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tatuchka [14]
3 years ago
10

The catapults on the uss george h.w. bush can launch aircraft from rest to a speed of 150 mph over a distance of 270 feet. find

the acceleration of the aircraft and express it in multiples of g, free fall acceleration.

Physics
1 answer:
vlabodo [156]3 years ago
8 0
Hope this helps, have a great day ahead!

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Particle A of charge 2.76 10-4 C is at the origin, particle B of charge -6.54 10-4 C is at (4.00 m, 0), and particle C of charge
Vanyuwa [196]

Answer:

a) F_net = 30.47 N ,   θ = 10.6º

b)  Fₓ = 29.95 N

Explanation:

For this exercise we use coulomb's law

          F₁₂ = k k \frac{ q_{1}  \  q_{2} }{ r^{2} }

the direction of the force is on the line between the two charges and the sense is repulsive if the charges are equal and attractive if the charges are different.

As we have several charges, the easiest way to solve the problem is to add the components of the force in each axis, see attached for a diagram of the forces

X axis

        Fₓ = F_{bc x}

Y axis  

       F_{y}Fy = F_{ab} - F_{bc y}

let's find the magnitude of each force

     F_{ab} = 9 10⁹ 2.76 10⁻⁴ 1.02 10⁻⁴ / 3²

      F_{ab} = 2.82 10¹ N

      F_{ab} = 28.2 N

   

      F_{bc} = 9 10⁹ 6.54 10⁻⁴ 1.02 10⁻⁴ / 4²

      F_{bc} = 3.75 10¹  N

       F_{bc} = 37.5 N

let's use trigonometry to decompose this force

      tan θ = y / x

      θ = tan⁻¹ and x

       θ= tan⁻¹ ¾

      θ = 37º

let's break down the force

      sin 37 = F_{bcy} / F_{bc}

      F_{bcy} = F_{bc} sin 37

      F_{bcy} = 37.5 sin 37

      F_{bcy} = 22.57 N

      cos 37 = F_{bcx} /F_{bc}

      F_{bcx} = F_{bc} cos 37

      F_{bcx} = 37.5 cos 37

      F_{bcx} = 29.95 N

let's do the sum to find the net force

X axis

        Fₓ = 29.95 N

Axis y

        Fy = 28.2 -22.57

        Fy = 5.63 N

we can give the result in two ways

a)  F_net = Fₓ i ^ + F_{y} j ^

    F_net = 29.95 i ^ + 5.63 j ^

b) in the form of module and angle

let's use the Pythagorean theorem

    F_net = \sqrt{ F_{x}^2 + F_{y}^2 }

    F_net = √(29.95² + 5.63²)

     F_net = 30.47 N

we use trigonometry for the direction

      tan θ= \frac{ F_{y}  }{  F_{x} }

       

      θ = tan⁻¹ \frac{ F_{y}  }{  F_{x} }

      θ = tan⁻¹ (5.63 / 29.95)

      θ = 10.6º

3 0
3 years ago
MIDDLE SCHOOL SCIENCE- <br> Please help, I will give brainliest to best answer.
tatyana61 [14]

Answer:

1.) Waves carry energy through empty space or through a medium without transporting matter. While all waves can transmit energy through a medium, certain waves can also transmit energy through empty space. ... When waves travel through a medium, the particles of the medium are not carried along with the wave.

2.) Mechanical Waves are waves which propagate through a material medium (solid, liquid, or gas) at a wave speed which depends on the elastic and inertial properties of that medium. There are two basic types of wave motion for mechanical waves: longitudinal waves and transverse waves. Longitudinal waves vibrating in the direction of propagation while Transverse waves vibrate at right angles to the direction of its propagation.

3.) They can carry a little energy or a lot of energy. They can be transverse or longitudinal. However, all waves have common properties—amplitude, wavelength, frequency, and speed. Amplitude describes how far the medium in a wave moves.

I hope this helps!

5 0
2 years ago
A hunter on a frozen, essentially frictionless pond uses a rifle that shoots 4.20g bullets at 965m/s. the mass of the hunter (in
weqwewe [10]

When the gun is fired horizontally :

m = mass of each bullet = 4.20 g = 0.0042 kg

v = velocity of the bullet after fire = 965 m/s

M = mass of the hunter including gun  = 72.5 kg

V = velocity of hunter including gun after fire = ?

V' = velocity of the combination of bullet , gun and hunter before fire = 0 m/s

Using conservation of momentum

m v + M V = (m + M) V'

(0.0042) (965) + (72.5) V = (0.0042 + 72.5) (0)

V = - 0.056 m/s

so recoil velocity comes out to be 0.056 m/s



When the gun is fired at angle 56.0⁰ above the horizontal :

m = mass of each bullet = 4.20 g = 0.0042 kg

v = velocity of the bullet after fire = 965 Cos56 = 539.62 m/s

M = mass of the hunter including gun  = 72.5 kg

V = velocity of hunter including gun after fire = ?

V' = velocity of the combination of bullet , gun and hunter before fire = 0 m/s

Using conservation of momentum

m v + M V = (m + M) V'

(0.0042) (539.62) + (72.5) V = (0.0042 + 72.5) (0)

V = - 0.031 m/s

so recoil velocity comes out to be 0.031 m/s




3 0
3 years ago
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