If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be 
.
<h3>What is an escape velocity?</h3>
The ratio of the object's travel distance over a specific period of time is known as its velocity. As a vector quantity, the velocity requires both the magnitude and the direction. the slowest possible speed at which a body can break out of the gravitational pull of a certain planet or another object.
The formula to calculate the escape velocity of earth is given below:-
Given that earth's mass was half its actual value but its radius stayed the same. The escape velocity will be calculated as below:-
.
Therefore, If the earth's mass were half its actual value but its radius stayed the same, the escape velocity of the earth would be .
To know more about escape velocity follow
brainly.com/question/14042253
#SPJ4
Thank you for posting your question here at brainly. Below is the solution. I hope the answer will help.
<span>Cl^- 1s^2 2s^2p^6 3s^2 3p^6 1s^2 2s^2p^6 S = 10; 3s^2 3p^6 S = 0 </span>
<span>Zeff = Z-S = 17- 10 =7 </span>
<span>K^+ 1s^2 2s^2p^6 3s^2 3p^6; 1s^2 2s^2p^6 S = 10; 3s^2 3p^6 S = 0 </span>
<span>Zeff = Z-S = 19- 10 = 9
</span>
S = 2 + 6.8 + 2.45 = 11.25
<span>Zeff(Cl^-) = 17 – 11.25 = 5.75 </span>
<span>K^+ 1s^2 2s^2p^6 3s^2 3p^6 same S as for Cl^- but Z increases by 2 hence </span>
<span>Zeff(K^+) = 19 - 11.25 = 7.75</span>
Answer:
because speed is the modulus of velocity which is a vector
the velocity to be zero it must be a round trip
Explanation:
This is because speed is the modulus of velocity which is a vector.
For the velocity to be zero it must be a round trip, therefore the resulting vector zero
On the other hand, the speed of the module is the same in both directions
<span>A measurement must include both a number and an unit.
</span>
