kinetic energy is given by 1 /2
then you have to replace the figures
A 0. 20-kg ball is attached to a vertical spring. the spring constant is 28 n/m. when released from rest, how far does the ball
1 answer:
A 0. 20-kg ball is attached to a vertical spring. The spring constant is 28 n/m. when released from rest, the ball will fall at a distance of x = 0.14 m before being brought to a momentary stop by the spring.
The spring constant is the force needed to stretch or compress a spring, divided by the distance that the spring gets longer or shorter. It's used to determine stability or instability in a spring
Elastic potential energy is Potential energy stored as a result of deformation of an elastic object, such as the stretching of a spring. It is equal to the work done to stretch the spring, which depends upon the spring constant k as well as the distance stretched.
When ball is released from rest with spring , it will loose some potential energy . Hence , loss in potential energy of the ball = gain in potential energy of spring
m*g*x = 1/2 * k *
0.20 * 9.8 * x = 1/2 * 28 *
x = (2 * 0.20 * 9.8) / 28
x = 0.14 m
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Answer:
a) V = 0.085 m^3
b) m = 17 kg
Explanation:
1) Data given
mb = 68 kg (mass for the block)
20% of the block volume is floating
100-20= 80% of the block volume is submerged
2) Notation
mb= mass of the block
Vw= volume submerged
mw = mass water displaced
V= total volume for the block
3) Forces involved (part a)
For this case we have two forces the buoyant force (B), defined as the weight of water displaced acting upward and the weight acting downward (W)
Since we have an equilibrium system we can set the forces equal. By definition the buoyant force is given by :
B = (mass water displaced) g = (mw) g (1)
The definition of density is :
If we solve for mw we got (2)
Replacing equation (2) into equation (1) we got:
(3)
On this case Vw represent the volume of water displaced = 0.8 V
If we replace the values into equation (3) we have
0.8 ρ_w V g = mg (4)
And solving for V we have
V = (mg)/(0.8 ρ_w g )
We cancel the g in the numerator and the denominator we got
V = (m)/(0.8 ρ_w)
V = 68kg /(0.8 x 1000 kg/m^3) = 0.085 m^3
4) Forces involved (part b)
For this case we have bricks above the block, and we want the maximum mass for the bricks without causing it to sink below the water surface.
We can begin finding the weight of the water displaced when the block is just about to sink (W1)
W1 = ρ_w V g
W1 = 1000 kg/m^3 x 0.085 m^3 x 9.8 m/s^2 = 833 N
After this we can calculate the weight of water displaced before putting the bricks above (W2)
W2 = 0.8 x 833 N = 666.4 N
So the difference between W1 and W2 would represent the weight that can be added with the bricks (W3)
W3 = W1 -W2 = 833-666.4 N = 166.6 N
And finding the mass fro the definition of weight we have
m3 = (166.6 N)/(9.8 m/s^2) = 17 Kg
Answer:
t = 4.0 min
Explanation:
given data:
diameter of rod = 2 cm
T_1 = 100 degree celcius
Air stream temperature = 20 degree celcius
heat transfer coefficient = 200 W/m2. K
WE KNOW THAT
copper thermal conductivity = k = 401 W/m °C
copper specific heat Cp = 385 J/kg.°C
density of copper = 8933 kg/m3
charateristic length is given as Lc
Biot number is given as
Bi = 0.0025
As Bi is greater than 0.1 therefore lumped system analysis is applicable
so we have
............1
where b is given as
b = 0.01163 s^{-1}
putting value in equation 1
solving for t we get
t = 4.0 min