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2 years ago

7

write a one or two summary paragraph discussing this experiment and the results use tye following questions 1 according to your

data was your hypothisis correct be sure to refer your data when answering this question 2 summarize ang difficulties or probloms you had in performing the experiment thaf might have affected the results list at least two real world examples that apply the findings for this experiment

1 answer:

Monica [59]2 years ago

8 0

Answer:heed

Explanation:

heed knows how to divide.

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If the electric potential at a point in space is zero, then the electric field at that point is:

icang [17]

E - impossible to determine based on the given information

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2 years ago

What is the full meaning of (i.p.s.m.n)

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Answer:

Intraductal Papillary Mucinous Neoplasm

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3 years ago

13. An object with a mass of 2.0 kg has a force of 4.0 newtons applied to it.

julsineya [31]

Answer:2m/s²

Explanation: Well F=MA so sice F=4N and M=2kg let's plug in the values

4N=2KG*A

A=4N/2KG

A=2m/s²

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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

In the equation for centripetal force, which expression represents the centripetal acceleration of the object? mv2 StartFraction

Sphinxa [80]

Answer: $\frac{V^{2}}{r}$

Explanation:

According to Newton's 2nd Law of motion the force $F$ is proportional to the mass $Fm$ and acceleration $a$ :

$F=m.a$ (1)

On the other hand, the equation for the Centripetal force is:

$F=\frac{mV^{2}}{r}$ (2)

Where:

$V$ is the velocity

$r$ is the radius of the circular motion

Making (1) and (2) equal:

$m.a=\frac{mV^{2}}{r}$ (3)

Hence:

$a=\frac{V^{2}}{r}$ This is the expression for the centripetal acceleration

It should be noted, this acceleration is directed toward the center of the circumference of the circular motion (that's why it's called centripetal acceleration).

3 0

3 years ago

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