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zvonat [6]
2 years ago
6

A truck slows from a velocity of 25 m/s to a stop in 70 m. What was the truck’s acceleration

Physics
1 answer:
igomit [66]2 years ago
4 0

v^2-u^2=2 x a x d

25^2-0^2=2 x a x 70

625-0=140 x a

625=140a

a=625/140

a=4.46 m/s^2

im not very sure but i think this is how you do this

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D. two positively charged objects

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A car accelerates from rest to 19 m/s in 6 seconds. What is the acceleration of the car?
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the car is moving so that how it gos so fast

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a wave in a rope is traveling at 6 m/s and at a frequency of 2 Hz. what is the wavelength of the wave producced
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2 years ago
When dots are placed on a page from a laser printer, they must be close enough so that you do not see the individual dots of ink
Dovator [93]

Answer:

 y <8 10⁻⁶ m

Explanation:

For this exercise, they indicate that we use the Raleigh criterion that establishes that two luminous objects are separated when the maximum diffraction of one of them coincides with the first minimum of the other.

 Therefore the diffraction equation for slits with m = 1 remains

             a sin θ = λ

in general these experiments occur for oblique angles so

             sin θ = θ

             θ = λ / a

in the case of circular openings we must use polar coordinates to solve the problem, the solution includes a numerical constant

           θ = 1.22 λ / a

The angles in these measurements are taken in radians, therefore

          θ = s / R

as the angle is small the arc approaches the distance s = y

          y / R = 1.22 λ / s

          y = 1.22 λ R / a

let's calculate

            y = 1.22 500 10⁻⁹ 0.42 / 0.032

            y = 8 10⁻⁶ m

with this separation the points are resolved according to the Raleigh criterion, so that it is not resolved (separated)

                 y <8 10⁻⁶ m

4 0
3 years ago
a water bomber flying with a horizontal speed of 85m/s at a height of 3000m drops a load on a fire below. How far in front of th
Andreyy89

Answer:

2081.65 m

Explanation:

We'll begin by calculating the time taken for the load to get to the target. This can be obtained as follow:

Height (h) = 3000 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

3000 = ½ × 10 × t²

3000 = 5 × t²

Divide both side by 5

t² = 3000 / 5

t² = 600

Take the square root of both side

t = √600

t = 24.49 s

Finally, we shall determine the distance from the target at which the load should be released. This can be obtained as follow:

Horizontal velocity (u) = 85 m/s

Time (t) = 24.49 s

Horizontal distance (s) =?

s = ut

s = 85 × 24.49

s = 2081.65 m

Thus, the load should be released from 2081.65 m.

3 0
3 years ago
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