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3 years ago

select and explain three features of the solar panel that maximises the final temperature of the water

Physics

1 answer:

MatroZZZ [7]3 years ago

7 0

Answer:

Irradiance (sunlight intensity or power), in Watts per square meter falling on a flat surface. ... Air Mass refers to “thickness” and clarity of the air through which the sunlight passes to reach the modules (sun angle affects this value). ... Cell temperature , which will differ from ambient air temperature

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Both Slicing a tomato and a chemical change has a burning toast cannot be reversed. however, why is slicing a tomato still consi

olchik [2.2K]

Because the tomato did not chemically changed it was only physically altered.

Hope this helps! :D

6 0

4 years ago

Read 2 more answers

A child whirls a ball in a vertical circle. Assuming the speed of the ball is constant (an approximation), when would the tensio

BARSIC [14]

Answer:

C. At the bottom of the circle.

Explanation:

Lets take

Radius of the circle = r

Mass = m

Tension = T

Angular speed = ω

The radial acceleration towards = a

a= ω² r

Weight due to gravity = mg

<h3>At the bottom condition</h3>

T - m g = m a

T = m ω² r + m g

<h3>At the top condition</h3>

T + m g = m a

T= m ω² r -m g

From above equation we can say that tension is grater when ball at bottom of the vertical circle.

Therefore the answer is C.

C. At the bottom of the circle.

8 0

3 years ago

PLEASE HELP Due today!

BigorU [14]

So i believe is exercise:)

7 0

3 years ago

Two blocks on a frictionless horizontal surface are on a collision course. one block with mass 0.6 kg moves at 0.8 m/s to the ri

Elanso [62]

First we can say that since there is no external force on this system so momentum is always conserved.

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

$0.6*0.8 + 1.2*0 = 0.6*v_{1f} + 1.2*v_{2f}$

$0.48= 0.6*v_{1f} + 1.2*v_{2f}$

$0.8 = v_{1f} + 2v_{2f}$

now by the condition of elastic collision

$v_{2f} - v_{1f} = 0.8 - 0[\tex]now add two equations[tex]3*v_{2f} = 1.6$

$v_{2f} = 0.533 m/s$

also from above equation we have

$v_{1f} = -0.267 m/s$

So ball of mass 0.6 kg will rebound back with speed 0.267 m/s and ball of mass 1.2 kg will go forwards with speed 0.533 m/s.

8 0

4 years ago

A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of

frosja888 [35]

1. 27.3 m/s

The velocity of the gazelle at any time is given by:

$v=u+at$

where

u is the initial velocity

a is the acceleration

t is the time

Here we have:

u = 0 (the gazelle starts from rest)

$a=4.2 m/s^2$

t = 6.5 s

Substituting the data, we find the gazelle's top speed:

$v=0+(4.2)(6.5)=27.3 m/s$

2. 3.8 s

The distance covered by the gazelle is

d = 30 m

We know that the gazelle accelerates during the first part of the motion and then it continues at constant speed. We need to find first if the gazelle completes the race during the first part of its motion (accelerated motion); to do this, we can calculate what would be the distance covered by the gazelle before reaching the top speed, after t = 6.5 s:

$d'=\frac{1}{2}at^2 = \frac{1}{2}(4.2)(6.5)^2=88.7 m$