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3 years ago

Thermodynamics

Physics

1 answer:

Akimi4 [234]3 years ago

7 0

Answer:

E = 3.8 kJ

Explanation:

Given that,

The mass of the object, m = 10 g = 0.01 kg

The heat of fusion of aluminum is 380 kJ/kg

We need to find the energy required to melt the mass of the aluminium. It can be calculated as follows:

E = mL

So,

E = 0.01 × 380

E = 3.8 kJ

So, the energy required to melt the mass is equal 3.8 kJ.

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A 150 kg uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. Calculate the magn

xenn [34]

Answer:

T = 2010 N

Explanation:

m = mass of the uniform beam = 150 kg

Force of gravity acting on the beam at its center is given as

W = mg

W = 150 x 9.8

W = 1470 N

T = Tension force in the wire

θ = angle made by the wire with the horizontal = 47° deg

L = length of the beam

From the figure,

AC = L

BC = L/2

From the figure, using equilibrium of torque about point C

T (AC) Sin47 = W (BC)

T L Sin47 = W (L/2)

T Sin47 = W/2

T Sin47 = 1470

T = 2010 N

6 0

3 years ago

A Carnot engine whose high-temperature reservoir is at 620 k takes in 550 J of heat at this temperature in each cycle and gives

yarga [219]

Answer:

(a) W=217 J

(b) Tc=378K

Explanation:

For part (a)

We to calculate the mechanical work W the engine does. By knowing QC and QH can obtain the work using equation

W = IQHI — IQcl .....................eq(1)

Put given values for QH and QC into equation (1) to get

the mechanical work of the engine

W = 550 - 335

W=217 J

For part (b)

We want to determine the temperature of low temperature reservoir which means Tc

IQc|/|Qh| =TC/TH

for Tc

Tc=(IQc|/|Qh|)*TH

Now we can put values

Tc= 620K (335/ 550.1)

Tc=378K

For part (c)

Here we want to find the thermal efficiency (e) of the cycle

e=1-TC/TH

e=1-(378/620)

e=0.39=39%

6 0

3 years ago

A 2290 kg car traveling at 10.5 m/s collides with a 2780 kg car that is initially at rest at the stoplight. The cars stick toget

avanturin [10]

Answer:

0.41

Explanation:

given,

mass of the car, m = 2290 Kg

initial speed = 10.5 m/s

mass of another car, M = 2780 Kg

distance moved = 2.80 m

coefficient of friction = ?

conservation of energy

m u = (M + m) V

2290 x 10.5 = (2290 + 2780) V

V = 4.74 m/s

using equation of motion

v² = u² + 2 a s

4.74² = 2 x a x 2.8

a = 4.02 m/s²

now using equation

a = μ g

4.02 = μ x 9.8

μ = 0.41

7 0

3 years ago

A 10-ohm resistor has a constant current. If 1200 C of charge flow through it in 4 minutes what

Amanda [17]

Answer:

B 5.0 A .

Explanation:

Hello.

In this case, since we know the charge (1200 C), time (4 min =240 s) and resistance (10Ω) which is actually not needed here, we compute the current as follows:

$I=\frac{Q}{t}$

Then, for the given data, we obtain:

$I=\frac{1200C}{4min}*\frac{1min}{60s}\\\\I=5A$

Therefore, answer is B 5.0 A .

Best regards!

4 0

3 years ago

During a very quick stop, a car decelerates at 28.4 rad/s?. Assume the tires initially rotated in the positive direction and rad

Damm [24]

Answer:

a) 24

b) 3.3 sec

c) 29.8 m/s

d) 48.85 m

Explanation:

α = angular acceleration = - 28.4 rad/s²

r = radius of the tire = 0.32 m

w₀ = initial angular velocity = 93 rad/s

w = final angular velocity = 0 rad/s

θ = angular displacement

Using the equation

w² = w₀² + 2αθ

0² = 93² + 2 (- 28.4) θ

θ = 152.3 rad

n = number of revolutions

Number of revolutions are given as

$n = \frac{\theta }{2\pi }$

$n = \frac{152.3 }{2(3.14) }$

$n = 24$

t = time taken to stop

using the equation

w = w₀ + αt

0 = 93 + (- 28.4) t

t = 3.3 sec

v₀ = initial velocity of the car

initial velocity of the car is given as

v₀ = r w₀ = (0.32) (93) = 29.8 m/s

v = final velocity = 0 m/s

a = linear acceleration = rα = (0.32) (- 28.4) = - 9.09 m/s²

d = distance traveled by car before stopping

Using the equation

v² = v₀² + 2 a d

0² = 29.8² + 2 (- 9.09) d

d = 48.85 m

8 0

4 years ago