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zavuch27 [327]
2 years ago
3

Find the final equilibrium temperature when 10.0 g of milk at 10.0°C is added to 160 g of coffee at 90.0°C.

Physics
1 answer:
GaryK [48]2 years ago
8 0

Answer:

The final equilibrium temperature is T = 85.3 °C.

Explanation:

Knowledge of specific heats and/or heat capacities and the fact that energy must be conserved allows us to determine the equilibrium temperature of two objects initially at different temperatures by demanding that,

                Heat lost by hot object +  Heat gained by cold object = 0

where we ignore heat gained or lost from/to the surroundings.

We can calculate the heat released or absorbed using the specific heat capacity C, the mass of the substance m, and the change in temperature ΔT in the equation:

q=m\times C \times \Delta T

where \Delta T=T_{final}-T_{initial}

Use the equation for heat transfer q=m\times C \times \Delta T to express the heat lost by the coffee

q_{hot}=160\times 1.00\times(T_f-90)

Express the heat gained by the milk

q_{cold}=10.0\times 1.00\times(T_f-10)

Use the law of conservation of energy and solve for T_f

                      Heat lost by hot object + Heat gained by cold object =0

160\times 1.00\times(T_f-90)+10.0\times 1.00\times(T_f-10)=0\\170T_f-14500=0\\T_f=\frac{14500}{170}= 85.29

The final equilibrium temperature is T = 85.3 °C.

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Snezhnost [94]

Explanation:

(a) Displacement of an object is the shortest path covered by it.

In this problem, a student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag.  She travels 0.3 km south to retrieve her item. She then travels 0.4 mi north to arrive at school.

0.4 miles = 0.64 km

displacement = 0.7-0.3+0.64 = 1.04 km

(b) Average velocity = total displacement/total time

t = 15 min = 0.25 hour

v=\dfrac{1.04\ km}{0.25\ h}\\\\v=4.16\ km/h

Hence, this is the required solution.

8 0
2 years ago
Two blocks, with masses m and 3m, are attached to the ends of a string with negligible mass that passes over a pulley, as shown
olasank [31]

Answer:

a)  v = √ g x , b)  W = 2 m g d , c)    a = ½ g

Explanation:

a) For this exercise we use Newton's second law, suppose that the block of mass m moves up

            T-W₁ = m a

            W₃ - T = M a

            w₃ - w₁ = (m + M) a

            a = (3m - m) / (m + 3m) g

            a = 2/4 g

            a = ½ g

the speed of the blocks is

          v² = v₀² + 2 ½ g x

          v = √ g x

b) Work is a scalar, therefore an additive quantity

light block s

           W₁ = -W d = - mg d

3m heavy block

             

            W₂ = W d = 3m g d

the total work is

             W = W₁ + W₂

             W = 2 m g d

c) in the center of mass all external forces are applied, they relate it is

                      a = ½ g

8 0
2 years ago
Read 2 more answers
Materials have unique properties because each one is made up of different kinds of which particle?
inysia [295]

D. Atoms.

Explanation:

All the matter is made of elementary particles called "atoms".

Further, an atom is made of electrons, protons and neutrons. The electrons & protons are again made of the fundamental sub-particles, electrons (leptons) and the protons(quarks).

The classification of particles is shown in the figure attached


7 0
2 years ago
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Merry-go-rounds are a common ride in park playgrounds. The ride is a horizontal disk that rotates about a vertical axis at their
Vera_Pavlovna [14]

Answer:

A = 2.36m/s

B = 3.71m/s²

C = 29.61m/s2

Explanation:

First, we convert the diameter of the ride from ft to m

10ft = 3m

Speed of the rider is the

v = circumference of the circle divided by time of rotation

v = [2π(D/2)]/T

v = [2π(3/2)]/4

v = 3π/4

v = 2.36m/s

Radial acceleration can also be found as a = v²/r

Where v = speed of the rider

r = radius of the ride

a = 2.36²/1.5

a = 3.71m/s²

If the time of revolution is halved, then radial acceleration is

A = 4π²R/T²

A = (4 * π² * 3)/2²

A = 118.44/4

A = 29.61m/s²

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2 years ago
A person drops a stone down a well and hears the echo 8.9 s later. if it takes 0.9 s for the echo to travel up the well, approxi
Temka [501]

Total time in between the dropping of the stone and hearing of the echo = 8.9 s

Time taken by the sound to reach the person = 0.9 s

Time taken by the stone to reach the bottom of the well = 8.9 - 0.9 = 8 seconds

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Acceleration due to gravity (g) = 9.8 m/s^2

Time taken (t) = 8 seconds

Let the depth of the well be h.

Using the second equation of motion:

h = ut + \frac{1}{2}\times a \times t^2

h = 0 \times 8 + \frac{1}{2} \times 9.8 \times 8^2

h = 313.6 m

Hence, the depth of the well is 313.6 m

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