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Nesterboy [21]
3 years ago
11

Brandon hits a golf ball with an initial velocity of 30 m/s at an angle of 30 above the horizontal. How long is it in the air?

Physics
1 answer:
siniylev [52]3 years ago
7 0

Given :

Brandon hits a golf ball with an initial velocity of 30 m/s at an angle of 30 above the horizontal.

To Find :

How long is it in the air.

Solution :

We know, the formula of time of flight is :

T = \dfrac{2usin\ \theta}{g}\\\\T = \dfrac{2\times 30\times sin\ 30^o}{9.8}\\\\T = 3.06\ seconds

Therefore, the ball is in air for 3.06 seconds.

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3 0
3 years ago
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A 220 g mass is on a frictionless horizontal surface at the end of a spring that has force constant of 7.0
Talja [164]

The concept of conservation of energy and harmonic motion allows to find the result for the power where the kinetic and potential energy are equal is:

        x = 0.135 cm

Given parameters

  • The mass m = 220 g = 0.220 kg
  • The spring cosntnate3 k = 7.0 N / m
  • Initial displacement A = 5.2 cm = 5.2 10-2 m

To find

  • The position where the kinetic and potential energy are equal

 

A simple harmonic movement is a movement where the restoring force is proportional to the displacement, the result of this movement is described by the expression.

          x = A cos wt + fi

          w² = \frac{k}{m}

Where x is the displacement from the equilibrium position, A the initial amplitude of the system, w the angular velocity t the time, fi a phase constant determined by the initial conditions, k the spring constant and m the mass.

The speed is defined by the variation of the position with respect to time.

       v = \frac{dx}{dt}

let's evaluate

       v = - A w sin (wt + Ф)

Since the body releases for a time t = 0 the velocity is zero, therefore the expression remains.

       0 = - A w sin Ф

For the equality to be correct, the sine function must be zero, this implies that the phase constant is zero

        x = A cos wt

Let's find the point where the kinetic and potential energy are equal.

        K = U

        ½ m v² = m g x

       

we substitute

        ½ A² w² sin² wt = g A cos wt

        sin² wt = \frac{2g}{A}  cos wt

let's calculate

      w = \sqrt{\frac{7}{0.220} }  

      w = 5.64 rad / s

      sin² 5.64t = 2 9.8 / 0.052 cos 5.64t

      sin² 5.64t = 376.92 cos 5.64 t

      1 - cos² 5.64t = 376.92 cos 5.64t

      cos² 5.64t -376.92 cos564t -1 = 0

we make the change of variable

       x = cos 5.64t

      x²- 376.92 x - 1 = 0

      x = 0.026

      cos 5.64t = 0.026

   

Let's find the displacement for this time

       x = 5.2 10-2 0.026

       x = 1.35 10-3 m

In conclusion Using the concepts of conservation of energy and harmonic motion we can find the result for the could where the kientic and potential enegies are equal is:

        x = 0.135 cm

Learn more here: brainly.com/question/15707891

8 0
3 years ago
Imagine you are in an open field where two loudspeakers are set up and connected to the same amplifier so that they emit sound w
mario62 [17]

Answer:

The distance we need to walk in other not to hear the speakers for a speaker separation distance of 1 m, while walking in front of one of the speaker is 1.875 meters

Explanation:

The wavelength of the wave is obtained from the formula, v = f × λ

Where;

v = The velocity of the wave = 344 m/s

f = The frequency of the wave = 688 Hz

λ = The wavelength of the wave

λ = v/f = (344 m/s)/(688 Hz) = 1/2 meters

Therefore, for one not to be able to here the speakers, there must be destructive interference and R₁ - R₂ = λ/2 = (1/2)/2 = 1/4 m

Where R₁ and R₂ are the distances from the person to the two speakers respectively

When the distance between the two speakers = 1 meter, we have

R₁ = √(x² + d²), R₂ = √((1 - x)² + d²)

R₁ - R₂ = √(x² + d²) - √((1 - x)² + d²) = 1/4

When we walk from directly in front of one of the speakers, we get;

R₁ - R₂ = √(1 + d²) - √((1 - 1)² + d²) = 1/4

√(1² + d²) - d = 1/4

√(1² + d²)  = 1/4 + d

Square both sides gives

1² + d² = 1/16 + d/2 + d²

1²  = 1/16 + d/2

d/2 = 1 - 1/16 = 15/16

d = 2 × 15/16 = 15/8

d = 15/8 = 1.875 meters.

3 0
3 years ago
A 7.75-l flask contains 0.482 g of hydrogen gas and 4.98 g of oxygen gas at 65°c. What is the partial pressure of oxygen in the
dimulka [17.4K]

Answer:

0.558 atm

Explanation:

We must first consider that both gases behaves like ideal gases, so we can use the following formula: PV=nRT

Then, we should consider that, whithin a mixture of gases, the total pressure is the sum of the partial pressure of each gas:

P₀ = P₁ + P₂ + ....

P₀= total pressure

P₁=P₂= is the partial pressure of each gass

If we can consider that each gas is an ideal gas, then:

P₀= (nRT/V)₁ + (nRT/V)₂ +..

Considering the molecular mass of O₂:

M O₂= 32 g/mol

And also:

R= ideal gas constant= 0.082 Lt*atm/K*mol

T= 65°C=338 K

4.98 g O₂ = 0.156 moles O₂

V= 7.75 Lt

Then:

P°O₂=partial pressure of oxygen gas=  (0.156x0.082x338)/7.75

P°O₂= 0.558 atm

3 0
4 years ago
A generator spins and does 64,000 J of work in 200 seconds. What is the power of the generator?
maria [59]

Answer:

L = V × I × t

P = V × I then

P = L/t

P = 64000J/200s= 320W

answer : P = 320 W

........

7 0
2 years ago
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