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denis23 [38]
3 years ago
6

How can you calculate the number of electrons in an atom

Physics
2 answers:
Serga [27]3 years ago
7 0
Convert 1 atom into 10 electrons (I think)
Gelneren [198K]3 years ago
6 0
The atomic number( number in the top left corner of the element’s box on the Periodic Table)
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If the average pitcher is releasing the ball from a height of 1.8 m above the ground, and the pitcher's mound is 0.2 m higher th
mina [271]

The catcher can catch the ball at a height of 0.96 m from the ground.

The distance between the pitcher's mound and the catcher's box is about 60'6", which translates to 18.44 m. An average pitcher can pitch with speeds ranging from 88 mph to 97 mph, which is from 39.3 m/s to 43.4 m/s.

Assume the pitcher pitches a ball horizontally with a speed of 40 m/s. If the catcher catches the ball in a time t, then the ball travels a horizontal distance x of 18.44 m and at the same time falls through a height y.

The horizontal motion of the ball is uniform motion since no force acts on the ball ( assuming no air resistance) and hence the acceleration of the ball along the horizontal direction is zero.

Therefore,

x=ut

Calculate the time t by substituting 18.44 m for x and 40 m/s for u.

t=\frac{x}{u} \\ =\frac{18.44 m}{40 m/s} \\ =0.461s

The ball is acted upon by the earth's gravitational attraction and hence it accelerates downwards with an acceleration equal to the acceleration due to gravity g.

Since a horizontal projection is assumed, the ball has no component of velocity in the downward direction.

Therefore, for vertical motion, which is an accelerated motion, the distance y, the ball falls in the time t taken by it to reach the catcher's box is given by the equation,

y=\frac{1}{2} gt^2

Substitute 9.8 m/s² for g and 0.461 s for t.

y=\frac{1}{2} gt^2\\ y=\frac{1}{2}(9.8 m/s^2)(0.461s)^2=1.04 m

The pitcher releases the ball at a height of 1.8 m from a mound which is at a height of 0.2 m. Thus, the ball is released at a height of 2.0 m from the ground. It falls through a distance of 1.04 m in the time it takes to reach the catcher.

Therefore, the height at which the catcher needs to keep his glove so as to catch the ball is given by,(2.0 m)-(1.04 m)=0.96 m

The catcher needs to hold his glove at a height of <u>0,96 m from the ground.</u>

8 0
3 years ago
A car traveling initially at 9.49 m/s accelerates at the rate of 0.988 m/s^2 for 3.05s. What is it’s velocity at the end of the
weeeeeb [17]

Answer:

12.50 m/s

Explanation:

Vi = 9.49 m/s

a = 0.988 m/s²

t = 3.05 s

Vf = ?

Vf = Vi + at

Vf = 9.49 + (0.988)(3.05)

Vf = 12.50 m/s

6 0
3 years ago
Read 2 more answers
57. Example of the law of force and acceleration
7nadin3 [17]

Answer:

Newton's Second Law of Motion says that acceleration (gaining speed) happens when a force acts on a mass (object). Riding your bicycle is a good example of this law of motion at work. Your bicycle is the mass. Your leg muscles pushing pushing on the pedals of your bicycle is the force.

Explanation:

6 0
2 years ago
ithout heat from the Sun, the water cycle would A) reverse. B) not work. C) slow down. D) not be affected.
lawyer [7]
B) Not work since decrease in temperature wouldnt cause evaporation or reverse in cycle.
7 0
3 years ago
Read 2 more answers
A 6.00-kg box sits on a ramp that is inclined at 37.0° above the horizontal. the coefficient of kinetic friction between the box
forsale [732]
We need to see what forces act on the box:

In the x direction:

Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.

In the y direction:

N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force. 

From N-Gcosα=0 we get: 

N=Gcosα, we will need this for the force of friction.

Now to solve for Fh:

Fh=ma + Ff + Gsinα,

Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²

Fh=ma + μmgcosα+mgsinα

Now we plug in the numbers and get:

Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N

The horizontal force for pulling the body up the ramp needs to be Fh=71 N.
8 0
3 years ago
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