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Oksana_A [137]
3 years ago
10

Fifteen grams of substance X at 95 degrees Celsius is mixed with 45 grams of substance Z at 85 degrees Celsius in a container wh

ich was originally at 25 degrees Celsius. Neglecting any transfer of heat to the environment, the temperature at the end of 15 minutes was 32 degrees Celsius. What is the equilibrium temperature?
A. 95 degrees Celsius
B. 85 degrees Celsius
C. 32 degrees Celsius
D. 25 degrees Celsius
Physics
1 answer:
Viefleur [7K]3 years ago
5 0
According to the Law of Conservation of Energy, energy is neither created nor destroyed. They are just transferred from one system to another. To obey this law, the energy of the substances inside the container must be equal to the substance added to it. The energy is in the form of heat. There can be two types of heat energy: latent heat and sensible heat. Sensible heat is energy added or removed when a substance changes in temperature. Latent heat is the energy added or removed at a constant temperature during a phase change. Since there is no mention of phase change, we assume the heat involved here is sensible heat. The equation for sensible heat is:

H = mCpΔT
where
m is the mass of the substance
Cp is the specific heat of a certain type of material or substance
ΔT is the change in temperature.

So the law of conservation of heat tells that:

Sensible heat of Z + Sensible heat of container = Sensible heat of X

Since we have no idea what these substances are, there is no way of knowing the Cp. We can't proceed with the calculations. So, we can only assume that in the duration of 15 minutes, the whole system achieves equilibrium. Therefore, the equilibrium temperature of the system is equal to 32°C. The answer is C.
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Answer:

Average net force, F = 15157.15 N

Explanation:

It is given that,

The mass of the car and riders is, m=3\times 10^3\ kg

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Time, t = 8.59 seconds

We need to find the  average net force exerted on the car and riders by the magnets. It can be calculated using second law of motion as :

F = m a

F=m(\dfrac{v-u}{t})

F=3\times 10^3\ kg\times (\dfrac{43.4\ m/s-0}{8.59\ s})

F = 15157.15 N

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A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, a
pantera1 [17]

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

120^2=130^2+50^2-2\times130\times50\cos\alpha

\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}

\alpha=\cos^{-1}(0.3846)

\alpha=67.38^{\circ}

We need to calculate the angle β

Using cosine law

50^2=130^2+120^2-2\times130\times120\cos\beta

\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}

\beta=\cos^{-1}(0.923)

\beta=22.63^{\circ}

We need to calculate the force on 130 cm side

Using formula of force

F_{130}=ILB\sin\theta

F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0

F_{130}=0

We need to calculate the force on 120 cm side

Using formula of force

F_{120}=ILB\sin\beta

F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63

F_{120}=0.1385\ N

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

F_{50}=ILB\sin\alpha

F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38

F_{50}=0.1385\ N

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

8 0
3 years ago
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Frequency is 1 over the period. Hence, the frequency is 1/18.
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