Time period of a simple pendulum if it makes 40 oscillations in 20 seconds.

You are considering developing a new food-chopping appliance which uses an Arduino with a touchscreen to control the motor. You

trasher [3.6K]

Answer:

Explanation:

The food chopping appliances in this scenario should adapt use of digital image processing systems to capture the images after the chopping begins. Certain variables can be defined like food particles size, area etc. This will help deciding a threshold for the variable and then the comparator used in the system will(run by real time operating system) compare the threshold size with the digital image size of particles. If the size is found small then the speed of motor should reduce meaning we have chopped the food as per our need. In case the chopping is not done aptly, the same can be enhanced by increasing the speed of motor till the final outcome is reached

4 0

3 years ago

The electric field strength E 0 is measured at a perpendicular distance R from an infinitely large, thin sheet that contains a u

Brilliant_brown [7]

Answer:

Electric Field Strength E₀ = E₀ = Constant

Explanation:

The Electric Filed Strength E₀ to an Infinite uniformly charge large sheet is constant how far is it i.e. it is independent of the distance away from the uniformly charge sheet.

Formula: E₀ = σ / 2 ε₀

7 0

3 years ago

A spring suspended vertically is 11.9 cm long. When you suspend a 37 g weight from the spring, at rest, the spring is 21.5 cm lo

Naddika [18.5K]

Answer:

Time period of oscillations is 0.62 s

Explanation:

Due to suspension of weight the change in the length of the spring is given as

$\Delta L = L_f - L_i$

$\Delta L = 21.5 - 11.9 = 9.6 cm$

now we know that spring is stretched due to its weight so at equilibrium the force due to weight is counter balanced by the spring force

$mg = kx$

$0.037 (9.81) = k(0.096)$

$k = 3.78 N/m$

Now the period of oscillation of spring is given as

$T = 2\pi \sqrt{\frac{m}{k}}$

Now plug in all values in it

$T = 2\pi \sqrt{\frac{0.037}{3.78}}$

$T = 0.62 s$

5 0

3 years ago

In a certain process, the energy change of the system is 250 \rm kJ. The process involves 480 \rm kJ of work done by the system.

alisha [4.7K]

Answer:

$\Delta Q=-230kJ$

Explanation:

Using the first law of thermodynamics:

$\Delta U=\Delta Q-W$

Where $\Delta U$ is the change in the internal energy of the system, in this case $\Delta U=250kJ$ , $\Delta Q$ is the heat tranferred, and $W$ is the work, $W=-480kJ$ with a negative sign since the work is done by the system.