Answer:
period of oscillations is 0.695 second
Explanation:
given data
mass m = 0.350 kg
spring stretches x = 12 cm = 0.12 m
to find out
period of oscillations
solution
we know here that force
force = k × x .........1
so force = mg = 0.35 (9.8) = 3.43 N
3.43 = k × 0.12
k = 28.58 N/m
so period of oscillations is
period of oscillations = 2π ×
................2
put here value
period of oscillations = 2π ×
period of oscillations = 0.6953
so period of oscillations is 0.695 second
Answer:
So the minimum force is
32.2Newton
Explanation:
To solve for the minimum force, let us assume it to be F (N)
So
F=mgsinA
But
=>>>> coefficient of static friction x (F + mgcosA
=>3 x 9.8 x sin35 = 0.3 x (F + 3 x 9.8 x cos35)
So making F subject of formula
F + 24.0 = 56.2
F = 32.2N
Answer:
Speed at which it will reach the ground is given as

Total time for which it will remain in air is given as
t = 6.3 s
Explanation:
As we know that the object is projected upwards with speed


now when it will reach the ground then we have

so we have


so we have

Now speed of the object when it reaches the ground is given as



Answer:
Explanation:
the one thrown below the horizontal is going straight down, while the one above the horizontal will experience a projectile motion which will makes it move farther away from the building where it was projected.
Answer:
a) t = 0.528 s
b) D = 1.62 m
Explanation:
given,
speed of the baseball = 3.75 m/s
angle made with the horizontal = 35°
height of the roof edge = 2.5 m
using equation of motion



4.9 t² + 2.15 t - 2.5 = 0
on solving the above equation
t = 0.528 s
b) D = v cos θ × t
D = 3.75 × cos 35° ×0.528
D = 1.62 m