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3 years ago

If I keep F constant in F=ma, what is the relationship between m and a?

Physics

1 answer:

Jobisdone [24]3 years ago

6 0

Answer:

If F is a constant, we can take f = 1

f = m×a

ma = 1

therefore we can say that force is hence proportinal to the product of mass and acceleration.

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Glycerin at a temperature of 30 degrees celcius flows at a rate of 8×10−6m3/s through a horizontal tube with a 30mm diameter. wh

MariettaO [177]

The pressure drop in pascal is 3.824*10^4 Pascals.

To find the answer, we need to know about the Poiseuille's formula.

<h3>How to find the pressure drop in pascal?</h3>

We have the Poiseuille's formula,

$Q=\frac{\pi Pr^4}{8\beta l}$

where, Q is the rate of flow, P is the pressure drop, r is the radius of the pipe, $\beta$ is the coefficient of viscosity (0.95Pas-s for Glycerin) and l being the length of the tube.

It is given that,

$Q=8*10^{-6}m^3/s\\diameter=30mm, thus,\\r=15mm\\l=100m\\\beta =0.95$

Thus, the pressure drop will be,

$P=\frac{8Q\beta l}{\pi r^4} =\frac{8*8*10^{-6}*0.95*100}{3.14*(15*10^{-3})^4} \\\\P=3.824*10^4Pascals.$

Thus, we can conclude that, the pressure drop in pascal is 3.824*10^4 Pascals.

Learn more about the Poiseuille's formula here:

brainly.com/question/13180459

#SPJ4

3 0

1 year ago

A 5.7 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 4.5 + 13.7 x − 1.5 x 2 , where

const2013 [10]

Answer:

The work done by the force on the particle is 29.85 J.

Explanation:

The work is given by:

$W = ^{x_{2}}_{x_{1}}\int F_{x} dx$

Where:

x₁: is the lower limit = 0 m

x₂: is the upper limit = 1.9 m

Fₓ: is the force in the horizontal direction = (4.5 + 13.7x - 1.5x²)N

$W = ^{1.9}_{0}\int (4.5 + 13.7x - 1.5x^{2}) dx$

$W = 4.5x|^{1.9}_{0} + \frac{13.7}{2}x^{2}|^{1.9}_{0} - \frac{1.5}{3}x^{3}|^{1.9}_{0}$

$W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3}$

$W = 29.85 J$

Therefore, the work done by the force on the particle is 29.85 J.

I hope it helps you!

6 0

3 years ago

Radioactive decay of 40k atoms in an igneous rock has resulted in a radio of 25 percent 40k atoms to 75 percent 40Ar and 40Ca at

Ivanshal [37]

In order to answer this, we would need to know
the half-life of ⁴⁰K. Perhaps when you look that up,
you'll be able to answer this on your own.

6 0

3 years ago

I NEED HELP LIKE RIGHT NOW!!!!... please

Maslowich

1.) TRUE

7.) I believe it is FALSE

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8 0

3 years ago

Read 2 more answers

Will someone please look at my other questions

Yuliya22 [10]

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3 years ago