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2 years ago

The value of specific heat for copper is 390 J/kg⋅C∘, for aluminun is 900 J/kg⋅C∘, and for water is 4186 J/kg⋅C∘.

Physics

1 answer:

abruzzese [7]2 years ago

5 0

Answer:

The equilibrium temperature is

21.97°c

Explanation:

This problem bothers on the heat capacity of materials

Given data

specific heat capacities

copper is Cc =390 J/kg⋅C∘,

aluminun Ca = 900 J/kg⋅C∘,

water Cw = 4186 J/kg⋅C∘.

Mass of substances

Copper Mc = 235g

Aluminum Ma = 135g

Water Mw = 825g

Temperatures

Copper θc = 255°c

Water and aluminum calorimeter θ1= 16°c

Equilibrium temperature θf =?

Applying the principle of conservation of heat energy, heat loss by copper equal heat gained by aluminum calorimeter and water

McCc(θc-θf) =(MaCa+MwCw)(θf-θ1)

Substituting our data into the expression we have

235*390(255-θf)=

(135*900+825*4186)(θf-16)

91650(255-θf)=(3574950)(θf-16)

23.37*10^6-91650*θf=3.57*10^6θf- +57.2*10^6

Collecting like terms and rearranging

23.37*10^6+57.2*10^6=3.57*10^6θf+91650θf

8.2*10^6=3.66*10^6θf

θf=80.5*10^6/3.6*10^6

θf =21.97°c

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A hydroelectric dam holds back a lake of surface area 3.0×106m2 that has vertical sides below the water level. The water level i

Ber [7]

Answer:

4.41 × 10¹² J, 2.72 × 10³ m³, 0.907 × 10 ⁻³ m

Explanation:

Gravitational potential energy = mgh

where m is mass in kg, g is acceleration due to gravity in m/s², and h is the distance from the base of the dam.

mass of the surface water = density of water × volume of water × 1 m = 1000 kg / m³ × 3.0 × 10⁶ m² × 1 m = 3 × 10⁹ kg

Gravitational potential energy = 3 × 10⁹ kg × 9.81 m/s² × 150 m = 4.41 × 10¹² J

b)what volume of water must pass through the dam to produce 1000 kw-hrs

1 000 kw-hr = 3.6 × 10 ⁹ J

the dam has mechanical energy conversion of 90% to electrical energy

Gravitational potential energy needed = 3.6 × 10 ⁹ J / 0.9 = 4 × 10⁹ J

mass of water needed = Energy required / g h = 4 × 10⁹ J / (9.81 m/s² × 150 m) = 2.718 × 10 ⁶ kg

density = mass / volume

volume = mass / density = 2.718 × 10 ⁶ kg / (1000 kg/ m³) = 2.72 × 10³ m³

the distance the level of the water in the lake fell = volume / area = 2.72 × 10³ m³ / (3.0×10⁶ m²) = 0.907 × 10 ⁻³ m

8 0

2 years ago

A box is sliding with a speed of 4.50 m/s on a horizontal surface when, at point P, it encounters a rough section. The coefficie

Rudiy27

Answer:

a) xf = 5.1 m

b) u = 0.304

c) x = 10.3 m

Explanation:

we will use the following formula:

u = 0.1 + A*x

Si x = 12.5 m, u = 0.6

Clearing A:

A = 0.5/12.5 = 0.04 m^-1

a) we have to:

W = Ekf - Eki

where Ekf = final kinetic energy

Eki = initial kinetic energy

9.8*(0.1xf + ((0.04*xf^2)/(2))) = (4.5^2)/(2)

Clearing xf, we have:

xf = 5.1 m

b) Replacing values for u:

u = 0.1 + (0.04*5.1) = 0.304

c) Wf = Ekf - Eki

-u*m*x*g = 0 - (m*v^2)/2

Clearing x:

x = v^2/(2*u*g) = (4.5^2)/(2*0.1*9.8) = 10.3 m

4 0

2 years ago

Read 2 more answers

Maximum voltage produced in an AC generator completing 60 cycles in 30 sec is 250V. (a) What is period of armature? (b) How many

Vlada [557]

Answer:

a. 2 Hz b. 0.5 cycles c . 0 V

Explanation:

a. What is period of armature?

Since it takes the armature 30 seconds to complete 60 cycles, and frequency f = number of cycles/ time = 60 cycles/ 30 s = 2 cycles/ s = 2 Hz

b. How many cycles are completed in T/2 sec?

The period, T = 1/f = 1/2 Hz = 0.5 s.

So, it takes 0.5 s to complete 1 cycles. At t = T/2 = 0.5/2 = 0.25 s,

Since it takes 0.5 s to complete 1 cycle, then the number of cycles it completes in 0.25 s is 0.25/0.5 = 0.5 cycles.

c. What is the maximum emf produced when the armature completes 180° rotation?

Since the emf E = E₀sinθ and when θ = 180°, sinθ = sin180° = 0

E = E₀ × 0 = 0

E = 0

So, at 180° rotation, the maximum emf produced is 0 V.

8 0

2 years ago

Some of the fastest dragsters (called "top fuel) do not race for more than 300-400m for safety reasons. Consider such a dragster

Masja [62]

Answer:

1.10261 times g

416.17506 mph

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 400=0\times 8.6+\frac{1}{2}\times a\times 8.6^2\\\Rightarrow a=\frac{400\times 2}{8.6^2}\\\Rightarrow a=10.81665\ m/s^2$

Dividing by g

$\dfrac{a}{g}=\dfrac{10.81665}{9.81}\\\Rightarrow \dfrac{a}{g}=1.10261\\\Rightarrow a=1.10261g$

The acceleration is 1.10261 times g

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 10.81665\times 1.6\times 10^3+0^2}\\\Rightarrow v=186.04644\ m/s$

In mph

$186.04644\times \dfrac{3600}{1609.34}=416.17506\ mph$

The speed of the dragster is 416.17506 mph

5 0

2 years ago

An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.

Alex17521 [72]

Answer:

The speed of the plane relative to the ground is 300.79 km/h.

Explanation:

Given that,

Speed of wind = 75.0 km/hr

Speed of plane relative to the air = 310 km/hr

Suppose, determine the speed of the plane relative to the ground

We need to calculate the angle

Using formula of angle

$\sin\theta=\dfrac{v'}{v}$

Where, v'=speed of wind

v= speed of plane

Put the value into the formula

$\sin\theta=\dfrac{75}{310}$

$\theta=\sin^{-1}(\dfrac{75}{310})$

$\theta=14.0^{\circ}$

We need to calculate the resultant speed

Using formula of resultant speed

$\cos\theta=\dfrac{v''}{v}$

Put the value into the formula

$\cos14=\dfrac{v''}{310}$

$v''=\cos14\times310$

$v''=300.79\ km/h$

Hence, The speed of the plane relative to the ground is 300.79 km/h.

6 0

2 years ago