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3 years ago

How many significant digits are in the measurement 4,032,010 m/s?

2 answers:

5 0

Since there is no decimal point in the number given above, the counting for the number of the significant figures will start from the left. Then, the first zero from the left is insignificant. Therefore, in this number there are 6 significant figures.

3241004551 [841]3 years ago

4 0

Answer:

Explanation:

The rules which we have to use to find the number of significant figures here are given below:

1. If a number does not contain a decimal point, then the zeros between two non zero digits are significant.

2. the trailing zeros are not significant.

The number is

4,032, 010 m/s

so, the zero between 4 and 3 is significant

the zero between 2 and 1 is significant

the zero at the last is not significant

So, the number of significant figures are 6.

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Match each example with the appropriate stage of technological design. drag each arrow to its correct match.

Akimi4 [234]

Answer:

Review each answer in the explanation part

Explanation:

Identify a problem or need = Aerospace engineers need a lightweight material to build a jet

Note: Identify a problem is related always with a need.

After identify the problem the next stage is the design

Design a solution = Engineers determine what features the material must have.

Note: The design process implies the choice of the materials and the proper numeric calculations.

Implement the solution = Engineers test a carbon - plastic compound in a wind tunnel.

Note: after the design, the following process is the built of the equipment and the different tests.

Evaluate the solution = The carbon-plastic compound is redesigned to save maximun energy.

Note: After testing, you enter a part of re-design that means making improvements to the already constructed.

7 0

3 years ago

Please answer of this question

Bogdan [553]

Answer:

$\frac{\pi }{15}$ m

Explanation:

At 10am, the minute hand and hour hand are ' 2 hours apart', since the minute hand is at 12pm and hour hand is at 10am.

Angle between the two hands = 2/12 * 360

= 60°

Arc Length = $2\pi r(\frac{60}{360} )$

= $2\pi (0.2)(\frac{1}{6} )\\= \frac{\pi }{15} m$

8 0

2 years ago

Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu

lora16 [44]

The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

$q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F$

So the force has increased by a factor 6.

8 0

3 years ago

Read 2 more answers

What is the difference between the equilibrium vector and the resultant vector

Tresset [83]

the difference between a resultant and equilibrant vector is that resultant vector is a direct quantity, one with both magnitude and direction, while the equilibrant vector is a force equal to, but opposite of, the resultant sum of vector forces, that force which balances other forces.

8 0

4 years ago

A 1369.4 kg car is traveling at 28.9 m/s when the driver takes his foot off the gas pedal. It takes 5.1 s for the car to slow do

Darya [45]

Answer:

F = 2389.603 N

Explanation:

Given:

Mass m = 1,369.4 kg

Initial velocity u = 28.9 m/s

Final velocity v = 20 m/s

Time t = 5.1 s

Find:

Net force

Computation:

a = (v - u)/t

a = (20 - 28.9)/5.1

a = -1.745 m/s²

F = ma

F = (1369.4)(1.745)

F = 2389.603 N

7 0

3 years ago