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Brilliant_brown [7]
2 years ago
15

What amount of heat is required to raise the temperature of 5 kg of water through 20° c?​

Mathematics
1 answer:
allochka39001 [22]2 years ago
7 0

Answer:

420,000 J/Kg

Step-by-step explanation:

energy= temperature change* mass* specific heat capacity

energy= 20*5*4200

energy= 420,000

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Rearrange the formula V = 1 3 πr2h for h. A) h = 3V πr2 B) h = V 3πr2 C) h = 3πr2 V D) h = πr2 3V
stiks02 [169]

Answer:

Option a) h=\frac{3V}{\pi r^{2} }

Step-by-step explanation:

The given formula seems to represent the volume of right circular cone. The correct formula is:

V=\frac{1}{3} \pi r^{2} h

We have to rearrange the formula for h. This means we have to move h on one side of the formula and all the other variables and constants on the other side of the equation. This can be done as shown below:

V=\frac{1}{3} \pi r^{2} h\\\\3V=\pi r^{2} h\\\\h=\frac{3V}{\pi r^{2} }

5 0
3 years ago
Factor the following expression completely. 16x^5-x^3<br> A. <br><br> B. <br><br> C. <br><br> D.
Ket [755]

Answer:

x^3(2x+1)(2x-1)

Step-by-step explanation:

first take common then use formula of a^2-b^2

6 0
3 years ago
Read 2 more answers
Help me asap thanks you.
liraira [26]
Ok i dont know the anwser but go to tiger algerbra and get the anwser there i hope you have a nice day
3 0
3 years ago
What is the area of the region bounded between the curves y=x and y=sqrt(x)?
erastova [34]

Answer:

\frac{1}{6}

Step-by-step explanation:

y = x     .....(1)

y=\sqrt{x}     .... (2)

By solving equation (1) and equation (2)

x = \sqrt{x}

\sqrt{x}\left ( \sqrt{x}-1 \right )=0

\sqrt{x}=0 or \left ( \sqrt{x}-1 \right )=0

x = 0, x = 1

y = 0, y = 1

A = \int_{0}^{1}ydx(curve)-\int_{0}^{1}ydx(line)

A = \int_{0}^{1}\sqrt{x}dx-\int_{0}^{1}xdx

A = \frac{2}{3}[x^\frac{3}{2}]_{0}^{1}-\frac{1}{2}[x^2]_{0}^{1}

A = \frac{2}{3}-\frac{1}{2}

A = \frac{1}{6}

8 0
3 years ago
15. Simplify sin(90° - O) cos 0 - sin(180° +0) sin e.​
dalvyx [7]

Answer:  cos²(θ) + sin(θ)sin(e)

<u>Step-by-step explanation:</u>

sin (90° - θ)cos(Ф) - sin(180° + θ) sin(e)

Note the following identities:

sin (90° - θ) = cos(x)

sin (180° + θ) = -sin(x)

Substitute those identities into the expression:

   cos(x)cos(x)  -  -sin(x)sin(e)

=  cos²(x) + sin(x)sin(e)

4 0
3 years ago
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