The imaginary point in the sky directly above the North Pole is ____

A particle moves along a straight path through displacement while force acts on it. (Other forces also act on the particle.) Wha

allsm [11]

a) c = 1.85

b) c = 0.8

c) c = 2.33

Explanation:

a)

The displacement of the particle is given by

$d=2.2i+cj$

While the force applied on the particle is

$F=3.2i-3.8 j$

So we have a problem in 2-dimensions.

The work done on the particle is given by the scalar product between force and displacement:

$W=F\cdot d$ (1)

Here the work done on the particle is zero, so

W = 0

Therefore from eq(1) we find:

$0=(3.2i-3.8j)\cdot (2.2i+cj)=7.04-3.8c\\3.8c=7.04\\c=\frac{7.04}{3.8}=1.85$

b)

In this problem, the work done on the particle is

$W=4.0 J$

The force and displacement are still

$d=2.2i+cj$ (displacement)

$F=3.2i-3.8 j$ (force)

Therefore, by calculting the scalar product between force and displacement and equating it to the work done (4.0 J), we find:

$W=F\cdot d$

$4.0 =(3.2i-3.8j)\cdot (2.2i+cj)=7.04-3.8c\\3.8c=3.04\\c=\frac{3.04}{3.8}=0.8$

c)

In this problem instead, the work done on the particle is negative:

$W=-1.8 J$

As before, the force and displacement are

$d=2.2i+cj$ (displacement)

$F=3.2i-3.8 j$ (force)

And so again, we calculate the scalar product between force and displacement and we equate it to the work done on the particle, -1.8 J.

Doing so, we find:

$W=F\cdot d$

$-1.8=(3.2i-3.8j)\cdot (2.2i+c)=7.04-3.8c\\3.8c=8.84\\c=\frac{8.84}{3.8}=2.33$

7 0

3 years ago

I will Mark Brainliest 1. ) Allena and Charissa were discussing whether Cl2 is an element or a compound. Allena said that it is

WITCHER [35]

The answer to question one is A.
The answer to question two is A.
The answer to question three is D.

7 0

4 years ago

Cual es la fuerza electrica sobre el electrón (-1.6 x 10¹⁹c) de un atomo de hidrógeno ejercida por el protón (1.6 x 10¹⁹c)? Supó

kkurt [141]

Answer:

La fuerza eléctrica es -8.2*10⁻⁸ N

Explanation:

El enunciado correcto es: ¿Cuál es la fuerza eléctrica sobre el electrón (-1.6 x 10⁻¹⁹c) de un átomo de hidrógeno ejercida por el protón (1.6 x 10⁻¹⁹c)? Supóngase que la distancia entre el electrón y el protón es de 5.3 x 10⁻¹¹ m

Entre dos o más cargas aparece una fuerza denominada fuerza eléctrica. Su valor depende del valor de las cargas y de la distancia que las separa, mientras que su signo depende del signo de cada carga. Las cargas del mismo signo se repelen entre sí, mientras que las de distinto signo se atraen.

La fuerza eléctrica con la que se atraen o repelen dos cargas puntuales en reposo es directamente proporcional al producto de las mismas e inversamente proporcional al cuadrado de la distancia que las separa:

$F=K*\frac{q1*q2}{d^{2} }$

donde:

F es la fuerza eléctrica de atracción o repulsión. En el Sistema Internacional (S.I.) se mide en Newtons (N).
q1 y q2 son lo valores de las dos cargas puntuales. En el S.I. se miden en Culombios (C).
d es el valor de la distancia que las separa. En el S.I. se mide en metros (m).
K es una constante de proporcionalidad llamada constante de la ley de Coulomb. Depende del medio en el que se encuentren las cargas. Para el vacío K tiene un valor aproximadamente de 9*10⁹ $\frac{N*m^{2} }{C^{2} }$ .

En este caso:

F=?

K= 9*10⁹ $\frac{N*m^{2} }{C^{2} }$

q1= -1.6*10⁻¹⁹ C

q2= 1.6*10⁻¹⁹ C

d= 5.3*10⁻¹¹ m

Reemplazando:

$F=9*10^{9} \frac{N*m^{2} }{C^{2} }*\frac{(-1.6*10^{19} C)*(1.6*10^{19} C)}{(5.3*10^{-11} )^{2} }$

Resolviendo:

F= -8.2*10⁻⁸ N

La fuerza eléctrica es -8.2*10⁻⁸ N

6 0

3 years ago

Suppose the experiment was conducted in the same manner, but the axle was now off center of the solid disk. Would you expect the

san4es73 [151]

Answer:

Explanation:

If Ig be moment of inertia about an axis through centre of mass and I be moment of inertia through any other axis parallel to earlier axis , then according to theory of parallel axis ,

I = Ig + Md²

where M is mass of the body and d is distance between two parallel axis.

So I is greater than Ig.

4 0

3 years ago

Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Sim

Nonamiya [84]

Answer:

The distance covered by puck A before collision is $z = 8.56 \ m$