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3 years ago

6

A lithium atom contains 3 protons, 4 neutrons and 3 electrons. What would be formed if one proton is added to this atom?

2 answers:

VARVARA [1.3K]3 years ago

8 0

I think the correct answer would be option C. Adding one proton to an atom of lithium with 3 protons, 4 neutrons and 3 electrons would form a beryllium ion. The new atom have 4 protons and 4 neutrons since Be has a mass number of 9 then it has to form an ion.

Alex Ar [27]3 years ago

5 0

Answer:the answer is D

Explanation:

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Both are at the side of the spectrum that has the lower frequency

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3 years ago

Mike walks 200 km in 6 hours.he then walks another 100km in 4 hours .what is his average speed?

kolbaska11 [484]

Average speed is defined as the ratio of total distance covered in total given time

$speed = \frac{distance}{time}$

here we know that total distance that man moved is

$d_1 = 200 km$

$d_2 = 100 km$

so total distance is

$d = d_1 + d_2$

$d = 200 km + 100 km$

$d = 300 km$

now here total time of the motion is

$t_1 = 6 hours$

$t_2 = 4 hours$

total time will be given as

$t = t_1 + t_2$

$t = 6 + 4 = 10 hours$

now by above formula

$v_{avg} = \frac{300}{10}$

$v_{avg} = 30 km/h$

so his average speed is 30 km/h

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4 years ago

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

$Q=Av$

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

$Q=1.20 m^3/s$

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

$A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2$

So we can re-arrange the equation to find the speed of the water:

$v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s$

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

$Q_1 = Q_2\\A_1 v_1 = A_2 v_2$

where we have

$A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s$

and where $A_2$ is the cross-sectional area of the pipe at the second point.

Solving for A2,

$A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2$

And finally we can find the radius of the pipe at that point:

$A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m$

6 0

3 years ago