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creativ13 [48]
2 years ago
11

Which describes the type of acceleration in the graph?

Physics
1 answer:
Vedmedyk [2.9K]2 years ago
6 0

Answer:

Constant positive acceleration

Explanation:

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What is the role of Langerhans cells?
boyakko [2]

     

Langerhans cells exist on the surface of the skin. They help to protect you by keeping dangerous microbes from entering your body. Langerhans cells do this in two very different ways: by protecting your skin from infection and also by stimulating allergic reactions.

 

4 0
3 years ago
Suppose that the design parameters of the satellite's control system require that the angular velocity of the satellite not exce
Stells [14]

Answer:

v=0.04m/s

Explanation:

To solve this problem we have to take into account the expression

\omega=\frac{v}{r}

where v and r are the magnitudes of the velocity and position vectors.

By calculating the magnitude of r and replacing w=0.02rad/s in the formula we have that

|r|=\sqrt{(-1.18m)^2+(-0.9m)^2}=2.01m\\\\v=\omega r=(0.02\frac{rad}{s})(2.01m)=0.04\frac{m}{s}

the maximum relative velocity is 0.04m/s

hope this helps!!

6 0
3 years ago
Read 2 more answers
A 6.0-kilogram cart initially traveling at 4.0 meters per second east accelerates uniformly at 0.50 meter per second squared eas
VMariaS [17]

Answer: 5.5m/s

Explanation:

vf=vi+at

vf= 4.0m/s + (0.50m/s^2)(3.0s)

3 0
3 years ago
What is the frequency of a wave traveling at 300,000,000 m/s with a wavelength of .0025 m/cycle? Please help me !!
Lelu [443]

Answer:

its 00.0035474

Explanation:

..... .. . . .. . . . . .. .

8 0
4 years ago
A​ DC-9 aircraft leaves an airport from a runway whose bearing is N4343degrees°E. After flying for one half 1 2 ​mile, the pilot
bearhunter [10]

Answer:

103.4° or S76.6°E

Explanation:

The direction N43°E is perpendicular to the direction south-east when the plane turn 90° and heads in the south-east direction.

Since the distance 1/2 mile N43°E is perpendicular to the distance 1 mile south-east, we have a right angled triangle.

So, the angle θ between the aircraft's new position and old position is gotten from tanθ = 1 ÷ 1/2 = 2

θ = tan⁻¹(2) = 63.43°

So, the total angle from North to its new position is 40° + 63.43° = 103.43°

Since we need the south-east bearing, the angle from south is 180° - 103.43° = 76.57° ≅ 76.6°

So, our bearing is 103.4° or S76.6°E

3 0
4 years ago
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