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3 years ago

5

Define the term pressure

2 answers:

Zielflug [23.3K]3 years ago

5 0

Pressure could possibly mean force on a certain type of area.....hope this helps!

Setler [38]3 years ago

3 0

Answer:

Pressure is the perpendicular force applied per unit area.

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Kathy 82 kg performer standing on a diving board at the carnival dive straight down into a small pool of water. Just before stri

mixas84 [53]

Solution :

Given weight of Kathy = 82 kg

Her speed before striking the water, $V_o$ = 5.50 m/s

Her speed after entering the water, $V_f$ = 1.1 m/s

Time = 1.65 s

Using equation of impulse,

$dP = F \times dT$

Here, F = the force ,

dT = time interval over which the force is applied for

= 1.65 s

dP = change in momentum

dP = m x dV

$= m \times [V_f - V_o]$

= 82 x (1.1 - 5.5)

= -360 kg

∴ the net force acting will be

$F=\frac{dP}{dT}$

$F=\frac{-360}{1.65}$

= 218 N

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

I think I know the answer but I am not sure

Romashka [77]

The answer is a

Hope this helps!

5 0

3 years ago

Read 2 more answers

Please help ASAP.

Bess [88]

Im not 100% sure you have to tell me if im wrong or not.

D

B

C

3 0

3 years ago

Read 2 more answers

PLS HELP FAST!!!!!!!

Elan Coil [88]

Answer:D

Explanation:

6 0

1 year ago

Read 2 more answers