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matrenka [14]
2 years ago
11

Joshua programs a robotic arm to place a block in a box by following the sequence of commands in the table. a 2-column table wit

h 4 rows. the first column labeled step has entries 1, 2, 3, 4. the second column labeled command has entries, grip block, lift block 10 centimeters, hold for 20 seconds, release block. during which step does the robotic arm do work on the block? step 1 step 2 step 3 step 4.
Physics
1 answer:
GarryVolchara [31]2 years ago
6 0

Step 2 does the robotic arm do work on the block. Joshua programs a robotic arm to place a block in a box. a four-row, two-column table.

<h3>What is a robotic arm?</h3>

A robotic arm is a type of artificial arm that may be programmed to do tasks similar to those of a human arm.

The arm can be the sum of the mechanism or part of a larger robot. Joints connect the manipulator's linkages, allowing rotational or translational motion.

By following the directions in the table, Joshua programs a robotic arm to place a block in a box. a four-row, two-column table.

The items in the first column labeled step are 1, 2, 3, and 4. Grip block, elevate block 10 cm, hold for 20 seconds, release block are the items in the second column labeled command.

Step 2 does the robotic arm do work on the block.

Hence,step 2 is the correct answer.

To learn more about the robotic arm refer:

brainly.com/question/14768119

#SPJ1

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i think its 470

Explanation:

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Question 4 (18 marks) (a) During a Physics Lab experiment, 1 st year SFY students analyzed the behavior of capacitors by connect
Nataly_w [17]

Answer:

1.) 274.5v

2.) 206.8v

Explanation:

1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.

The potential difference and charge across EACH capacitor will be

V = Voe

Where Vo = initial voltage

e = natural logarithm = 2.718

For the first capacitor 2.50 µF,

V = Vo × 2.718

746 = Vo × 2.718

Vo = 746/2.718

Vo = 274.5v

To calculate the charge, use the below formula.

Q = CV

Q = 2.5 × 10^-6 × 274.5

Q = 6.86 × 10^-4 C

For the second capacitor 6.80 µF 

V = Voe

562 = Vo × 2.718

Vo = 562/2.718

Vo = 206.77v

The charge on it will be

Q = CV

Q = 6.8 × 10^-6 × 206.77

Q = 1.41 × 10^-3 C

B.) Using the formula V = Voe again

165 = Vo × 2.718

Vo = 165 /2.718

Vo = 60.71v

Q = C × 60.71

Q = C

4 0
3 years ago
Which starting material is the limiting reagent in this procedure? which reagent is used in excess? how great is the molar exces
alexandr1967 [171]

The limiting reagent in this procedure is the substance that is completely used in the reaction. The substance which is not completely used is called excess.

The beginning cloth in a chemical reaction is known as the reactant. each chemical response could have one or more reactants.

In a chemical response limiting reagent is the reactant this is fed on first and prevents any similar reaction from happening. the quantity of product shaped during the response is decided by means of the limiting reagent. as an example, allow us to recall the response of solution and chlorine

Materials used during the manufacture of the lively substance (e.g., tradition media, growth factors, and so on) and that are not meant to form a part of the active substance shall be taken into consideration as raw substances. cloth forming a critical part of the lively materials element shall be considered as the beginning substance.

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2 years ago
If the 3rd harmonic has a frequency of 600Hz, the frequency of the fundamental is ____________?
castortr0y [4]

Answer:

Fundamental frequency is 200 Hz.

Explanation:

It is given that,

Frequency of the 3rd harmonic is 600 Hz.

Let f is the fundamental frequency. We need to find the value of f. The frequency of third harmonic is given by :

3f=600\ Hz

So, fundamental frequency f is equal to :

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So, the fundamental frequency of the harmonics is 200 Hz. Hence, this is the required solution.

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