All categories

2 years ago

Which of the four spheres is this pictures/scene?

Physics

2 answers:

Verizon [17]2 years ago

8 0

Answer:

D) atmosphere

Explanation:

but the correct answer is troposhare

Zigmanuir [339]2 years ago

7 0

I believe the correct answer is atmosphere (D).

You might be interested in

Two point charges, initially 2.0 cm apart, experience a 1.0 N force. If they are moved to a new separation of 0.25 cm, what is t

den301095 [7]

Explanation:

Th electric force between charges is inversely proportional to the square of distance between them. It means,

$F\propto \dfrac{1}{r^2}$

Initial distance, r₁ = 2 cm

Final distance, r₂ = 0.25 cm

Initial force, F₁ = 1 N

We need to find the electric force between charges if the new separation of 0.25 cm. So,

$\dfrac{F_1}{F_2}=(\dfrac{r_2}{r_1})^2\\\\F_2=\dfrac{F_1r_1^2}{r_2^2}\\\\F_2=\dfrac{1\times 2^2}{(0.25)^2}\\\\F_2=64\ N$

So, the new force is 64 N if the separation between charges is 64 N.

7 0

3 years ago

What are the coordinates of the point that is 1/5 of the way from A(-7,-4) to B(3, 6)?

Olegator [25]

Answer:

coordinate will be

r = (1.33,4.33)

Explanation:

As we know that if a point will divide two given coordinates in m : n ratio

then in that case the point is given as

$x = \frac{mx_1 + nx_2}{m+n}$

$y = \frac{my_1 + ny_2}{m+n}$

here we know that two points are

(-7, -4) and (3, 6)

and the ratio is given as 1 : 5

now from above formula we will have

$x = \frac{1(-7) + 5(3)}{1 + 5} = 1.33$

$y = \frac{1(-4) + 5(6)}{1 + 5} = 4.33$

so coordinate will be

$r = (1.33, 4.33)$

7 0

3 years ago

Two very small spherical metal objects, each with 1 coulomb of charge, are brought together in a vacuum so that the separation d

DaniilM [7]

Answer: 9*10^15 N

Force=kqq/r^2

F=[(9*10^9)(1)(1)]/.001^2=9.0*10^15

8 0

3 years ago

Read 2 more answers

Three charges 1.5*10-6, 3*10-6, -3*10-6 are placed at three vertices of an equilateral triangle of side 30cm. Find the net force

gulaghasi [49]

Answer:

F = 0N

Explanation:

The force between two charges is given by

$F=k\frac{q_1q_2}{r^2}$

where r is the distance between the charges and K is the Coulomb's constant

(k=8-89*10^9Nm^2/C^2)

The force in the first charge is only the sum of the forces due to the other charges. Hence we have

$F_T=F_1+F_2=k\frac{q_2q_1}{r^2}+k\frac{q_3q_1}{r^2}$

$F_T=(8.89*10^9\frac{Nm^2}{C^2})\frac{(3*10^{-6}C)(1.5*10^{-6}C)}{(0.3m)^2}+(8.89*10^9\frac{Nm^2}{C^2})\frac{(-3*10^{-6}C)(1.5*10^{-6}C)}{(0.3m)^2}\\\\F_T=0.445N-0.445N=0N$

Ft=0N

Hope this helps!!

5 0

3 years ago

Read 2 more answers

900 x 60050 / 820 x 5632 + 88973 x 903 = what?

andre [41]

451539497.049 is the answer

7 0

3 years ago