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3 years ago

47. Calculate the final steady temperature obtained when

Physics

1 answer:

Rashid [163]3 years ago

4 0

Answer:

the final steady temperature of the mixture is 46.67 ⁰C

Explanation:

Given;

mass of first water,m₁ = 6 g = 0.006 kg

initial temperature, t₁ = 50 ⁰C

mass of the second water, m₂ = 3.0 g = 0.003 kg

initial temperature of the second water, t₂ = 40 ⁰C

specific heat capacity of water, C = 4,200 J/kg.K

Let the final temperature of the mixture = T

Based on Law of conservation of energy, the temperature of the final mixture is calculated as follows;

m₁c(t₁ - T) = m₂c(T - t₂)

0.006 x 4200 x (50 - T) = 0.003 x 4200 x (T - 40)

1,260 - 25.2T = 12.6T - 504

1,260 + 504 = 12.6T + 25.2T

1764 = 37.8T

T = 1764/37.8

T = 46.67 ⁰C

Therefore, the final steady temperature of the mixture is 46.67 ⁰C

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The 68-kg crate is stationary when the force P is applied. Determine the resulting acceleration of the crate if (a) P = 0, (b) P

bogdanovich [222]

Explanation:

Mass of the crate, m = 68 kg

We need to find the resulting acceleration if :

(a) Force, P = 0

P = m a

⇒ a = 0

(b) P = 181 N

$a=\dfrac{P}{m}$

$a=\dfrac{181\ N}{68\ kg}$

$a=2.67\ m/s^2$

(c) P = 352 N

$a=\dfrac{P}{m}$

$a=\dfrac{352\ N}{68\ kg}$

$a=5.17\ m/s^2$

Hence, this is the required solution.

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3 years ago

oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student

12345 [234]

Answer:

Mass, m = 6.18 kg

Explanation:

Given the following data;

Frequency, F = 10 Hz

Spring constant, k = 250 N/m

We know that pie, π = 22/7

To find the mass, we would use the following formula;

F = 1/2π√(k/m)

Where;

F is the frequency of oscillation.

k is the spring constant.

m is the mass of the spring.

Substituting into the formula, we have;

10 = 1/2 * 22/7 * √250/m

10 = 22/14 * √250/m

Cross-multiplying, we have;

140 = 22 * √250/m

Dividing both sides by 22, we have;

140/22 = √250/m

6.36 = √250/m

Taking the square of both sides, we have;

6.36² = (√250/m)²

40.45 = 250/m

Cross-multiplying, we have;

40.45m = 250

Mass, m = 250/40.45

Mass, m = 6.18 kg

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3 years ago

A speeder passes a parked police car at a constant speed of 23.3 m/s. At that instant, the police car starts from rest with a un

KonstantinChe [14]

Answer:

32s

Explanation:

We must establish that by the time the police car catches up to the speeder, both have travelled a certain distance during the same amount of time. However, the police car experiences accelerated motion whereas the speeder travels at a constant velocity. Therefore we will establish two formulas for distance starting with the speeder's distance:

$x=vt=23.3\frac{m}{s}t$

and the police car distance:

$x=vt+\frac{at^{2}}{2}=0+\frac{2.75\frac{m}{s^{2}} t^{2}}{2}=0.73\frac{m}{s^{2}}$

Since they both travel the same distance x, we can equal both formulas and solve for t:

$0 = 0.73\frac{m}{s^{2}}t^{2}-23.3\frac{m}{s} t\\\\0=t(0.73\frac{m}{s^{2}}t-23.3\frac{m}{s} )\\\\$

Two solutions exist to the equation; the first one being $t=0$

The second solution will be:

$0.73\frac{m}{s^{2}}t=23.3\frac{m}{s}\\\\t=\frac{23.3\frac{m}{s}}{0.73\frac{m}{s^{2}}}=32s$

This result allows us to confirm that the police car will take 32s to catch up to the speeder

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3 years ago

The tongue weight of a trailer should be what percent of the gross trailer weight rating

mario62 [17]

Answer:

between 10 and 15 percent

Explanation:

How to put your load

- First load the heavy

The safe trailer starts loading correctly. Uneven weight can affect steering, brakes and swing control.

In general, 60% of the weight of the load should be in the front half of the trailer and 40% in the rear half (unless the manufacturer indicates something different). When you place the load, you want it to be balanced from side to side, keeping the center of gravity near the ground and on the axle of the trailer.

- Hold your load

After balancing the load, you must hold it in place. An untapped load can move when the vehicle is moving and cause trailer instability.

- Trailer weight

To avoid overloading the trailer, look for the recommended weight rating. It is located on the VIN plate in the trailer chassis, usually on the tongue. Confirm the Gross Vehicle Weight Classification (GVWR) before towing.

GVWR: is the total weight that the trailer can support, including its weight. You can also find this number as the Gross Trailer Weight (GTW). The weight of the tongue should be 10-15% of the GTW.

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3 years ago

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Leno4ka [110]

Answer:

Explanation:

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3 years ago