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Alexandra [31]
2 years ago
5

Potassium has an atomic number

Physics
2 answers:
Fantom [35]2 years ago
8 0

Answer:

20 neutron

Explanation:

neutron=atomic mass-atomic no.

=39-19

=20

Lera25 [3.4K]2 years ago
5 0
Answer: C) 20

Detailed Explanation:

No. of Neutrons = Mass No. - Atomic No.

Therefore,
= 39 - 19
=> 20

No. Of Neutrons = 20
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2. A new racecar track is being created, and the owners want the cars to complete 10
jeka57 [31]

Answer:

0.45 miles

Explanation:

since, car will be running on the edge of track, distance covered in one lap by car will be equal to circumference of track.\

we will use value of pi as 22/7

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Hence, in one lap \pi D distance will be covered

in 7 laps 7 \pi D distance will be covered.

but is given that 7 laps should be equal to 10 miles

so 10 miles will be equal to 7\pi D

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3 0
3 years ago
A baseball is projected horizontally with an initial speed of 18.1 m/s from a height of 1.85 m. at what horizontal distance will
gtnhenbr [62]
1) The motion of the baseball consists of two separate motions along the horizontal (x) and vertical (y) axis. The position on the two directions at time t is given by:
x(t)=v_0t
y(t)=h- \frac{1}{2}gt^2
where the motion on the x-axis is a uniform motion with constant speed v_0=18.1 m/s, while the motion on the y-axis is a uniformly accelerated motion with constant acceleration g=9.81 m/s^2, and initial height h=1.85 m.

First of all, we can find the time t at which the ball reaches the ground by requiring y(t)=0:
0=h- \frac{1}{2}gt^2
t= \sqrt{ \frac{2h}{g} }= \sqrt{ \frac{2(1.85 m)}{9.81 m/s^2} }=  0.61 s

and if now we substitute this time into the equation for x(t), we find the horizontal distance covered during this time interval, which is the horizontal distance covered by the ball before hitting the ground:
x(t)=v_0 t=(18.1 m/s)(0.61 s)=11.04 m

2) Speed of the ball as it hits the ground
We need to find the components of the velocity on the two directions, at the istant when the ball hits the ground.
On the x-axis, the velocity is the same as the initial one: 
v_x=18.1 m/s
On the y-axis, the velocity is given by
v_y(t)=gt
If we substitute t=0.61 s (the time at which the ball reaches the ground), we find
v_y =gt=(9.81 m/s^2)(0.61 s)=6.0 m/s
And the speed of the ball is the magnitude of the resultant of the two components:
v= \sqrt{v_x^2+v_y^2}= \sqrt{(18.1 m/s)^2+(6.0 m/s)^2}=19.1 m/s
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