Answer:
0.114 kg or 114 g
Explanation:
From the diagram attaches,
Taking the moment about the fulcrum,
sum of clockwise moment = sum of anticlockwise moment.
Wd = W'd'
Where W = weight of the mass, W' = weight of the meter rule, d = distance of the mass from the fulcrum, d' = distance of the meter rule.
make W' the subject of the equation
W' = Wd/d'................ Equation 1
Given: W = mg = 0.0515(9.8) = 0.5047 N, d = (39.2-16) = 23.2 cm, d' = (49.7-39.2) = 10.5 cm
Substitute these values into equation 1
W' = 0.5047(23.2)/10.5
W' = 1.115 N.
But,
m' = W'/g
m' = 1.115/9.8
m' = 0.114 kg
m' = 114 g
Light year is light in a year so 4 light years are 4 Earth years
I believe it'd accelerate at 1.25 m/s^2 instead of 1, as it lost 1/4 of its mass (.25), so now it is .25 of 1 faster.
Answer:
a) 0.036 J b) 0.036J c) 0.036 d) 1.9m/s e) 0.18 m
Explanation:
Mass of the dart = 0.02kg, the spring was compressed to 6cm
Work needed to compress the spring = 1/2*k*x ^2 where k is the force constant of the spring in N/m, x is the distance it was compressed in m
Work needed to compress the spring = 0.5 * 20* 0.06^2 since 6cm = 6 / 100 = 0.06 m
Work needed to compress the spring = 0.036J
b) the total energy stored in the spring = the work done to compress the spring = 0.036J
c) kinetic energy of the dart as it leaves the the spring = elastic potential energy stored in the spring = the work done in compressing the = 0.036J using the law of conservation of energy; energy is neither created nor destroyed but transformed from one form to another.
d) 1/2mv^2 = 0.036
mv^2 = 0.036*2
v^2 = 0.036*2 / 0.02 = 3.6
v = √3.6 = 1.897 approx 1.9m/s
e) kinetic energy of the dart = work done against gravity to get the body to height h
Work done against gravity = potential energy conserved at height = -mgh g is negative because the motion is upward while gravity acts downward
0.036 = 0.02 * 9.81 * h
0.036 / ( 0.02*9.81) = h
h = 0.18 m
Answer:
answer is option 4
Explanation:
you have to use option 4 because u need to find out initial velocity (Vi)
