1. Two students have volunteered to explore the galaxy in a

Two upwards forces act: one of 34N which is 3.5m to the left of the axle, the other is 25N and is 2.6m to the right of the axle.

KiRa [710]

Answer:

See explanation below

Explanation:

In this case, you want to know if you put an object between these forces, which direction would go.

To know this, we need to calculate the moment of an object, which is defined as the product of a force and it's distance. In other words:

M = F * d (1)

And, in order to reach equilibrium the force will exert a direction in clockwise or anticlosewise, and these moments, should be even:

anticlockwise moment = clockwise moment.

The clockwise would be the forces to the right, and anticlock would the only force to the left of the axle.

Clockwise moment = (10 * 0.8) + (25 * 2.6) = 73 Ns

Anticlockwise moment = 34 * 3.5 = 119 Ns.

As we can see, the moment in the anticlockwise is higher than the actual clockwise moment, therefore, we can assume that the object will move anticlockwise, or simply move to the left.

Hope this helps

5 0

3 years ago

-In the Bohr model, as it is known today, the electron is imagined to move in a circular orbit about a stationary proton. The fo

vampirchik [111]

Answer:

$4.3859007196\times 10^{-9}\ m$

Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

v = Velocity of electron = $2.4\times 10^5\ m/s$

q = Charge of electron = $1.6\times 10^{-19}\ C$

m = Mass of electron = $9.11\times 10^{-31}\ kg$

r = Radius

The electrical and centripetal force will balance each other

$\dfrac{kq^2}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow r=\dfrac{kq^2}{mv^2}\\\Rightarrow r=\dfrac{8.99\times 10^9\times (1.6\times 10^{-19})^2}{9.11\times 10^{-31}\times (2.4\times 10^5)^2}\\\Rightarrow r=4.3859007196\times 10^{-9}\ m$