = calor sensível do bloco =  (120g).(0,094 cal/g°C).(42-100)


substituindo na expressão para calcular c, o calor específico do líquido :


(6 cal/°C).(42°C-18°C) +  (85g).c.(42°C-18°C) + (120g).(0,094 cal/g°C).(42°C-100°C) = 0


144cal + 2040c - 654,2cal = 0


 2040c = 510cal


c = 510cal/2040g.°C


c = 0,25cal/g.°C


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Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are
Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

d_{12} = \sqrt{3^{2}+3^{2}  }  = \sqrt{18} m : distance from q₁ to q₂

(d₁₂)² = 18 m²

d_{13} =\sqrt{1^{2}+3^{2}  } = \sqrt{10} m  : distance from q₁ to q₃

(d₁₃)² = 10 m²

d_{14} =\sqrt{3^{2}+2^{2}  } = \sqrt{13} m  : distance from q₁ to q₄

(d₁₄)² = 13 m²

K=  8.98755 × 10⁹ N *m²/C²

q₁=  7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45°  ,  β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N  

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }

F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }

Fn₁= 23.95*10⁻⁹ N

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Which of the following solutions should be chosen to help restore an ecosystem naturally? A. reintroducing an animal to the ecos
jolli1 [7]

Answer:

A. reintroducing an animal to the ecosystem

Explanation:

As generally, all know that for restoring an ecosystem naturally, it requires reintroduction of an animal to the ecosystem. As though it helps in reimposing the ecosystem back, and also helps to improve our ecosystem in natural surroundings, natural terrain, and population density. Basically reintroducing an animal is also required for the balancing of the ecosystem. As everything requires a properly balanced nature.

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2 years ago
Rachel and Sarah are on a bus travelling at 5 mph past John who is standing on the sidewalk. Rachel then throws a ball
oksian1 [2.3K]

Answer:

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{C - V}.

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2 years ago
1. Mr. Ure has a mass of 65 kg, due to the fact that he is WAY too skinny! What is the force of Earth's gravity on him?
shepuryov [24]

1.

m = mass of Mr. Ure = 65 kg

g = acceleration due to gravity = 9.8 m/s²

force of earth's gravity on Mr. Ure is given as

F = mg

F = 65 x 9.8

F = 637 N


2.

F = force of gravity on car = 3050 N

m = mass of the car = ?

g = acceleration due to gravity = 9.8 m/s²

force of gravity on car is given as

F = mg

3050 = m (9.8)

m = 3050/9.8

m = 311.22 kg


3.

m = mass of Mr. Rees = 90 kg

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force of earth's gravity on Mr. Rees is given as

F = mg

F = 90 x 9.8

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Whats 6 3/7 ×1 5/9.
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