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Sladkaya [172]
3 years ago
12

there is only one basic format for a professional report. a. audience b. All of these should be considered when deciding on a re

port format c. length d.purpose
Physics
1 answer:
Anna007 [38]3 years ago
5 0

Answer:

The answer is letter b. All of these should be considered when deciding on a report format.

Explanation:

A Professional Report is a type of formal document about a topic or information that is intended for a specific audience or purpose. The report's style of writing needs a lot of knowledge from the writer. Oftentimes, it involves the following important elements: <em>Title, Summary, Body, Discussion, Conclusion and Recommendation. </em>

The writer should write according to his target audience and purpose. He also needs to consider the length of his report, as well as the suitable words and sentences that he should use.

Thus, all of the choices are important in writing a professional report. So, the answer is letter b.

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An experiment that produces waves on a string is analyzed, and the experimental value for the wave speed is found to be 66 m/s.
Alexandra [31]

4.8 is the correct answer

7 0
3 years ago
Read 2 more answers
An object that has lost its electrons become?​
Andrej [43]
Answer: A positively charged ion/ cation

Explanation: When an atom loses all of its electrons, it becomes a positively charged ion. Also referred to as the cation, the object will have a shortage of electrons, leading to a positive charge. Keep in mind that an atom will become positive if it loses electrons, and will become negative if it gains electrons.


Hope this helps! Comment below for more questions.
7 0
3 years ago
A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.820 m. W
Vikentia [17]

Answer:

1.0752 kgm/s

Explanation:

Considering when the drop was dropped from rest from a height,

mass of the ball, m = 0.120 kg

height, h = - 1.25 m

the initial velocity, u = 0 m/s

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

                            V^{2} = U^{2} + 2gh

Substituting the values,

                             V^{2} = 0^{2} + 2(-9.8 m/s^{2})(-1.25 m)

                             V^{2} = 24.5 m/s

                             V = \sqrt{24.5} \ m/s

                             V = 4.95 \ m/s

                            V = ± 4.95 m/s

                            V = - 4.95 m/s

Since the ball is moving downward, the final velocity of the ball when it hits the floor is  V = - 4.95 m/s  

Considering when the ball rebounds from the floor,

assume the mass of the ball still remain, m = 0.120 kg

height, h = 0.820 m

the final velocity, v = 0 m/s  

the acceleration due to gravity, g = - 9.8 m/s²

From equation of motion

                            V^{2} = U^{2} + 2gh

Substituting the values,

                            0^{2} = U^{2} + 2(-9.8 m/s^{2})(0.820 m)

                            0 = U^{2} - 16.072 m/s

                            U^{2} = 16.072 m/s

                            U = \sqrt{16.072} \ m/s

                           U = ± 4.01 m/s

                          U = + 4.01 m/s

Since the ball is moving upward, the initial velocity of the ball from the bounce from the floor is  U = + 4.01 m/s                        

From Newton's second law of motion, applied force is directly proportional to the rate of change in momentum.

                            F = \frac{mv - mu}{t}

                          F.t = m(v - u)

       ⇒      Impulse = Change in momentum

To calculate the impulse, the moment before the ball hits the ground will be the initial momentum while the moment the ball rebounces will be the final velocity,                        

          ∴          F.t = 0.120  kg(4.01  m/s - (-4.95  m/s) )

                      F.t = 0.120  kg(4.01  m/s + 4.95  m/s) )

                      F.t = 0.120  kg × 8.96  m/s

                      Impulse  = 1.0752 kgm/s

The impulse given to the ball by the floor is 1.0752 kgm/s

                             

6 0
3 years ago
Resistance of a light bulb with 0.33 A of current flowing from a 12V battery?
Sergio039 [100]

Resistance = (voltage) / (current)

Resistance = (12v) / (0.33 A)

Resistance = (12/0.33) ohms

<em>Resistance = 36.4 ohms</em>

8 0
3 years ago
A body has masses of 0.013kg and 0.012kg in oil and water respectively, if the relative density of oil is 0.875, calculate the m
konstantin123 [22]

Answer:

the mass of the body is 0.02 kg.

Explanation:

Given;

relative density of the oil, \gamma _0 = 0.875

mass of the object in oil, M_o = 0.013 kg

mass of the object in water, M_w = 0.012 kg

let the mass of the object in air = M_a

weight of the oil, W_0 = M_a - 0.013

weight of the water, W_w = M_a - 0.012

The relative density of the oil is given as;

\gamma_0 = \frac{density \ of \ oil }{density \ of \ water} = \frac{W_0}{W_w} = \frac{M_a -0.013}{M_a -0.012} \\\\0.875 = \frac{M_a -0.013}{M_a -0.012}\\\\0.875(M_a - 0.012) = M_a - 0.013\\\\0.875M_a - 0.0105 = M_a -0.013\\\\0.875M_a - M_a = 0.0105 - 0.013\\\\-0.125 M_a = -0.0025\\\\M_a = \frac{0.0025}{0.125} \\\\M_a = 0.02 \ kg

Therefore, the mass of the body is 0.02 kg.

6 0
3 years ago
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