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Vladimir [108]
3 years ago
14

Match the type of heat transfer with its description

Physics
1 answer:
VikaD [51]3 years ago
5 0

Answer:

1. Convection

2. Radiation

3. Conduction

Hope this helps!

Explanation:

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If my final exam is worth 30% and my class average is 80, what do I need to get to pass the exam?
kumpel [21]
You haven't told us what the passing percentage is on the exam,
or what the passing percentage is for the semester, or any of that.


5 0
3 years ago
Select the decimal that is equivalent to 29/37​
morpeh [17]

Answer:

0.78

Explanation:

6 0
2 years ago
In a digital multimeter, what changes inside the meter when you push the button to change from 200v range to the 20v range? by w
charle [14.2K]
So we want to know what changes inside the multimeter when we change the voltage range from 200 V to  20 V, by what factor and does it increase or decrease. What we want when trying to measure the voltage with a multimeter is that a minimal current passes trough the mulitmeter so when we change the voltage range, we decrease the resistance by a factor of 10 because the voltage is decreased by a factor of 10. 
6 0
3 years ago
A concrete piling of 50 kg is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. How much will the wire stretch?
pentagon [3]

Explanation:

It is given that,

Mass of concrete pilling, m = 50 kg

Diameter of wire, d = 1 mm

Radius of wire, r = 0.0005 m

Length of wire, L = 11.2

Young modulus of steel, Y=20\times 10^{10}\ N/m^2

The young modulus of a wire is given by :

Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}

Y=\dfrac{F.L}{A\Delta L}

\Delta L=\dfrac{F.L}{A.Y}

\Delta L=\dfrac{50\ kg\times 9.8\ m/s^2\times 11.2\ m}{\pi (0.0005\ m)^2\times 20\times 10^{10}\ N/m^2}

\Delta L=0.034\ m

So, the wire will stretch 0.034 meters. Hence, this is the required solution.

8 0
3 years ago
The energy released from condensation in thunderstorms can be very large. Calculate the energy (in J) released into the atmosphe
makkiz [27]

Answer:

7.19 * 10^14J

Explanation:

Given that

Density of water Pwater= 1000kg/m3

R=2.1km = 2.1*10^3m

H= 2.3cm. = 2.3*10^-2m

Lv water= 2256 * 10^3J/kg

First, mass of water need to be calculated, using an imaginary cylinder

Density= Mass /Volume

Mass= Density* Volume

Volume of a cylinder= πR2h

Therefore mass= PπR2H

= 1000 * π * (2.1 *10^3)^2 * (2.3 * 10^-2)

= 3.18 *10^8

Heat Released Qv = mLV

= 3.18*10^8 * 2236*10^3

= 7.19 * 10^14J

7 0
3 years ago
Read 2 more answers
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