Answer:
Keq = 2k₃
Explanation:
We can solve this exercise using Newton's second one
F = m a
Where F is the eleatic force of the spring F = - k x
Since we have two springs, they are parallel or they are stretched the same distance by the object and the response force Fe is the same for the spring age due to having the same displacement
F + F = m a
k₃ x + k₃ x = m a
a = 2k₃ x / m
To find the effective force constant, suppose we change this spring to what creates the cuddly displacement
Keq = 2k₃
Answer:
88.34 N directed towards the center of the circle
Explanation:
Applying,
F = mv²/r................... Equation 1
F = Force needed to keep the mass in a circle, m = mass of the mass, v = velocity of the mass, r = radius of the circle.
But,
v = 2πr/t................... Equation 2
Where t = time, π = pie
Substitute equation 2 into equation 1
F = m(2πr/t)²/r
F = 4π²r²m/t²r
F = 4π²rm/t²............. Equation 3
From the question,
Given: m = 0.8 kg, r = 0.7 m, t = 0.5 s
Constant: π = 3.14
Substitute these values into equation 3
F = 4(3.14²)(0.7)(0.8)/0.5²
F = 88.34 N directed towards the center of the circle
Answer:
v = 79.2 m/s
Solution:
As per the question:
Mass of the object, m = 250 g = 0.250 kg
Angle, 
Coefficient of kinetic friction, 
Mass attached to the string, m = 0.200 kg
Distance, d = 30 cm = 0.03 m
Now,
The tension in the string is given by:
(1)
Also
T = m(g + a)
Thus eqn (1) can be written as:
Now, the speed is given by the third eqn of motion with initial velocity being zero:
where
u = initial velocity = 0
Thus
Less than because a mile is 1600 meters
