$W = \int{F(x)} \, dx = \int\limits^0_{-20} {(-kx)} \, dx = \frac{-kx^2}{2}\left \{ {{x=0} \atop {x=-20}} \right. = \frac{-60*(-20)^2}{2} \\W = -12000J$

Explanation:

Alternatively, the work done by the object is equal to the elastic potantial energy done by the spring.

$U = \frac{1}{2}kx_2^2 - \frac{1}{2}kx_1^2 =0 - \frac{1}{2}60(-20)^2 = -12000J$

6 0

3 years ago

Which of the following is equivalent to 2.5 meters

blsea [12.9K]

B is the correct answer for sure bro

5 0

3 years ago

Read 2 more answers

Oscilloscopes have parallel metal plates inside them to deflect the electron beam. These plates are called the deflecting plates

Alina [70]

Answer:

The capacitance of the deflecting plates is $C=1.59 pF$ .

Explanation:

The expression for the capacitance of the capacitor in terms of area and distance is as follows;

$C=\frac{\varepsilon _{0}A}{d}$

Here, C is the capacitance, A is the area, d is the distance and $\varepsilon _{0}$ is the absolute permittivity.

Convert the side of the square from cm to m.

s= 3.0 cm

s= 0.030 m

Calculate the area of the square.

$A= s_^{2}$

Put s= 0.030 m.

$A=(0.030)_^{2}$

$A=9\times10^{-4}m^{-2}$

Convert distance from mm to m.

d= 5.0 mm

$d=5\times10^{-3}m$

Calculate the capacitance of the deflecting plates.

$C=\frac{\varepsilon _{0}A}{d}$

Put $d=5\times10^{-3}m$ , $A=9\times10^{-4}m^{-2}$ and $\varepsilon _{0}=8.85\times 10^{-12}Fm^{-1}$ .

$C=\frac{(8.85\times 10^{-12})(9\times10^{-4})}{5\times10^{-3}}$

$C=1.59\times 10^{-12}F$

$C=1.59 pF$

Therefore, the capacitance of the deflecting plates is $C=1.59 pF$ .

4 0

3 years ago