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8_murik_8 [283]
2 years ago
12

Rebecca heated 50mL of water from 0 degrees Celsius to 60 degrees Celsius. How much energy did she use to heat the water? Rememb

er: cal=m*deltaT
(WRITE THE WORK!/Give explanation.)
Physics
1 answer:
GuDViN [60]2 years ago
7 0

Answer:

Q = 12540  J

Explanation:

It is given that,

Mass of water, m = 50 mL = 50 g

It is heated from 0 degrees Celsius to 60 degrees Celsius.

We need to find the energy required to heat the water. The formula use to find it as follows :

Q=mc\Delta T

Where c is the specific heat of water, c = 4.18 J/g°C

Put all the values,

Q=50\times 4.18\times (60-0)\\Q=12540\ J

So, 12540 J of energy is used to heat the water.

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1. A signal source has an open-circuit voltage of 1V, and a short-circuit current of10mA. What is the source resistance
LiRa [457]

Answer:

100 Ω

Explanation:

Given that

Open circuit voltage, V = 1 V

Short circuit current, I = 10 mA

Source resistance R, = ?

This is rather a straight forward question. Remember Ohms Law? Current being directly proportional to the voltage and inversely proportional to the resistance?

Yeah, that's the formula we'd be using.

Ohms Law states that V = IR, and thus, if we make R subject of the formula, we have

R = V / I, on substituting the values, we have

R = 1 / 10*10^-3

R = 1 / 0.01

R = 100 Ω

8 0
3 years ago
The terminal velocity is not dependent on which one of the following properties? the drag coefficient 1 the force of gravity 2 c
ahrayia [7]
<h2>Answer: the falling time</h2>

Explanation:

When a body or object falls, basically two forces act on it:  

1. The force of air friction, also called<em> </em><u><em>"drag force"</em></u> D:  

D={C}_{d}\frac{\rho V^{2} }{2}A  (1)

Where:  

C_ {d} is the drag coefficient  

\rho is the density  of the fluid (air for example)

V is the velocity  

A is the transversal area of the object

So, this force is proportional to the transversal area of ​​the falling element and to the square of the velocity.  

2. Its <u>weight </u>due to the gravity force W:  

W=m.g

(2)

Where:  

m is the mass of the object

g is the acceleration due gravity  

So, at the moment <u>when the drag force equals the gravity force, the object will have its terminal velocity:</u>

D=W (3)

{C}_{d}\frac{\rho V^{2} }{2}A=m.g  (4)

V=\sqrt{\frac{2m.g}{\rho A{C}_{d}}}  (5) This is the terminal velocity

As we can see, there is no "falling time" in this equation.

Therefore, the terminal velocity is not dependent on the falling time.

6 0
2 years ago
How many significant digits are measurement 0.00210 mg?
jek_recluse [69]
There are 3 significant figures, if that answers the question.
8 0
3 years ago
Read 2 more answers
A uniform-density 7 kg disk of radius 0.21 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is
GREYUIT [131]

Answer:

\omega = 22.67 rad/s

Explanation:

Here we can use energy conservation

As per energy conservation conditions we know that work done by external source is converted into kinetic energy of the disc

Now we have

W = \frac{1}{2}I\omega^2

now we know that work done is product of force and displacement

so here we have

W = F.d

W = (44 N)(0.9 m) = 39.6 J

now for moment of inertia of the disc we will have

I = \frac{1}{2}mR^2

I = \frac{1}{2}(7 kg)(0.21^2)

I = 0.154 kg m^2

now from above equation we will have

39.6 = \frac{1}{2}(0.154)\omega^2

\omega = 22.67 rad/s

5 0
3 years ago
Read 2 more answers
A need the answers pleaseeeee
IRISSAK [1]

It's just asking you to sit down and COUNT the little squares in each sector.

It'll help you keep everything straight if you take a very sharp pencil and make a tiny dot in each square as you count it.  That way, you'll be able to see which ones you haven't counted yet, and also you won't count a square twice when you see that it already has a dot in it.

(If, by some chance, this is a picture of the orbit of a planet revolving around the sun ... as I think it might be ... then you should find that both sectors jhave the same number of squares.)  

7 0
3 years ago
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