Answer:
Explanation:
a) Using the equation of motion
S = ut + 1/2gt²
S is the distance of fall
g is the acceleration due to gravity
t is the time taken
Given S = 12.0m, g = 9.81m/s^2, un= 0m/s
12 = 0+1/2(9.81)t²
12 = 4.905t²²²
t² = 12/4.905
t² = 2.446
t = √2.446
t = 1.56secs
b) To determine how fast is the frog falling at this point, we need to calculate the speed of the frog. Using the equaton v = u+gt
v = 0+9.81(1.56)
v = 15.34m/s
Hence the frog is falling at the rate of 15.34m/s
Answer:
The value of the distance is 
.
Explanation:
The velocity of a particle(v) executing SHM is
where, 
is the angular frequency, 
is the amplitude of the oscillation and 
is the displacement of the particle at any instant of time.
The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e., 
.
The maximum velocity(
) is
Divide equation (1) by equation(2).
Given, 
and 
. Substitute these values in equation (3).
The amount of heat in the body in joule
Answer:
f1/f2 =W1/W2 = 1/3
.0 f2 = 3f1
As ,
1/F= 1/f1 +1/f2
...1/40 = 1/f1 - 1/3f1
f1=> 80/3 cm
... f2 = 2f1 = 3 x 80/3 = 80 cm
