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N76 [4]
3 years ago
14

Does adding electrons change the mass

Physics
1 answer:
Lostsunrise [7]3 years ago
8 0
“Gaining or losing electrons has no impact on the number of protons and neutrons an atom has, which is another way of saying that the mass number is not affected by changes in the net charge.”
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A skateboarder traveling with an initial velocity 9.0 meters per second,
meriva

Answer:

25m/s

Steps:

<em> First, The equation v= u + a * t shows us what we need to find, (the finale velocity). </em>

<em />

Second, we substitute the values given:

v= 9m/s + 4m/s2 * 4s

Last, We calculate the values:

Multiply 4m/s2 * 4s = 16m/s  

Add 9m/s + 16m/s

<u></u>

<u>Answer:  25m/s</u>

Hope this helps :)

4 0
3 years ago
A baseball pitcher throws a fastball horizontally at a speed of 43.0 m/s. Ignoring air resistance, how far does the ball drop be
Nady [450]
Based on your problem where as ask for the distance of the ball drop between the pitchers mound and the home plate and with a given of the speed of ball is 43m/s and the homeplates is 60.6ft away. Based on my step by step procedure and also considering the value of gravity by 9.8m/s^2 i came up with the distance of 144m away
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3 years ago
Read 2 more answers
1. What do you notice or wonder about ultrasound imaging? (black and white pictures)
a_sh-v [17]

Answer:

the shape how it involve into a picture

Explanation:

3 0
3 years ago
Which of these is an uneven rhythm movement? Question 5 options:
Musya8 [376]

Answer:

Sliding would be an uneven rhythm because Galloping and skipping has a constant flow of the same movement and same noise.

Explanation:

8 0
3 years ago
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A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

Finally:

x=14.49 m

5 0
3 years ago
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