Answer:
Explanation:
The law of conservation of mass states that the mass of the elements at the beginning of the reaction(reactants) will equal the mass at the end of the reaction (product) .
In the chemical equation above,the total mass of the reactants is 80g(16+64) and the total mass of the products is also 80g(44+36).therefore the mass remained constant and that's how the equation represents the law of conservation of mass
Answer:
The height at point of release is 10.20 m
Explanation:
Given:
Spring constant : K= 5 x 10 to the 3rd power n/m
compression x = 0.10 m
Mass of block m= 0.250 kg
Here spring potential energy converted into potential energy,
mgh = 1/2 kx to the 2 power
For finding at what height it rise,
0.250 x 9.8 x h = 1/2 x 5 x 10 to the 3 power x (0.10)to the 2 power) - ( g= 9.8 m/8 to the 2 power
h= 10.20
Therefore, the height at point of release is 10.20 m
Answer:
a) k = 2231.40 N/m
b) v = 0.491 m/s
Explanation:
Let k be the spring force constant , x be the compression displacement of the spring and v be the speed of the box.
when the box encounters the spring, all the energy of the box is kinetic energy:
the energy relationship between the box and the spring is given by:
1/2(m)×(v^2) = 1/2(k)×(x^2)
(m)×(v^2) = (k)×(x^2)
a) (m)×(v^2) = (k)×(x^2)
k = [(m)×(v^2)]/(x^2)
k = [(3)×((1.8)^2)]/((6.6×10^-2)^2)
k = 2231.40 N/m
Therefore, the force spring constant is 2231.40 N/m
b) (m)×(v^2) = (k)×(x^2)
v^2 = [(k)(x^2)]/m
v = \sqrt{ [(k)(x^2)]/m}
v = \sqrt{ [(2231.40)((1.8×10^-2)^2)]/(3)}
= 0.491 m/s
Answer
given,
speed of motorist = 25 m/s
distance = 100 m
time to apply break = 0.5 s
a) distance traveled during the reaction time
D = s x t
D = 25 x 0.5
D = 12.5 m
b) the velocity of the motorist at the end of the reaction time is equal to 0 m/s
c) using equation of motion
v² = u² + 2 a s
v = 0 m/s
u = 25 m/s
0² = 25² + 2 x a x 100
a = -3.125 m/s²
