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3 years ago

What will happen if an electron is added to an atom

1 answer:

5 0

Atoms can gain or lose electrons. When an atom gains an electron, energy is given off and is known as the electron affinity. Electron affinity is defined as the energy released when an electron is added to a gaseous atom or ion.

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A vertical spring (spring constant =160 N/m) is mounted on the floor. A 0.340-kg block is placed on top of the spring and pushed

AleksandrR [38]

(a) 3.5 Hz

The angular frequency in a spring-mass system is given by

$\omega=\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

Here in this problem we have

k = 160 N/m

m = 0.340 kg

So the angular frequency is

$\omega=\sqrt{\frac{160 N/m}{0.340 kg}}=21.7 rad/s$

And the frequency of the motion instead is given by:

$f=\frac{\omega}{2\pi}=\frac{21.7 rad/s}{2\pi}=3.5 Hz$

(b) 0.021 m

The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at

x = A

where A is the amplitude of the motion.

The maximum displacement is given by Hook's law:

$F=kA$

where

F is the force applied initially to the spring, so it is equal to the weight of the block:

$F=mg=(0.340 kg)(9.81 m/s^2)=3.34 N$

k = 160 N/m is the spring constant

Solving for A, we find

$A=\frac{F}{k}=\frac{3.34 N}{160 N/m}=0.021 m$

3 0

3 years ago

A 0.25 kg ball is suspended from a light 0.65 m string as shown. The string makes an angle of 31° with the vertical. Let U = 0 w

steposvetlana [31]

Explanation:

a) The height of the ball h with respect to the reference line is

$h = L - L\cos{31°} = L(1 - \cos{31°})$

so its initial gravitational potential energy $U_0$ is

$U = mgh = mgL(1 - \cos{31°})$

$\:\:\:\:\:=(0.25\:\text{kg})(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31})$

$\:\:\:\:\:=0.23\:\text{J}$

b) To find the speed of the ball at the reference point, let's use the conservation law of energy:

$\Delta{K} + \Delta{U} = 0 \Rightarrow K_0 + U_0 = K + U$

We know that the initial kinetic energy $K_0,$ as well as its final gravitational potential energy $U$ are zero so we can write the conservation law as

$mgL(1 - \cos{31°}) = \frac{1}{2}mv^2$

Note that the mass gets cancelled out and then we solve for the velocity v as

$v = \sqrt{2gL(1 - \cos{31°})}$

$\:\:\:\:\:= \sqrt{2(9.8\:\text{m/s}^2)(0.65\:\text{m})(1 - \cos{31°})}$

$\:\:\:\:\:= 1.3\:\text{m/s}$

5 0

3 years ago

Read 2 more answers

What is true about the direction of the force of friction? A. It always points straight down. B. It always opposes the direction

Bezzdna [24]

The direction of the force of static friction is along the plane of contact, and is opposite to the direction in which there would be relative motion if there was no friction (for example, if one of the surfaces suddenly turned to ice). Hope this helped!

8 0

3 years ago

In a supermarket, you place a 22.3-N (around 5 lb) bag of oranges on a scale, and the scale starts to oscillate at 2.7 Hz. What

allsm [11]

Answer:

Force constant, k = 653.3 N/m

Explanation:

It is given that,

Weight of the bag of oranges on a scale, W = 22.3 N

Let m is the mass of the bag of oranges,

$m=\dfrac{W}{g}$

$m=\dfrac{22.3}{9.8}$

m = 2.27 kg

Frequency of the oscillation of the scale, f = 2.7 Hz

We need to find the force constant (spring constant) of the spring of the scale. We know that the formula of the frequency of oscillation of the spring is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$