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DaniilM [7]
3 years ago
13

When you turn on the hot water to wash dishes, the water pipes have to heat up. How much heat is absorbed by a copper water pipe

with a mass of 2.3 kg when its temperature is raised from 20.0C to 80.0 C?
Physics
1 answer:
bearhunter [10]3 years ago
7 0

Answer:

53,130 J

Explanation:

When a certain substance absorbs heat, the temperature of the substance increases according to the equation:

Q=mC\Delta T

where

Q is the amount of heat absorbed

m is the mass of the substance

C the specific heat capacity

\Delta T the change in temperature of the substance

In this problem:

m = 2.3 kg is the mass of copper

\Delta T=80.0C-20.0C=60.0^{\circ}C is the increase in temperature

C=385J/kgC is the specific heat of copper

So, the amount of heat absorbed is:

Q=(2.3)(385)(60)=53,130 J

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Which of the following is required for work to be done on an object?
ad-work [718]
I think that in order for work to be done, the object must move in the direction of the force and move over a distance.
7 0
3 years ago
A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calcu
galina1969 [7]

Question:

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

(a)60 (b)90 (c)120

Answer:

(a)5.42 N (b)6.26 N (c)5.42 N

Explanation:

From the question

Length of wire (L) = 2.80 m

Current in wire (I) = 5.20 A

Magnetic field (B) = 0.430 T

Angle are different in each part.

The magnetic force is given by

F=I \times B \times L \times sin(\theta)

So from data

F = 5.20 A \times 0.430 T \times 2.80 sin(\theta)\\\\F=6.2608 sin(\theta) N

Now sub parts

(a)

\theta=60^{o}\\\\Force = 6.2608 sin(60^{o}) N\\\\Force = 5.42 N

(b)

\theta=90^{o}\\\\Force = 6.2608 sin(90^{o}) N\\\\Force = 6.26 N

(c)

\theta=120^{o}\\\\Force = 6.2608 sin(120^{o}) N\\\\Force = 5.42 N

3 0
3 years ago
A mass m = 14 kg is pulled along a horizontal floor with NO friction for a distance d =5.7 m. Then the mass is pulled up an incl
frosja888 [35]

Answer:

W ≅ 292.97 J

Explanation:

1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)

Workdone by the tension before the block goes up the incline on the horizontal surface can be calculated using the expression;

W = (Fcosθ)d

Given that:

Tension of the force = 62 N

angle of incline θ =  34°

distance d =5.7 m.

Then;

W = 62 × cos(34) × 5.7

W = 353.4 cos(34)

W = 353.4 × 0.8290

W = 292.9686 J

W ≅ 292.97 J

Hence,  the work done by tension before the block goes up the incline = 292.97 J

8 0
3 years ago
Two forces one of 12N and another of 5N act on a body in such away that they makes an angle of 90 digre with each other,what is
ELEN [110]

Answer:

Sorry I didn't know the answer

6 0
3 years ago
B. Jerome plays middle linebacker for South's varsity football team. In a game against
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Answer:

Option D

670 Kg.m/s

Explanation:

Initial momentum is given by mv=82*5.6=459.2 Kg.m/s (taking eastward as positive)

Final momentum is also mv but v being westward direction, we take it negative

Final momentum=82*-2.5= -205 Kg.m/s

Change in momentum=Final momentum-Initial momentum=-205-459.2=-664.2 Kg.m/s

Impulse=change in momentum=664.2 Kg.m/s rounded off as 670 Kg.m/s

4 0
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