Answer:
acceleraions 5.76g and 20.55g
Explanation:
This constant acceleration exercise can be solved using the kinematic equations in one dimension
Vf = Vo + a t
As part of the rest Vo = 0
a = Vf / t
a = 282/5
a = 56.4 m / s2
In relation to the acceleration of gravity
a ’= a / g = 56.4 / 9.8
a ’= 5.76g
To calculate the acceleration to stop we use the same formula
a2 = 282 / 1.40
a2 = 201.4 m / s2
This acceleration of gravity acceleration function is
a2 ’= 201.4 / 9.8
a2 ’= 20.55g
A. 4% you divide 2000 to 500 and you get 4 which is your answer
1. Fertile soil.
2. Ore deposits.
3. Fossil fuels
The constant velocity that the spacecraft must travel is : 3.49 * 10⁸ m/s
<u>Given data :</u>
Distance of star from earth = 4.3 light years
Observers time = 3.7 years
<h3>Determine the constant velocity the spacecraft must travel </h3>
Observers time = 3.7 * 365 * 24 * 60 * 60
Distance of star from earth = 4.3 * 9.46 * 10¹⁵
The velocity the spacecraft must travel will be calculated using the equation
V = distance / time
= ( 4.3 * 9.46 * 10¹⁵ ) / ( 3.7 * 365 * 24 * 60 * 60 )
= 3.49 * 10⁸ m/s
Hence we can conclude that The constant velocity that the spacecraft must travel is : 3.49 * 10⁸ m/s
Learn more about space travelling : brainly.com/question/1344685
<em />
<em>Attached below is the missing detail related to the question </em>
Answer:
Explanation:
3.75 * 10^-7
=3.75 * 1/10^7
=3.75/10000000
=3/800000000
any base which has it's power negative do it's reciprocal then the power will be positive.