Answer:
50J
Explanation:
At the top you have(A)
KE_a = O
PE_a = 100J
KE + PE = 100J
At the bottom you have (C)
KE_c= 100J
PE_c=0J
KE+PE = 100J
At point C:
You are at half the height.
We know that at H, PE =100J
PE_c = mgH
At C,
PE_c= mg (H/2) *at half the height
*m and g stay the same
Intuitively, the higher you are, the more potential energy you have.
If you decrease the height by a half, your PE will also decrease
At A:
PE_a / (mg) = H
At B:
PE_b / (mg) = H/2
to also get H on the right hand side, multiply by 2
2 (PE_b/ (mg))= H
2PE_b / (mg) = H
Ok, now that we have set up 2 equations (where H is isolated), find PE at B
AT A = AT B *This way you are saying that H = H (you compare both equations)
PE_a / (mg) = 2x PE_b / (mg)
*mg are the same for both cancel them (you can do that because of the = sign)
PE_a = 2PE_b
We know that PE_a = 100J
100J/2 = PE_b
PE at b = 50J
**FIND KE at b
We know that
KE_b + PE_b is always 100J
100J = 50J + KE_b
KE_b = 50J
Apparent magnitude means how bright a star APPEARS to us on Earth. It can be affected by ...
... how bright the star really is
... how far the star is from us
... how much gas and dust is between us
... how much of the star's total output is in visible light
Answer:
Option D - 0.2 s
Explanation:
We are given;
Initial velocity; u = 7 m/s
Height of table; h = 1.8m
Now,since we want to find the time the car spent in the air, we will simply use one of Newton's equation of motion.
Thus;
h = ut + ½gt²
Plugging in the relevant values, we have;
1.8 = 7t + ½(9.8)t²
4.9t² + 7t - 1.8 = 0
Using quadratic formula to find the roots of the equation gives us;
t = -1.65 or 0.22
We can't have negative t value, thus we will pick the positive one.
So, t = 0.22 s
This is approximately 0.2 s
Answer:
a) 
b) 
Explanation:
a) The displacement of the first object is 22.5 m, so we can use the next equation:
positive acceleration.
b) Using the same equation we can find the second value of the acceleration:
positive acceleration.
I hope it helps you!
