"which of newton's laws could we have used to predict that the forces in parts a and b are equal and opposite?"

A negatively charged balloon has 4 μC of charge. How many excess electrons are on this bal- loon? The elemental charge is 1.6 ×

bagirrra123 [75]

Data:

The charge of a body depends on the amount of electrons it gains or loses. Q = n * e, where "Q" is charge, "n" is the number of plus or minus electrons, and "e" is the fundamental charge of an electron $1,6 * 10 ^{-19}C$ . To know if the body has gained or lost, we look at the signal of its charge, remembering that the electron is negative. The charge of the body is 4 μC (positive), so there is a lack of electrons!

Q = 4 μC → $Q = 4*10^{-6}$
$e = 1,6 * 10 ^{-19}C$
$n = ?$ 

We have:
 $Q = n*e$
$n = \frac{Q}{e}$
$n = \frac{4*10^{-6}}{1,6 * 10 ^{-19}}$
$n = 2,5*10^{-6-(-19)}$
$n = 2,5*10^{-6+19}$
$\boxed{n = 2,5*10^{13}electrons}$

7 0

3 years ago

When was the copernican treatise published

LiRa [457]

They were published in 1542.

5 0

4 years ago

It takes 20s for 3x10^6 electrons to flow through a wire ,what is the current?

Elina [12.6K]

Answer:

Explanation:

The variables we know and are given are:

time, t = 20s

Charge, Q = 3x1-^-6 electrons, which is just 3x10^-6C (C stands for Coulombs, which is the unit for Charge)

We need to find the current, I, and since we know Q and t we can substitute these values into the given equation:

I=Q/t (which if you look at what the RHS is saying, its Charge over time, or more literally means the amount of charge passing a point over a period of time)

If we substitute these values, we will get I as:

I = Q / t

I = 3x10^-6 / 20

I = 1.5x10^-7 A

Hope this helps!

6 0

3 years ago

Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with

Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

Part A)

Let's use the parabolic motion equation to solve it. Let's define the variables:

y(i) is the initial height, it is a constant.
y(f) is the final height, in our case is 0
v(i) is the initial velocity (v(i)=16 m/s)
θ1 is the first angle, 30°
θ2 is the first angle, -30°

For the first stone

$y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}$

$0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}$

$0=y_{i1}+8t_{1}-4.905*t_{1}^{2}$ (1)

For the second stone

$0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}$

$0=y_{i2}-8t_{2}-4.905t_{2}^{2}$ (2)

If we solve the equation (1) we will have:

$t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can do the same procedure for the equation (2)

$t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

Part B)

We can use the equation of the horizontal position here.

First stone

$x_{f1}=x_{i1}+vcos(30)t_{1}$