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1 year ago

What is one advantage of using primary sources when doing research on an

1 answer:

5 0

One advantage of using primary sources when doing research on an experiment is, it contain more details about the experiment. Thus, option D is right .

To find the answer, we have to know more about the Primary sources.

<h3>What is the advantages of primary sources?</h3>

Primary sources are the direct or first-hand excerpts that scientists have independently written based on their experiments in order to support their research.
Utilizing original sources encourages analytical and critical thought in relation to the research.

It aids in exploring from various angles, which leads to the discovery of extra facts.
Primary data aids in navigating the conflicts. Since it is a direct resource related to the experiment, it serves as evidence for the data.

Thus, we can conclude that, Primary sources contain more details about the experiment.

Learn more about the Primary sources here:

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Answer:

Explanation:

because they are easily reduced and oxidised

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2 years ago

Order, from smallest to largest, the following objects:

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2 years ago

Read 2 more answers

Please help me with questions 1, 2 and 3. <br> i need a step by step explanation

kifflom [539]

Answer:

1) d

2) 5 m/s

3) 100

Explanation:

The equation of position x for a constant acceleration a and an initial velocity v₀, initial position x₀, time t is:

(i) $x=\frac{1}{2}at^2+v_0t+x_0$

The equation for velocity v and a constant acceleration a is:

(ii) $v=at+v_0$

1) Solve equation (ii) for acceleration a and plug the result in equation (i)

(iii) $a = \frac{v -v_0}{t}$

(iv) $x = \frac{v-v_0}{2t}t^2+v_0t + x_0$

Simplify equation (iv) and use the given values v = 0, x₀ = 0:

(v) $x=-\frac{v_0}{2}t + v_0t= \frac{v_0}{2}t$

2) Given v₀= 3m/s, a=0.2m/s², t=10 s. Using equation (ii) to get the final velocity v: $v=at+v_0=0.2\frac{m}{s^2} * 10s+3\frac{m}{s}=2\frac{m}{s}+3\frac{m}{s}=5\frac{m}{s}$

3) Given v₀=0m/s, t₁=10s, t₂=1s and x₀=0. Looking for factor f = x(t₁)/x(t₂) using equation(i) to calculate x(t₁) and x(t₂):

$f=\frac{x(t_1)}{x(t_2)}=\frac{\frac{1}{2}at_1^2 }{\frac{1}{2}at_2^2}=\frac{t_1^2}{t_2^2}=\frac{10^2}{1^2}=\frac{100}{1}$

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3 years ago

A driver averaged 64 mph and took 3½ hours to travel from st. Louis to chicago. Based on this, what is the distance between st.

sammy [17]

Average speed of the driver is given as

$v = 64 mph$

if he moved for total time t = 3.5 hours

so the distance between the tow is given as

$d = v* t$

$d = 64 * 3.5$

$d = 224 miles$

so the distance between St. Louis and Chicago is 224 miles

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3 years ago

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3 years ago