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3 months ago

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What will be the final temperature if a 4.00 g silver ring at 41.0◦C if it gives off 18.0 J of heat to the surroundings? The spe

cific heat capacity of silver is 0.235 J/(g C) Express your answer in ◦C.

1 answer:

kicyunya [3.2K]3 months ago

5 0

Final temperature to determine: Given the following details, the calculations proceed as follows: Mass of the silver ring is m = 4 g, initial temperature is presented, and the heat released is Q = -18 J (indicating heat loss). The specific heat of silver is considered next to find the final temperature.

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Point m is located a distance 2d from the midpoint between the two wires. find the magnitude of the magnetic field b1m created a

Sav [3153]

Reminder: The illustration mentioned in the inquiry is included here as a file.

Response:

The intensity of the magnetic field is $B = \frac{0.071 \mu I}{d}$

Clarification:

The magnetic field can be calculated using the following equation:

$B = \frac{\mu I}{2\pi R}$ ...............(1)

Where I refers to the current flowing through the wires

To find the distance R from point 1 to m, apply the Pythagorean theorem

$R = \sqrt{d^{2} + (2d)^{2} }$

$R = \sqrt{5d^{2} } \\R = d\sqrt{5}$

Inserting R into equation (1)

$B = \frac{\mu I}{2\pi d\sqrt{5} }$

$B = \frac{0.071 \mu I}{d}$

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2 months ago

A 0.050-kg lump of clay moving horizontally at 12 m/s strikes and sticks to a stationary 0.15-kg cart that can move on a frictio

Ostrovityanka [3204]

Answer:

Explanation:

According to the parameters provided,

mass of the clay lump, m₁ = 0.05 kg

initial velocity of the lump, u₁ = 12 m/s

mass of the cart, m₂ = 0.15 kg

initial speed of the cart, u₂ = 0

As the clay adheres to the cart, we have an inelastic collision scenario. Let v represent the combined speed of both the cart and lump post-collision. Given that momentum is conserved, we have:

$m_1u_1+m_2u_2=(m_1+m_2)v$

$v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}$

$v=\dfrac{0.05\ kg\times 12\ m/s+0}{0.05\ kg+0.15\ kg}$

The resultant speed is v = 3 m/s.

Thus, the final speed of both cart and lump following the collision is 3 m/s. This concludes the solution.

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3 months ago

A water jet that leaves a nozzle at 60 m/s at a flow rate of 120 kg/s is to be used to generate power by striking the buckets lo

Ostrovityanka [3204]

Response:

216000 W or 216 kW

Details:

Power: This refers to the rate at which energy is utilized or consumed. The standard unit for power is the Watt (W).

In general,

Power = Energy/time

P = E/t........................ Equation 1.

However,

E = 1/2mv²..................... Equation 2

with m representing mass, and v indicating velocity.

Substituting equation 2 into equation 1,

P = 1/2mv²/t...................... Equation 3

Assuming flow rate (Q) = m/t,

Q = m/t................ Equation 4

Insert equation 4 into equation 3,

P = Qv²/2........................ Equation 5

Where Q stands for flow rate, v is velocity, and P denotes power.

Given: Q = 120 kg/s, v = 60 m/s

Plug these values into equation 5,

P = 120(60)²/2

P = 60(60)²

P = 60×3600

P = 216000 W.

Thus, the potential power generation of the water jet is 216000 W or 216 kW.

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4 months ago

A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 108.7 g of water at 60.2°C with 108

Ostrovityanka [3204]

Answer: The calorimeter's heat capacity is $6.72J/g^oC$

Explanation:

This scenario assumes the amount of heat lost by the hot object equals the amount of heat gained by the cold object.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat capacity of water = $4.184J/g^oC$

$c_2$ = specific heat capacity of calorimeter =?

$m_1$ = mass of water = 108.7 g

$m_2$ = mass of calorimeter = 108.7 g

$T_f$ = final temperature of the mixture = $35.0^oC$

$T_1$ = initial temperature of the water = $60.2^oC$

$T_2$ = initial temperature of calorimeter = $19.3^oC$

Now substituting all provided values into the formula, we obtain

$(108.7g)\times (4.184J/g^oC)\times (35.0-60.2)^oC=-(108.7g)\times c_2\times (35.0-19.3)^oC$

$c_2=6.72J/g^oC$

Hence, the heat capacity of the calorimeter is $6.72J/g^oC$

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4 months ago