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2 years ago

A student estimated a mass to be 325 g, but the actual mass is 342 g. What is the percent error?

Physics

1 answer:

Dovator [93]2 years ago

3 0

Answer:

Answer:

5.2307 %

Explanation:

(acutal mass- estimated mass) / ( estimated mass)

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How did the social structure of ancient Rome affect the lives of its people?

Neporo4naja [7]

Answer:

Explanation:

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3 years ago

The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr

MA_775_DIABLO [31]

Answer:

$q=49Q/64$

and

$x =16R/15$

Explanation:

See attached figure.

$E_{Q}=$ E due to sphere

$E_{q}=$ E due to particule

$E_{total}=E_{Q}-E_{q}=0$ (1)

according to the law of gauss and superposition Law:

$E_{Q}=E_{1}+E_{2}=E_{2}$ ; electric field due to the small sphere with r1=R/4

$E_{Q}=kq_{2}/(r_{1}^{2})=$

$q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}$

then: $E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})$ (2)

on the other hand, for the particule:

$E_{q}=kq/(r_{p}^{2})$

$r_{p}=2R-R/4=7R/4$ ⇒ $E_{q}=16kq/(49R^{2})$ (3)

We replace (2) y (3) in (1):

$E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})$

$q=49Q/64$

--------------------

if R<x<2R AND $E_{total}=E_{Q}-E_{q}=0$

$E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})$

remember that $q=49Q/64$

then:

$Q(2R-x^{2})=49/64*x^{2}$

solving:

$x_{1} =16R/15$

$x_{2} =16R$

but: R<x<2R

so : $x =16R/15$

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2 years ago

Soil can best be described as the

Rom4ik [11]

<span>The correct option is D. Soil can best be described as the loose covering of weathered rocks and decaying organic matter. There are different types of soil, the type of soil formed depends majorly on the type of parent rock from which the soil is formed and the amount of organic matter present in the soil.</span>

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2 years ago

Observe: Air pressure is equal to the weight of a column of air on a particular location. Airpressure is measured in millibars (

lesya [120]

Answer:

a) When moving towards a high pressure center the pressure values increase in the equipment

b) This area is called high prison since the weight of the atmosphere on top is maximum

Explanation:

A) A high atmospheric pressure system is an area where the pressure is increasing the maximum value is close to 107 Kpa, the other side as low pressure can have small values 85.5 kPa.

When moving towards a high pressure center the pressure values increase in the equipment

B) This area is called high prison since the weight of the atmosphere on top is maximum

in general they are areas of good weather

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3 years ago

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3 years ago