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3 years ago

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when the mass of an object is increased it would decrease its acceleration in the force is left alone

1 answer:

mojhsa [17]3 years ago

4 0

Answer:

See the explanation below.

Explanation:

This analysis can be easily deduced by means of Newton's second law which tells us that the sum of the forces or the total force on a body is equal to the product of mass by acceleration.

∑F = m*a

where:

F = total force [N]

m = mass [kg]

a = acceleration [m/s²]

We must clear the acceleration value.

$a=\frac{F}{m}$

We see that the term of the mass is in the denominator, so that if the value of the mass is increased the acceleration decreases, since they are inversely proportional.

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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

If a victim of sexual harassment asks a supervisor not report it, the supervisor should respect his or her wishes.

inn [45]

Answer:

False

Explanation:

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3 years ago

Read 2 more answers

Can someone please help me match these up

yaroslaw [1]

Answer:

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Explanation:

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3 years ago

Can any one help me on these two please

Svetach [21]

I THINK 5 is A and 6 is C

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1. Your 15-year-old daughter, Fatamorgana, has been ill for 3 days with a fever, lassitude, anorexia, sore anus and elbows, and

aksik [14]

Answer:

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Explanation:

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