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3 years ago

A LETTER FROM THE LORAX

Physics

2 answers:

KengaRu [80]3 years ago

5 0

Answer:

Explanation:

You could try to say how helpful they are what they are and what they do

Kitty [74]3 years ago

5 0

Answer:

you could say how helpful they are and how they help you

Explanation:

hope this helps

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Creating plasma can be dangerous because of the high amount of ___ needed to create it.

Allushta [10]

<span>Creating plasma can be dangerous because of the high amount of ENERGY needed to create it.</span>

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3 years ago

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What is the speed of a wave that has a frequency of 125 Hz and a wavelength of 1.25 meters? Express your answer to the nearest w

shtirl [24]

156 is the answer. so 156.25 is almost the same thing, you just round. It's not hard. Thank you!!!

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3 years ago

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A 20 kg truck drives in a circle of radius 4 m at 10m/s. What is the

Tems11 [23]

Mass=m=20kg
Radius=r=4m
Velocity=10m/s=v

We know

$\boxed{\sf F_c=\dfrac{mv^2}{r}}$

$\\ \sf\longmapsto F_c=\dfrac{20(10)^2}{4}$

$\\ \sf\longmapsto F_c=\dfrac{20(100)}{4}$

$\\ \sf\longmapsto F_c=\dfrac{2000}{4}$

$\\ \sf\longmapsto F_c=500N$

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3 years ago

A Brayton cycle has air into the compressor at 95 kPa, 290 K, and has an efficiency of 50%. The exhaust temperature is 675 K. Fi

motikmotik

Answer:

The specific heat addition is 773.1 kJ/kg

Explanation:

from table A.5 we get the properties of air:

k=specific heat ratio=1.4

cp=specific heat at constant pressure=1.004 kJ/kg*K

We calculate the pressure range of the Brayton cycle, as follows

n=1-(1/(P2/P1)^(k-1)/k))

where n=thermal efficiency=0.5. Clearing P2/P1 and replacing values:

P2/P1=(1/0.5)^(1.4/0.4)=11.31

the temperature of the air at state 2 is equal to:

P2/P1=(T2/T1)^(k/k-1)

where T1 is the temperature of the air enters the compressor. Clearing T2

11.31=(T2/290)^(1.4/(1.4-1))

T2=580K

The temperature of the air at state 3 is equal to:

P2/P1=(T3/T4)^(k/(k-1))

11.31=(T3/675)^(1.4/(1.4-1))

T3=1350K

The specific heat addition is equal to:

q=Cp*(T3-T2)=1.004*(1350-580)=773.1 kJ/kg

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3 years ago

How do you graph distance and time for an object that moves at a constant speed?

LekaFEV [45]

Answer:

It would be a straight line

Explanation:

On a distance-time graph, an object that moves at constant speed would be represented by a straight line.

In fact, in a distance-time graph, the slope of the line corresponds to the speed of the object. We can demonstrate that. In fact:

- The speed of the object is equal to the ratio between the distance covered $(\Delta s)$ and the time taken ( $\Delta t$ ):

$v=\frac{\Delta s}{\Delta t}$

On a distance-time graph, the distance is on the y-axis while the time is on the x-axis. The slope of the line is defined as:

$m=\frac{\Delta y}{\Delta x}$

But the variation on the y-axis ( $\Delta y$ ) is equal to the distance covered ( $\Delta s$ ), while the variation on the x-axis $(\Delta x)$ corresponds to the time taken ( $\Delta t$ ), so the slope can also be rewritten as

$m=\frac{\Delta s}{\Delta t}$

which is equal to the speed of the object. Therefore, an object moving at constant speed would be represented by a line with constant slope, which means a straight line.

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