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3 years ago

a water heater has a power rating of 1 kW. how many seconds will this heater take to boil 1 liter of water?

1 answer:

5 0

Answer:
4.65 minutes.

Explanation:
Theoretical Assumptions (not realistic): 1) ALL of the heat energy is transferred to the water, and 2)the water is uniformly heated. 3) The water density is 1.0g/mL.
Because we are only heating it TO the “boiling point”, ONLY the energy to heat the water from 20 to 100’C using the heat capacity or specific heat of water is required. A “Watt” is 1 Joule/second. The specific heat of water is 4.184 J/g-’C. Water mass is 1000g. The energy required to heat it is therefore
1000
g
⋅
4.184
(
J
g
’
C
)
⋅
(
100
−
20
)
’
C
=
334720
J

The heater supplies 1200 J/s, so the time to provide this heat to the water is
334720
J
1200
(
J
s
)
=
278.9
s
or 4.65 minutes.

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During a circus act, an elderly performer thrills the crowd by catching a cannon ball shot at him. The cannon ball has a mass of

olasank [31]

Answer:

3.416 m/s

Explanation:

Given that:

mass of cannonball $m_A$ = 72.0 kg

mass of performer $m_B$ = 65.0 kg

The horizontal component of the ball initially $\mu_{xA}$ = 6.50 m/s

the final velocity of the combined system v = ????

By applying the linear momentum of conservation:

$m_A \mu_{xA}+m_B \mu_{xB} = (m_A+m_B) v$

$72.0 \ kg \times 6.50 \ m/s+65.0 \ kg \times 0 = (72.0 \ kg+65.0 \ kg) v$

$468 kg m/s + 0 = (137 kg)v$

$v = \dfrac{468\ kg m/s }{137 \ kg}$

v = 3.416 m/s

8 0

3 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago

What is endurance? a). the ability to run faster b). a combination of balance and coordination c). how much you can stretch d).

Alex787 [66]

Answer:

D. the ability to exercise for longer periods of time

Explanation:

For example, when someone does endurance training, they are stretching their body's ability to do a certain exercise for longer times as opposed to increasing strength.

8 0

3 years ago

In what two ways can we increase the potential difference?

vredina [299]

Yes I hope this helps

4 0

3 years ago

A group of runners complete a 26.2 mile marathon in 3.4 hours. The distance between the start and finish lines is 12.2 miles. Wh

NNADVOKAT [17]

26.2/3.4 would be the average velocity for the run.

7.7 miles/hr

8 0

3 years ago

Read 2 more answers