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Harlamova29_29 [7]
3 years ago
14

Sound is an example of ?

Physics
2 answers:
wlad13 [49]3 years ago
8 0
Aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
valentina_108 [34]3 years ago
3 0
A. Longitudinal wave
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A 12kg cheetah accelerates 24 m/s". What is the force the cheetah needed to run?
Kobotan [32]

Answer:

288N

Explanation:

Given parameters:

Mass of Cheetah = 12kg

Acceleration  = 24m/s²

Unknown:

Force needed by the cheetah to run  = ?

Solution:

The force needed by the Cheetah to run is the net force.

According to Newton's law;

    Force  = mass x acceleration

Insert the given parameters and solve;

   Force  = 12 x 24  = 288N

7 0
3 years ago
A 57-kg woman holds a 6-kg package as she stands within an elevator which briefly accelerates upward at a rate of 0.15g. Determi
Temka [501]

Answer:

R = 710.7N

L = 67.689 N

During gravity fall L = R = 0 N

Explanation:

So the acceleration that the elevator is acting on the woman (and the package) in order to result in a net acceleration of 0.15g is

g + 0.15g = 1.15g

The force R that the elevator exerts on her feet would be product of acceleration and total mass (Newton's 2nd law):

a(m + M) = 1.15g(57 + 6) = 1.15*9.81*63 = 710.7N

The force L that she exerts on the package would be:

am = 1.15g *6 = 1.15*9.81*6 = 67.689N

When the system is falling, all have a net acceleration of g. So the acceleration that the elevator exerts on the woman (and the package) is 0, and so are the forces L and R.

7 0
3 years ago
Which of the following formulas is used to solve for current flowing through an electrical circuit? A. current = voltage ÷ resis
Alex787 [66]
The answer to your question is a

8 0
3 years ago
Read 2 more answers
A round object of mass 10 kg and radius 0.5 m rolls without slipping down a hill from a height of 4.5 m. If its speed at the bot
Mice21 [21]

Answer:

moment of inertia is 2.72 kg m²

Explanation:

given data

mass m = 10kg

height h = 4.5 m

radius r  = 0.5 m

speed  v = 6.5 m/s

to find out

moment of inertia

solution

we apply here conservation of energy

that is

mgh = 1/2 ×mv² + 1/2 × Iω²

here I is moment of inertia we find and

we know ω = Velocity / radius = 6.5 / 0.5 = 13

and g = 9.8

so put here all these value

10 (9.8) 4.5 = 1/2 ×(10)(6.5)² + 1/2 × I(13)²

441 = 211.25 + 1/2 × I( 169 )

I = 2.72

so moment of inertia is 2.72 kg m²

7 0
3 years ago
3. A model rocket is launched straight upward at 58.8 m/s.
kolezko [41]

Answer:

a). 6 seconds

b). 12 seconds

c). 176.4 meters

Explanation:

a). Equation to be applied to calculate the time taken by the rocket to reach at the peak height,

   v = u - gt

where v = final velocity

u = initial velocity = 58.8 m per sec

g = gravitational pull = 9.8 m per sec²

t = duration of the flight

At the peak height,

v = 0

Therefore, 0 = 58.8 - (9.8)(t)

t =  \frac{58.8}{9.8}

 = 6 seconds

b). Total time of flight = 2(Time taken to go up)

                                    = 2×6

                                    = 12 sec

c). Formula to get the peak height is,

   h=ut-\frac{1}{2}gt^2

   h = (58.8)6 - \frac{1}{2}(9.8)(6)^2

      = 352.8 - 176.4

      = 176.4 meters

8 0
3 years ago
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