Answer:
0.6 m
Explanation:
When a spring is compressed it stores potential energy. This energy is:
Ep = 1/2 * k * x^2
Being x the distance it compressed/stretched.
When the spring bounces the ice cube back it will transfer that energy to the cube, it will raise up the slope, reaching a high point where it will have a speed of zero and a potential energy equal to what the spring gave it.
The potential energy of the ice cube is:
Ep = m * g * h
This is vertical height and is related to the distance up the slope by:
sin(a) = h/d
h = sin(a) * d
Replacing:
Ep = m * g * sin(a) * d
Equating both potential energies:
1/2 * k * x^2 = m * g * sin(a) * d
d = (1/2 * k * x^2) / (m * g * sin(a))
d= (1/2 * 25 * 0.1^2) / (0.05 * 9.81 * sin(25)) = 0.6 m
Answer: b is sedimentary. c is metamorphic. and a is igneous.
Explanation:
Answer: 2.94×10^8 J
Explanation:
Using the relation
T^2 = (4π^2/GMe) r^3
Where v= velocity
r = radius
T = period
Me = mass of earth= 6×10^24
G = gravitational constant= 6.67×10^-11
4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]
= 0.9865 x 10^-13
Therefore,
T^2 = (0.9865 × 10^-13) × r^3
r^3 = 1/(0.9865 × 10^-13) ×T^2
r^3 = (1.014 x 10^13) × T^2
To find r1 and r2
T1 = 120min = 120*60 = 7200s
T2 = 180min = 180*60= 10800s
Therefore,
r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m
r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m
Required Mechanical energy
= - GMem/2 [1/r2 - 1/r1]
= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]
= (2001 x 10^7)/2 * (0.1239 - 0.0945)
= (1000.5 × 10^7) × 0.0294
= 29.4147 × 10^7 J
= 2.94 x 10^8 J.
Burning of gases is one the example of chemical change
