20. For each improvement in glider design, engineers follow

I need help plzzz 10 points

Yakvenalex [24]

Explanation:

The numbers do not increase much more then this, so it must be the max

3 0

3 years ago

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A uniform ladder is 10 m long and weighs 400 n. it rests with its upper end against a frictionless vertical wall. its lower end

Nikolay [14]

<h3><u>Answer;</u></h3>

200 N

<h3><u>Solution;</u></h3>

Given ;

Weight = 400 N

Length of the ladder = 10 m

Using the equation;

Torque = d x F, where F is the perpendicular force and d is the distance

Calculating torque about the lower end,

Tweight = Twall

400N(5cos30) = F (10sin30)

F = 200 N

Therefore, the magnitude of force exerted on the peg by the ladder is 200 N

5 0

4 years ago

( 45 points)

WINSTONCH [101]

The British physicist Joseph John (J. J.) Thomson (1856–1940) performed a series of experiments in 1897 designed to study the nature of electric discharge in a high-vacuum cathode-ray tube, an area being investigated by many scientists at the time. Thomson's model showed the atom as a positively charged ball of matter with negatively changed electrons floating freely around inside of it. This model showed the atom having no structure. There are also no protons and neutrons in this model. Thomson knew that the atom had positively and negatively charges particles in it he just didn't know how they were arranged. <span>Today's model gives us a much clearer picture of the atom. There is a positively charged center of the atom that is denser than the rest of it called the nucelus. This dense center is made up of positively charged protons and neutrally charged neutrons. Around the outside of the nucleus the electrons are organized on rings. These electrons are arranged in a certain pattern that is the same for all atoms.</span>

6 0

4 years ago

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Examples of uniform velocity

slava [35]

Explanation:

When a truck travels in equal distances in equal intervals of time then we say that the body has got a uniform velocity. In the above example a truck is traveling at 5 miles in all the positions at A, B, and C and all in the intervals of 5 minutes each.

8 0

3 years ago

A treasure map directs you to start at palm tree

masya89 [10]

The direction of the 90° turns are the possible directions used for the calculations

The four distances and directions are;

<u>24.2 m, 65.6° West of North</u>

<u>29.7 m, 42.7° East of North</u>

<u>24.2 m, 65.6° East of North</u>

<u>29.7m, 47.7° West of North</u>

Reason:

Let point A represent the motion of the treasure hunter, we have:

Turning West then West;

Walking due north to location (0, 15)

Turn 90° West and walk 22.0 m to the location (-22, 15)

Turn 90° West again and walk 5.00 m. to the location (-22, 10)

The location of point A = (-22, 10)

$Direction \ of \ point \ A = arctan \left(\dfrac{10}{-22} \right) \approx -24.4^{\circ}$

Direction to North = 90° - 24.4° ≈ 65.6°

Distance = √((-22)² + 10²) ≈ 24.2

Therefore, we have;

24.2 m, 65.6° West of North

Turning East then West:

Turn 90° East and walk 22.0 m to the location (22, 15)

Turn 90° West again and walk 5.00 m. to the location (22, 20)

The location of point A = (22, 20)

$Direction \ of \ point \ A = arctan \left(\dfrac{20}{22} \right) \approx 42.7^{\circ}$

Direction to North = 90° - 42.3° ≈ 47.7° East

Distance = √((-22)² + 20²) ≈ 29.7

Therefore, we have;

29.7 m, 42.7° East of North

Turning East then East:

Turn 90° East and walk 22.0 m to the location (22, 15)

Turn 90° East again and walk 5.00 m. to the location (22, 10)

The location of point A = (22, 10)

$Direction \ of \ point \ A = arctan \left(\dfrac{10}{22} \right) \approx 24.4^{\circ}$

Direction to North = 90° - 24.4° ≈ 65.6° East

Distance = √((-22)² + 10²) ≈ 24.2

Therefore, we have;

24.2 m, 65.6° East of North

Turning West then East:

Turn 90° West and walk 22.0 m to the location (-22, 15)

Turn 90° East and walk 5.00 m. to the location (-22, 20)

The location of point A = (-22, 20)

$Direction \ of \ point \ A = arctan \left(\dfrac{20}{-22} \right) \approx -42.7^{\circ}$

Direction to North = 90° - 42.7° ≈ 47.7° West

Distance = √((-22)² + 20²) ≈ 29.7

Therefore, we have;

29.7m, 47.7° West of North

Learn more here:

brainly.com/question/11489059

6 0

2 years ago