When does spring wardrobe start selling in stores?

soon, like the end of February.

What is the spring stiffness constant of the spring, and how do you calculate the formula?

the formula for energy in a spring is 1/2kx^2 using conservation of energy, calculate the kinetic energy of the car =1/2 m v^2 =1/2 *1200kg * (65km/hr*60min/hr *60sec/min*1000m/km)^2 the energy of the spring is equal to that =1/2 *k *2.2^2 k=1.36x10^19 that's a strong spring j

What is the spring constant of the spring and what is the period of the oscillation?

The equation for the force of a spring is: F = -k*(dx) k - spring constant dx - change in elongation of spring The force of gravity on the ball has to be the same as the force on the spring (when at rest), so mg = k*(dx) (you can drop the negative sign since you know the forces are in opposite directions) When you plug in all the numbers, you get k = 19.6 N/m (pay attention to units... the easiest thing to do is change everything to meters and kilograms) For the second part, the frequency of the motion is: T = 2*pi*(m/k)^0.5 You now have both m (mass) and k, so you can plug in and get T = 0.401 s

What is the spring stiffness constant of a horizontal spring struck by a 1200-kg car?

KE = SE ½m*V² = ½k*X² → k = m*V²/X² where V = (k/h)/3.6 = 13.89 m/s k = 1200*13.89²/2.4² = 40188 N/m

How do i determine the spring constant for this question about two springs?

well there is the spring equation F=kx, right? in this case the force will be the force due to gravity so F=mg

What is the potential energy stored in the spring when the spring is compressed by 20 cm?

The potential energy stored in a spring is given by this formula: U = 1/2 • k • x^2 Where U is the potential energy, k is the spring constant, and x is the distance that the spring is compressed or stretched. Thus, in this case: U = 1/2 • k • x^2 U = 1/2 • 100 N/m • (0.2 m)^2 U = 50 N/m • 0.04 m^2 U = 2 Joules The kinetic energy of the block will be equal to the potential energy of the spring. This happens because all of the energy stored as elastic potential energy in the spring is converted into kinetic energy when the spring is released. Since KE = 1/2 • m • v^2, we know that: 1/2 • m • v^2 = U m • v^2 = 2U v^2 = 2U/m v = SQRT (2U/m) By substituting m = 0.5 kg and U = 2 Joules, you get: v = SQRT (4 J / 0.5 kg) v = SQRT (8 m^2/s^2) v = 2.828 m/s So, our final answer is that the potential energy in the spring is 2 Joules, and the velocity of the block is 2.828 m/s. Hope this helps!

What is the spring constant and the maximum compression?

a) k(avg)= (k1 + k2)/2 k(avg)= (4+ 6)/2=5 N/m b) Ps=.5k(avg)X^2 (potential energy of the spring) Ke=.5mV^2 (kinetic energy of the mass) Pe=Ke then X(max)=sqrt(mV^2/k(avg)) X(max)=sqrt(1.0 (0.5)^2 / 5) X(max)=0.316 m or 31.6 cm

When does spring start informally in your hometown?

I'd say in Missouri Spring starts around mid-March, but Missouri weather is really unpredictable. It could be snowing or 80 degrees....or both in a day.

A spring is suspended from the ceiling. What is the spring constant and maximum velocity of the object?

a) k = F/∆ = mg/∆ = 10*9.8/.4 = 245 N/m b) ω = √(k/m) = √(245/10) = 1.565 rad/s; Vmax = .20*ω = .313 m/s


i'll bling up the day i have money