Cells & Cellular Processes

A student pushes a 0.2 kg box against a spring causing the spring to compress 0.15 m. When the spring is released, it will launc

german

Answer:

The maximum height the box will reach is 1.72 m

Explanation:

F = k·x

Where

F = Force of the spring

k = The spring constant = 300 N/m

x = Spring compression or stretch = 0.15 m

Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N

Mass of box = 0.2 kg

Work, W, done by the spring = $\frac{1}{2} kx^2$ and the kinetic energy gained by the box is given by KE = $\frac{1}{2} mv^2$

Since work done by the spring = kinetic energy gained by the box we have

$\frac{1}{2} mv^2$ = $\frac{1}{2} kx^2$ therefore we have v = $\sqrt{\frac{kx^2}{m} }$ = $x\sqrt{\frac{k}{m} }$ = $0.15\sqrt{\frac{300}{0.2} }$ = 5.81 m/s

Therefore the maximum height is given by

v² = 2·g·h or h = $\frac{v^2}{2g}$ = $\frac{5.81^{2} }{2*9.81}$ = 1.72 m

6 0

3 years ago

A car is traveling south is 200 kilometers from it’s starting point after 2 hours. What is the average velocity of the car

Lesechka [4]

Answer:

100

Explanation:

take note that v=d/t (velocity is distance over(divided by) time, so in this case it would be 200 (distance) divided by 2 (time) = 100

6 0

3 years ago

Ted William drops a ball from 14.5 meter to a desk that is 1.9 meters tall. What is the final speed of the ball right before it

makkiz [27]

Answer:

15.7m/s

Explanation:

To solve this problem, we use the right motion equation.

Here, we have been given the height through which the ball drops;

Height of drop = 14.5m - 1.9m = 12.6m

The right motion equation is;

V² = U² + 2gh

V is the final velocity

U is the initial velocity = 0

g is the acceleration due to gravity = 9.8m/s²

h is the height

Now insert the parameters and solve;

V² = 0² + 2 x 9.8 x 12.6

V² = 246.96

V = √246.96 = 15.7m/s

4 0

3 years ago

A spring with spring constant 33N/m is attached to the ceiling, and a 4.8-cm-diameter, 1.5kg metal cylinder is attached to its l

mylen [45]

Answer:

0.423m

Explanation:

Conversion to metric unit

d = 4.8 cm = 0.048m

Let water density be $\who_w = 1000 kg/m^3$

Let gravitational acceleration g = 9.8 m/s2

Let x (m) be the length that the spring is stretched in equilibrium, x is also the length of the cylinder that is submerged in water since originally at a non-stretching position, the cylinder barely touches the water surface.

Now that the system is in equilibrium, the spring force and buoyancy force must equal to the gravity force of the cylinder. We have the following force equation:

$F_s + F_b = W$

Where $F_s = kx$ N is the spring force, $F_b = W_w = m_wg = \rho_w V_s g$ is the buoyancy force, which equals to the weight $W_w$ of the water displaced by the submerged portion of the cylinder, which is the product of water density $\rho_w$ , submerged volume $V_s$ and gravitational constant g. W = mg is the weight of the metal cylinder.