Which describes why people on earth can see light from the stars in the sky that are so far away?

According to Coulomb’s Law, the force between two charged objects is related to _____.

Answer:

A.) the inverse of the square of the distance separating them

Explanation:

Coulombs law states that "the force of attraction between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them."

Mathematically, F = kq1q2/r²

Where q1 and q2 are the charges

r is the distance between the charges.

According to the law, the force between two charged objects is related to the inverse of the square of the distance separating them.

5 0

3 years ago

A group of students prepare for a robotic competition and build a robot that can launch large spheres of mass M in the horizonta

jeyben [28]

Answer:

$V_0$

Explanation:

Given that, the range covered by the sphere, $M$ , when released by the robot from the height, $H$ , with the horizontal speed $V_0$ is $D$ as shown in the figure.

The initial velocity in the vertical direction is $0$ .

Let $g$ be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. $V_0$ will remain constant throughout the projectile motion.

So, if the time of flight is $t$ , then

$D=V_0t\; \cdots (i)$

Now, from the equation of motion

$s=ut+\frac 1 2 at^2\;\cdots (ii)$

Where $s$ is the displacement is the direction of force, $u$ is the initial velocity, $a$ is the constant acceleration and $t$ is time.

Here, $s= -H, u=0,$ and $a=-g$ (negative sign is for taking the sigh convention positive in $+y$ direction as shown in the figure.)

So, from equation (ii),

$-H=0\times t +\frac 1 2 (-g)t^2$

$\Rightarrow H=\frac 1 2 gt^2$

$\Rightarrow t=\sqrt {\frac {2H}{g}}\;\cdots (iii)$

Similarly, for the launched height $2H$ , the new time of flight, $t'$ , is

$t'=\sqrt {\frac {4H}{g}}$

From equation (iii), we have

$\Rightarrow t'=\sqrt 2 t\;\cdots (iv)$

Now, the spheres may be launched at speed $V_0$ or $2V_0$ .

Let, the distance covered in the $x-$ direction be $D_1$ for $V_0$ and $D_2$ for $2V_0$ , we have

$D_1=V_0t'$

$D_1=V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_1=\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_1=1.41 D$ (approximately)

This is in the $3$ points range as given in the figure.

Similarly, $D_2=2V_0t'$

$D_2=2V_0\times \sqrt 2 t$ [from equation (iv)]

$\Rightarrow D_2=2\sqrt 2 D$ [from equation (i)]

$\Rightarrow D_2=2.82 D$ (approximately)

This is out of range, so there is no point for $2V_0$ .

Hence, students must choose the speed $V_0$ to launch the sphere to get the maximum number of points.

7 0

3 years ago

A speaker vibrates at a frequency of 200 hz what is its period

Norma-Jean [14]

Period = (1) / (frequency)

Period = (1) / (200 per second) = 0.005 second = 5 milliseconds

5 0

3 years ago

Read 2 more answers

Which is the correct definition for sea floor spreading

jok3333 [9.3K]

Seafloor spreading is a geologic process in which tectonic plates—large slabs of Earth's lithosphere—split apart from each other.

6 0

2 years ago

Read 2 more answers

Select the correct answer.

kvasek [131]

Answer:

8.37×10⁻⁴ N/C

Explanation:

Electric Field: This is the ratio of electrostatic force to electric charge. The S.I unit of electric field is N/C.

From the question, the expression for electric field is given as,

E = F/Q.......................... Equation 1

Where E = Electric Field, F = force experienced by the charged balloon, Q = Charge on the balloon.

Given: F = 8.2×10⁻² Newton, Q = 9.8×10 Coulombs = 98 Coulombs

Substitute these values into equation 1

E = 8.2×10⁻² /98

E = 8.37×10⁻⁴ N/C

Hence the Electric Field of the charged balloon = 8.37×10⁻⁴ N/C

4 0

2 years ago

Read 2 more answers