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alina1380 [7]
3 years ago
13

How are chargeable cells different from ordinary dry cells​

Physics
1 answer:
topjm [15]3 years ago
6 0
Ordinary cells can convert chemical energy to electrical energy only, but rechargeable cells can also store electrical energy into chemical energy and vice versa. You will study more about it in your higher classes. secondary cells can be recharged and used again but dry cells cannot be recharged.
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Using a schematic diagram, explain the steps of the laser technique.
nignag [31]

The steps of the use of the laser technique is explained below:

  • A laser beam is emitted from the diode in the unit and spread into a laser plane (1).
  • The laser plane, appearing as a line on the sphere (2), is reflected and collected by dual CCD arrays (1).
  • The resulting 2D profile is digitized and as the unit travels along the x-axis of the object, multiple profiles are collected yielding a 3D coordinate point cloud of the surface (3).

<h3>What is a Laser Technique?</h3>

This refers to the type of surgery that makes use of special light beams in order to cut open the human body in a surgical procedure.

Hence, we can see that the laser technique is considered safer than conventional surgical methods.

Laser techniques include:

  • laser vaporization,
  • laser resection-enucleation,  
  • coagulation.

Read more about laser technique here:

brainly.com/question/14091177

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3 0
2 years ago
HELPP PLEEEAAAAASSSEEEEWKKKKKKK!!!!!
nalin [4]
1. make good decisions

2. explosive

3. favorable

answered these on edge. also, you marked this as physics even though its english. might wanna watch out for that lol, so youll get a quicker answer
8 0
3 years ago
Read 2 more answers
A solid circular shaft and a tubular shaft, both with the same outer radius of c=co = 0.550 in , are being considered for a part
Norma-Jean [14]

Answer:

The power for circular shaft is 7.315 hp and tubular shaft is 6.667 hp

Explanation:

<u>Polar moment of Inertia</u>

(I_p)s = \frac{\pi(0.55)4}2

      = 0.14374 in 4

<u>Maximum sustainable torque on the solid circular shaft</u>

T_{max} = T_{allow} \frac{I_p}{r}

         =(14 \times 10^3) \times (\frac{0.14374}{0.55})

         = 3658.836 lb.in

         = \frac{3658.836}{12} lb.ft

        = 304.9 lb.ft

<u>Maximum sustainable torque on the tubular shaft</u>

T_{max} = T_{allow}( \frac{Ip}{r})

          = (14 \times10^3) \times ( \frac{0.13101}{0.55})

          = 3334.8 lb.in

          = (\frac{3334.8}{12} ) lb.ft

          = 277.9 lb.ft

<u>Maximum sustainable power in the solid circular shaft</u>

P_{max} = 2 \pi f_T

          = 2\pi(2.1) \times 304.9

          = 4023.061 lb. ft/s

          = (\frac{4023.061}{550}) hp

          = 7.315 hp

<u>Maximum sustainable power in the tubular shaft</u>

P _{max,t} = 2\pi f_T

            = 2\pi(2.1) \times 277.9

            = 3666.804 lb.ft /s

            = (\frac{3666.804}{550})hp

            = 6.667 hp

7 0
3 years ago
When a single source of light shines through an extremely thin rectangular slit and projects on a far away viewing screen, a sin
saveliy_v [14]
True
If it helps you plz brainlest me
7 0
3 years ago
Car A hits car B (initially at rest and of equal mass) from behind while going 15 m/s Immediately after the collision, car B mov
Mamont248 [21]

Given :

Initial speed of car A is 15 m/s and initial speed of car B is zero.

Final speed of car A is zero and final speed of car B is 10 m/s.

To Find :

What fraction of the initial kinetic energy is lost in the collision.

Solution :

Initial kinetic energy is :

K.E_i = \dfrac{15^2m}{2} + 0\\\\K.E_i = \dfrac{225 m}{2}

Final kinetic energy is :

K.E_f = \dfrac{10^2m}{2} + 0\\\\K.E_f = \dfrac{100m}{2}

Now, fraction of initial kinetic energy loss is :

Loss = \dfrac{\dfrac{225m}{2}-\dfrac{100m}{2}}{\dfrac{100m}{2}}\\\\Loss = \dfrac{125}{100}\\\\Loss = 1.25

Therefore, fraction of initial kinetic energy loss in the collision is 1.25 .

6 0
3 years ago
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