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4 years ago

How much heat is absorbed by a 47g iron skillet when its temperature rises from 12oC to 20oC?

Physics

1 answer:

jolli1 [7]4 years ago

6 0

Answer

169.2 J

Explanation

Given in the question,

mass of iron = 47g

specific heat capacity of iron = 0.450 (J/g 0C)

initial temperature = 12° C

final temperature = 20° C

The energy q needed to increase an object of mass m and specific heat capacity c by a temperature θ is given by:

q = mcΔt

q = 47(0.45)(20-12)

q = 169.2 J

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Từ độ cao 100 m người ta thả một vật thẳng đứng xuống với v = 10 m/s, g = 10 m/s2 . a. Sau bao lâu vật chạm đất. b. Tính vận tốc

Bas_tet [7]

Answer:

10 points dapat yawa aman daw gud

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3 years ago

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Suppose a 4.0-kg projectile is launched vertically with a speed of 8.0 m/s. What is the maximum height the projectile reaches?

eduard

Answer:

h = 3.3 m (Look at the explanation below, please)

Explanation:

This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy

Kinetic energy = $\frac{1}{2}$ m $v^{2}$

Plug in the numbers = $\frac{1}{2}$ (4.0)( $8^{2}$ )

Solve = 2(64) = 128 J

Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.

Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)

Kinetic energy = Potential Energy

128 J = 39.2h

h = 3.26 m

h= 3.3 m (because of significant figures)

7 0

3 years ago

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400

liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, $f_{ra} = \mu mg\\$ ............(1)

Since frictional force is a type of force, we are safe to say $f_{ra} = ma$ .......(2)

Equating (1) and (2)

$ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}$

a) Speed of A at the start of the sliding

$d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s$

b) Speed of B at the start of the sliding

$d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s$

Let the speed of car B before collision = $v_{B1}$

Momentum of car B before collision = $m_{B} v_{B1}$

Momentum after collision = $m_{A} v_{A} + m_{B} v_{B2}$

Applying the law of conservation of momentum:

$m_{B} v_{B1} = m_{A} v_{A} +m_{B} v_{B2}$

$m_{A} = 1100 kg\\m_{B} = 1400 kg$

$(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s$

3 0

4 years ago

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The mass of the bicycle and rider is 60 kg and the total area of the tyres in contact with

AleksAgata [21]

Answer:

a little

Explanation:

First of all, it's not how you spell "tyres", it is tires.

Second of all, you already know the Mass so what you need to find out now is contact the road. You Know that your number and letter are squared so that would turn into 6m x 2.4. Then you do the math do continue on to finish it. Have a great day!! Good luck with the answer!!

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3 years ago

Which of the following is equal to an impulse of 15 units?

strojnjashka [21]

Answer:

B) Force = 7.5, Time = 2 is equal to an impulse of 15 units

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3 years ago

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