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Anna11 [10]
3 years ago
8

When balancing an equation, you can

Physics
1 answer:
attashe74 [19]3 years ago
7 0
D. write down the coefficients
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4. Two forces act on a 2 kg object as shown. What is the magnitude of the acceleration of the object?
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The resultant force on the object is

∑ <em>F</em> = 〈0, 8〉 N + 〈6, 0〉 N = 〈6, 8〉 N

which has a magnitude of

<em>F</em> = √((6 N)² + (8 N)²) = √(100 N²) = 10 N

By Newton's second law, the acceleration has magnitude <em>a</em> such that

<em>F</em> = <em>m a</em>

10 N = (2 kg) <em>a</em>

<em>a</em> = (10 N) / (2 kg)

<em>a</em> = 5 m/s²

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3 years ago
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A container of gas is at a pressure of 1.3x10^5 Pa and a volume of 6 m^3. How much work is done by the gas if it expands at a co
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In thermodynamics, work of a system at constant pressure conditions is equal to the product of the pressure and the change in volume. It is expressed as follows:

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3 years ago
One end of a metal rod is in contact with a thermal reservoir at 745. K, and the other end is in contact with a thermal reservoi
Masteriza [31]

Answer:

a)S_1=-9.65}\ J/K

b)S_2=71.18\ J/K

c) 0 J/K

d)S= 61.53 J/K

Explanation:

Given that

T₁ = 745 K

T₂ = 101 K

Q= 7190 J

a)

The entropy change of reservoir 745 K

S_1=-\dfrac{7190}{745}\ J/K

Negative sign because heat is leaving.

S_1=-9.65}\ J/K

b)

The entropy change of reservoir 101 K

S_2=\dfrac{7190}{101}\ J/K

S_2=71.18\ J/K

c)

The entropy change of the rod will be zero.

d)

The entropy change of the system

S= S₁ + S₂

S = 71.18 - 9.65 J/K

S= 61.53 J/K

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