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horsena [70]
3 years ago
11

PLEASE HELP ME!! I'LL GIVE BRAINLEST, this also​

Physics
2 answers:
svetoff [14.1K]3 years ago
8 0

75 kg/ the mass doesnt differ

kumpel [21]3 years ago
8 0

Answer:

mass of the rock 75*8.87 = 665.25

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A small aircraft accelerated down a runway at 4.0 m/s²
Radda [10]

Given data in the problem :-

  • Acceleration (a) = 4.0 m/s^2
  • Initial velocity (u) = 0 m/s
  • Final velocity (v) = 34 m/s
  • Distance travelled by aircraft (S) =  ?

From Newton's Laws of Motion we know that ,

v = u + at  [t = Time taken by aircraft to cover the distance]

⇒ 34 = 0 + 4t

⇒ t = 34/4 s

∴  t = 8.5 s

From Newton's Laws of Motion we also know that ,

S = u.t + 1/2a.t^2

⇒ S = 0×8.5 + 1/2 × 4 × (8.5)^2 m

∴  S = 144.50 m

Thus the distance travelled by the aircraft while accelerating is 144.50 meter .

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2 years ago
A sound mixer is impressed by the new equipment that was just installed in his recording studio. He says that now he will be abl
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3 0
4 years ago
a brick of mass of 1 kg and density 2.5gm/cm³ is immersed in water how much mass of water is displace by it.​
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6 0
3 years ago
Describe two methods a researcher can use to avoid obtaining a nonrepresentative sample
exis [7]

Answer:

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3 0
4 years ago
Young Jeffrey is bored, and decides to throw some things out of the window for fun. But since he is also very curious, he decide
Inga [223]

Answer:

a) Stone            v_{y} = - 7.25 ft / s ,  vₓ = 0.362 ft / s

b) tennis ball    v_{y} =  -3.16 ft / s ,   vₓ = 0.634 ft / s

c) golf ball        v_{y} = - 1,536 ft / s, vx = 0.634 ft / s

2) golf ball

Explanation:

1) The average speed is defined with the displacement interval in the given time interval

           v =( x_{f}-x₀) / Δt

let's use this expression for each object

a) Stone

  It tells us that it is released from y₀ = 10 ft and reaches the floor at

t = 0.788 s, but in the problem they tell us that the calculation is for t = 1.38 s

           v_{y} = (0-10) / 1.38

           v_{y} = - 7.25 ft / s

 in this interval a distance of x_{f} = 0.500 ft was moved away from the building (x₀ = 0 ft)

          vₓ = (0.500- 0) / 1.38

          vₓ = 0.362 ft / s

In my opinion it makes no sense to keep measuring the time after the stone has stopped.

b) tennis ball

It leaves the building at a height of y₀= 10ft and at the end of the period it is at a height of y_{f} = 5.63 ft, all this in a time of t = 0.788 + 0.591 = 1.38 s

       

the average vertical speed is

            v_{y} = (5.63 - 10) / 1.38

            v_{y} = -3.16 ft / s

for horizontal velocity the ball leaves the building xo = 0 reaches the floor

x₁ = 0.500 foot and when bouncing it travels x₂ = 0.375 foot, therefore the distance traveled

         x_{f} = x₁ + x₂

         x_{f} = 0.500 + 0.375

         x_{f} = 0.875 ft

we calculate

         vₓ = (0.875 - 0) / 1.38

         vₓ = 0.634 ft / s

c) The golf ball

the vertical displacement y₀ = 10 ft, and y_{f} = 7.88 ft

          v_{y} = (7.88 - 10) / 1.38

          v_{y} = - 1,536 ft / s

the horizontal displacement x₀ = 0 ft to the point xf = 0.875 ft

          vₓ = (0.875 -0) / 1.38

          vₓ = 0.634 ft / s

2) in this part we are asked for the instantaneous speed at the end of the time interval

a) the stone is stopped so its speed is zero

          v_{y} = vₓ = 0

b) the tennis ball

It is at its maximum height so its vertical speed is zero

        v_{y} = 0

horizontal speed does not change

          vₓ = 0.634 ft / s

c) The golf ball

they do not indicate that it is still rising. Therefore its vertical speed is greater than zero

         v_{y} > 0

horizontal speed is constant

         vₓx = 0.634 ft / s

the total velocity of the object can be found with the Pythagorean theorem

         v = √ (vₓ² + v_{y}²)

When reviewing the results, the golf ball is the one with the highest instantaneous speed at the end of the period

4 0
3 years ago
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