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3 years ago

This is data from a visible satellite image. What types of weather conditions are associated with the image?

Physics

2 answers:

babunello [35]3 years ago

8 0

It look like a storm is forming out into the ocean and building pressure and maybe become a hurricane.
I hope this helps :)!!!

ser-zykov [4K]3 years ago

3 0

From the picture we can conclude that there's a hurricane forming somewhere above the sea.

In the picture, we can easily see the circular movement and arrangement of the air masses, and that is a sign that a cold (heavier) and warm (lighter) air masses have collided, thus mixing and creating a circular movement. Also we can notice the ''eye'' in the middle, which is a trademark of the hurricanes. So wherever this image may be taken, there's extremely strong winds and intense rainfall.

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How much energy does a pair of 800W hair straighteners transfer every second

Natali5045456 [20]

800 J since watts is a measurement of joules per second

3 0

4 years ago

A muscle inserts 1.5 cm from the joint axis and exerts 300 N of force at an angle of pull of 60 degrees. How much torque is prod

Amanda [17]

Answer:

torque = 3.897 N-m

Explanation:

given data

force = 300 N

angle = 60 degree

distance = 15 cm

to find out

torque

solution

we will apply here torque formula that is given below

torque = force × sinθ × distance ...................1

put here all these value in equation 1

we get torque

torque = force × sinθ × distance

torque = 300 × sin60 × 1.5 × $10^{-2}$

torque = 300 × 0.8660 × 1.5 × $10^{-2}$

torque = 259.80 × 1.5 × $10^{-2}$

torque = 389.711 × $10^{-2}$

torque = 3.897 N-m

8 0

3 years ago

The circumference of a sphere was measured to be

professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

$\phi = 76cm$

$Error (dr) = 0.5cm$

The radius then would be

$\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm$

And

$\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr$

PART A ) For the Surface Area we have that,

$A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}$

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

$\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)$

Maximum error:

$dA = 4(\frac{38}{\pi})(0.5)$

$dA = \frac{76}{\pi}cm^2$

The relative error is that between the value of the Area and the maximum error, therefore:

$\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}$

$\frac{dA}{A} = 0.01315 = 1.31\%$

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

$V = \frac{4}{3} \pi r^3$

$V = \frac{4}{3} \pi (\frac{38}{\pi})^3$

$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

$\frac{dV}{dr} = \frac{4}{3} 3\pi r^2$

$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

$dV = \frac{2888}{\pi^2} cm^3$

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

$\frac{dV}{V} = 3.947\%$

3 0

3 years ago

Si units are the modern form of the<br> Reset<br> Next

Zanzabum

Answer:

Reset

Explanation:

The International System of Units (SI, abbreviated from the French Système international (d'unités)) is the modern form of the metric system and is the most widely used system of measurement.

7 0

4 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

3 0

3 years ago