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Licemer1 [7]
3 years ago
9

Why does cold water have less thermal energy than hot water?

Physics
1 answer:
beks73 [17]3 years ago
8 0
This is because the water molecules have less kinetic energy.
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An object with a charge of 0.9 x 10^-5 C is separated from a second object with a chare of 2.5 x 10^-4 C by a distance of 0.5 m.
meriva

Answer: Force = 81 N

Explanation:

from Columbs law,

F = k(q1*q2)/r²

k = 9 x 10^9 Nm²/C²

F = (9 x 10^9)x (0.9x10^-5 x 2.5x10^-4)/(0.5)²

F = 81 Newtons

4 0
3 years ago
A force of 25 newtons moves a box a distance of 4 meters in 5 seconds.
Simora [160]

By definition, we have that the work done is given by:

W = F * d

Where,

F: force in the direction of displacement

d: displaced distance

Substituting values we have:

W = (25) * (4)\\W = 100 Nm

Then, the power is given by:

P = \frac{W}{t}

Where,

t: time

Substituting values we have:

P = \frac{100}{5}

P = 20\frac{Nm}{s}

Answer:

The work done on the box is 100 Nm, and the power is 20 Nm/s.

5 0
3 years ago
Read 2 more answers
Which of the following instrument measure time most accurately
valkas [14]
C. Quarts watches is the correct answer
8 0
2 years ago
Which type of heat transfer takes place in gases by the movement of particles through a medium?
Sergeu [11.5K]

Answer: Convention

Explanation: Convention is pretty well known for being a process that transmits heat from one place to another place with the movements of heated particles. I got this answer from my notebook during my chemistry class.

Hope this answer helps!

4 0
2 years ago
Read 2 more answers
When the ambulance passes the person (switching from moving towards him to moving away from him), the perceived frequency of the
dlinn [17]

To develop this problem we will apply the considerations made through the concept of Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. At first the source is moving towards the observer. Than the perceived frequency at first

F_1 = F \frac{{343}}{(343-V)}

Where F is the actual frequency and v is the velocity of the ambulance

Now the source is moving away from the observer.

F_2 = F\frac{343}{(343+V)}

We are also so told the perceived frequency decreases by 11.9%

F_2 = F_1 - 9.27\% \text{ of } F_1

F_2 = F_1-0.0927F_1

F_2 = 0.9073F_1

Equating,

F\frac{343}{(343+V)}= 0.9073(F\frac{343}{(343-V)})

\frac{1}{(343+V)}= 0.9073\frac{1}{(343-V)}

0.9073(343+V) = 343-V

(0.9073)(343)+(0.9073)V = 343-V

V+0.9073V = 343-(0.9073)(343)

Solving for V,

V = 16.67 m/s

5 0
2 years ago
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