<h2>
Value of g on the distant planet is 16 m/s²</h2>
Explanation:
We have equation of motion v² = u² + 2as
Here initial velocity, u = 2 m/s
Final velocity, v = 10 m/s
Displacement, s = 3 m
We need to find acceleration, a.
Substituting
v² = u² + 2as
10² = 2² + 2 x a x 3
6a = 96
a = 16 m/s²
Value of g on the distant planet is 16 m/s²
Answer: Your code returns a number of 99.123456789 +0.00455679
Ok, you must see where the error starts to affect your number.
In this case, is in the third decimal.
So you will write 99.123 +- 0.004 da da da.
But you must round your results. In the number you can see that after the 3 comes a 4, so the 3 stays as it is.
in the error, after the 4 comes a 5, so it rounds up.
So the final presentation will be 99.123 +- 0.005
you are discarding all the other decimals because the error "domains" them.
Answer:
Sound energy to electric energy - a person talking into a microphone
Radiant energy to electric energy - sunlight falling on solar panels
Gravitational potential energy to motion energy - a ball dropped from a height
Explanation:
A person talking is the sound energy and going into an electric phone
Sunlight or Radiant energy falls onto the solar panels creating electric energy
The ball is being pulled down by gravity from a certain height, going down to the ground, it’s motion, falling
10 grams of salt is the result
Answer:
Explanation:
Given that
F=ax^3/2. a is a constant
The force does a work of
W=2.01KJ from x=0 to x=15.2m
We need to find a
Work is give as,
W=∫F.ds
But this is in x direction only then,
W=∫Fdx. from x=0 to x=15.2m
W=∫ax^3/2dx from x=0 to x=15.2m
W=ax^(3/2+1)/(3/2+1).
W=ax^(5/2)/5/2
W=ax^(2/5)/2.5 from x=0 to x=15.2m
Cross multiply
2.5W=ax^2.5. from x=0 to x=15.2m
2.5W= a (15.2^2.5-0)
W=2.01KJ=2010J
2.5×2010=a×900.76
Therefore,
a=5.56