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vodka [1.7K]
2 years ago
10

after an investigation kuri determines that her hypothesis was wrong what is the best thing for kuri to do next​

Physics
2 answers:
aksik [14]2 years ago
7 0
: You could make slight changes to the process or consider wheather the experiment was carried out correctly
Lina20 [59]2 years ago
3 0

Answer:

Its b :)

Explanation:

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A space probe has two engines. Each generates the same amount of force when fired, and the directions of these forces can be in-
Leto [7]

Answer:

t = 39.60 s

Explanation:

Let's take a careful look at this interesting exercise.

In the first case the two motors apply the force in the same direction

            F = m a₀          

           a₀ = F / m

with this acceleration it takes t = 28s to travel a distance, starting from rest

           x = v₀ t + ½ a t²

           x = ½ a₀ t²

           t² = 2x / a₀

           28² = 2x /a₀          (1)

in a second case the two motors apply perpendicular forces

we can analyze this situation as two independent movements, one in each direction

           

in the direction of axis a, there is a motor so its force is F/2

               

the acceleration on this axis is

          a = F/2m

          a = a₀ / 2

so if we use the distance equation

             x = v₀ t + ½ a t²

as part of rest v₀ = 0

             x = ½ (a₀ / 2) t²

             

let's clear the time

             t² = (2x / a₀)  2

we substitute the let of equation 1

             t² = 28² 2

             t = 28 √2

             t = 39.60 s

4 0
3 years ago
A transverse standing wave is set up on a string that is held fixed at both ends. The amplitude of the standing wave at an antin
ZanzabumX [31]

Answer:

a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Explanation:

Given the data in the question;

as the equation of standing wave on a string is fixed at both ends

y = 2AsinKx cosωt

but k = 2π/λ and ω = 2πf

λ = 4 × 0.150 = 0.6 m

and f =  v/λ = 260 / 0.6 = 433.33 Hz

ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so V_{max1} = A × ω

V_{max1} = 2.2×10⁻³ × 2722.69 m/s

V_{max1} =  5.9899 m/s

therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b)

A' = 2AsinKx

= 2.20sin( 2π/0.6 ( 0.075) rad )

= 2.20 sin(  0.7853 rad ) mm

= 2.20 × 0.706825 mm

A' = 1.555 mm = 1.555×10⁻³

so

V_{max2} = A' × ω

V_{max2} = 1.555×10⁻³ × 2722.69

V_{max2} = 4.2338 m/s

Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

8 0
2 years ago
Through which gas in the table will sound waves travel faster than they would
Leni [432]

Answer: Methane

Explanation: I just took the AP€X quiz and Methane was the correct answer!

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Coordinates, displacement

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