Ask question

Ask question

All categories

3 years ago

11

Can an electron be found in an exact spot within an atom

1 answer:

jolli1 [7]3 years ago

3 0

No, it is impossible to determine the exact location of an electron. This is because electrons don't have a definite position, and direction of motion, at the same time and its movements are unpredictable

You might be interested in

How does compression (the force) work? Not tension and compression. Just compression.

Eddi Din [679]

<span>What allows an arch bridge to span greater distances than a beam bridge, or a suspension bridge to stretch over a distance seven times that of an arch bridge? The answer lies in how each bridge type deals with the important forces of compression and tension.</span>

6 0

3 years ago

Which two factors affect the amount of thermal energy an object has?

Scrat [10]

Answer:

Explanation:

1) The average kinetic energy of the particles of the object or the temperature of the substance.

2) The mass of the object.

8 0

3 years ago

What will be the acceleration of an 8 kg object when something hits it with an 12 N force?

liraira [26]

$\sf Force = mass \times acceleration\\\\12~N = 8~kg \times acceleration\\\\ acceleration = \frac{12~N}{8~kg} \\\\\boxed {\sf a = 1.5~ \frac{m}{s^2} }$

3 0

4 years ago

A baseball approaches home plate at a speed of 44.0 m/s, moving horizontally just before being hit by a bat. The batter hits a p

Luda [366]

Explanation:

It is given that,

Speed of the baseball, u = 44 m/s

Speed of the baseball, v = 53 m/s

Mass of the ball, m = 145 g = 0.145 kg

Time of contact between the ball and the bat, t = 2.2 ms = 0.0022 s

$F=ma$

$F=\dfrac{mv}{t}$

$F_1=\dfrac{0.145\ kg\times 44\ m/s}{0.0022\ s}$

F₁ = 2900 N...........(1)

$F=ma$

$F=\dfrac{mv}{t}$

$F_2=\dfrac{0.145\ kg\times 53\ m/s}{0.0022\ s}$

F₂ = 3493.18 N.........(2)

In average vector form force is given by :

$F=F_1+F_2$

$F=(2900i+(-3493.18)\ N$

$F=(2900i-3493.18j)\ N$

Hence, this is the required solution.

6 0

4 years ago

For fully developed laminar pipe flow in a circular pipe, the velocity profile is u(r) = 2(1-r2 /R2 ) in m/s, where R is the inn

sdas [7]

Answer:

a) $v_{max} = 2\ \textup{m/s}$

b) $v_{avg} = 1\ \textup{m/s}$

c) Q = 1.256 × 10⁻³ m³/s

Explanation:

Given:

The velocity profile as:

$u(r) = 2(1-\frac{r^2}{R^2} )$

Now, the maximum velocity of the flow is obtained at the center of the pipe

i.e r = 0

thus,

$v_{max}=u(0) = 2(1-\frac{0^2}{R^2} )$

or

$v_{max} = 2\ \textup{m/s}$

Now,

$v_{avg} = \frac{v_{max}}{2}\ \textup{m/s}$

or

$v_{avg} = \frac{2}}{2}\ \textup{m/s}$

or

$v_{avg} = 1\ \textup{m/s}$

Now, the flow rate is given as:

Q = Area of cross-section of pipe × $v_{avg}$

or

Q = $\frac{\pi D^2}{4}\times v_{avg}$

or

Q = $\frac{\pi 0.04^2}{4}\times 1$

or

Q = 1.256 × 10⁻³ m³/s

7 0

3 years ago