When electrons in an atom in an excited state fall to lower energy levels, energy is?

A baseball rolls off a 1.20m high desk and strikes the floor 0.50m away from the base of the desk . How fast was it rolling?

noname [10]

The initial velocity of the ball is 1.01 m/s

Explanation:

The motion of the ball rolling off the desk is a projectile motion, which consists of two independent motions:

- A uniform horizontal motion with constant horizontal velocity

- A vertical accelerated motion with constant acceleration ( $g=9.8 m/s^2$ , acceleration due to gravity)

We start by analyzing the vertical motion: we can find the time of flight of the ball by using the following suvat equation

$s=ut+\frac{1}{2}gt^2$

where

s = 1.20 m is the vertical displacement (the height of the desk)

u = 0 is the initial vertical velocity

$g=9.8 m/s^2$

t is the time of flight

Solving for t,

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(1.20)}{9.8}}=0.495 s$

Now we analyze the horizontal motion. We know that the ball covers a horizontal distance of

d = 0.50 m

in a time

t = 0.495 s

Therefore, since the horizontal velocity is constant, we can calculate it as

$v_x = \frac{d}{t}=\frac{0.50}{0.495}=1.01 m/s$

So, the ball rolls off the table at 1.01 m/s.

Learn more about projectile motion:

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4 0

3 years ago

Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p

Marta_Voda [28]

Answer:

Acceleration, $a=9.91\times 10^{15}\ m/s^2$

Explanation:

It is given that,

Separation between the protons, $r=3.73\ nm=3.73\times 10^{-9}\ m$

Charge on protons, $q=1.6\times 10^{-19}\ C$

Mass of protons, $m=1.67\times 10^{-27}\ kg$

We need to find the acceleration of two isolated protons. It can be calculated by equating electric force between protons and force due to motion as :

$ma=\dfrac{kq^2}{r^2}$

$a=\dfrac{kq^2}{mr^2}$

$a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.67\times 10^{-27}\times (3.73\times 10^{-9})^2}$

$a=9.91\times 10^{15}\ m/s^2$

So, the acceleration of two isolated protons is $9.91\times 10^{15}\ m/s^2$ . Hence, this is the required solution.

3 0

3 years ago

What property is a characteristic of a pure substance that can be observed with out changing the substance into something else

WARRIOR [948]

physical property is a characteristics of a pure substance that can be observed without changing the substance into someone else.

4 0

3 years ago

Comment l'azote entre dans l'atmosphère

Kruka [31]

Answer:

shark puppet

Explanation:

8 0

4 years ago

A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a

EastWind [94]

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

$p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2}$ (1)

Where:

$p_{S,1}$ , $p_{S,2}$ - Initial and final momemtums of the small object, measured in kilogram-meters per second.

$p_{B,1}$ , $p_{B,2}$ - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that $p_{S,1} = 7\,\frac{kg\cdot m}{s}$ , $p_{B,1} = 0\,\frac{kg\cdot m}{s}$ and $p_{S, 2} = 4\,\frac{kg\cdot m}{s}$ , then the final momentum of the big object is: