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iris [78.8K]
3 years ago
13

A ball is shot straight up into the air from the ground with initial velocity of 44ft/sec44ft/sec. assuming that the air resista

nce can be ignored, how high does it go?\
Physics
1 answer:
solniwko [45]3 years ago
3 0
Convert the given in SI units.

         (44 ft/sec)(1 m/ 3.28 ft) = 13.41 m/sec

The distance traveled and the initial velocity can be related through the equation,
  
   d = (Vf)² - (Vi)²/ 2a

where d is the distance, Vf is the final velocity, Vi is the initial velocity, a is the acceleration due to gravity. Substituting the known values from the given above,

    d = ((0 m/s)² - (13.41 m/s)²)/ 2(-9.8 m/s²)

The value of d from the equation,

    d = 9.17 meters

Convert this to feet,

    d = (9.17 m)(3.28 ft / 1 m) = 30 ft

Answer: 30 ft
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The concept of critical density is just like the idea of escape velocity for a projectile launched from the earth. An object lau
ikadub [295]

Answer:

v = √ 2 G M/ R_{e}

Explanation:

To find the escape velocity we can use the concept of mechanical energy, where the initial point is the surface of the earth and the end point is at the maximum distance from the projectile to the Earth.

Initial

        Em₀ = K + U₀

Final

        Em_{f} =  U_{f}

The kinetic energy is k = ½ m v²

The gravitational potential energy is U = - G m M / r

r is the distance measured from the center of the Earth

How energy is conserved

       Em₀ =  Em_{f}

       ½ mv² - GmM / R_{e} = -GmM / r

       v² = 2 G M (1 /  R_{e} – 1 / r)

       v = √ 2GM (1 / R_{e} – 1 / r)

The escape velocity is that necessary to take the rocket to an infinite distance (r = ∞), whereby 1 /∞ = 0

        v = √ 2GM /  R_{e}

5 0
3 years ago
When light of wavelength 160 nm falls on a gold surface, electrons having a maximum kinetic energy of 2. 66 ev are emitted. Find
enyata [817]

When the light of wavelength is falling on gold surface, the electrons begin to exchange energies.

a)The work function in eV is Φ =5.097 eV.

b) The cut-off wavelength is λ₀ = 243.71 nm

c) The frequency is ν₀  =1.231 × 10¹⁵ Hz

<h3>What is work function?</h3>

The energy needed for a particle to escape and break through the surface.

The kinetic energy of the light emitted is 2.66 eV and wavelength of the light is 160 nm = 160 × 10⁻⁹ m.

a) The work function of the gold for given maximum kinetic energy is

Φ = hc / λ  - K.Emax

Substituting 6.626 × 10⁻³⁴ J.s for h, 3 × 10⁸ m/s for c and 2.66 eV for K.Emax, work function will be

Φ =8.16 × 10⁻¹⁹ J

1 eV = 1.6 × 10⁻¹⁹

The work function in eV is Φ =5.097 eV.

b) The cutoff wavelength is related to work function as

λ₀ = hc / Φ

Substitute the corresponding values into the equation, we get the cut off wavelength

λ₀ = 243.71 nm

c) The frequency corresponding to the cut-off wavelength is

ν₀ = c / λ₀

Substitute the corresponding values into the equation, we get the frequency,

ν₀  =1.231 × 10¹⁵ Hz

Therefore, the values for the following are

a)The work function in eV is Φ =5.097 eV.

b) The cut-off wavelength is λ₀ = 243.71 nm

c) The frequency is ν₀  =1.231 × 10¹⁵ Hz

Learn more about wave function.

brainly.com/question/17484291

#SPJ4

3 0
2 years ago
A baby bounces up and down in her crib. Her mass is 13.0 kg, and the crib mattress can be modeled as a light spring with force c
Julli [10]

Answer:

a) 0.804second

b) 0.162m

Explanation:

In simple harmonic motion;

Period T = 2π√m/k

Where m is the mass of the object

K is the force constant

Given:

M = 13.0kg

K = 788N/m

T = 2π√13/788

T = 2π√0.0165

T = 2π×0.128

T = 0.804second

Frequency is the reciprocal of the period. F = 1/T

F = 1/0.804

F = 1.244Hertz

b) To get the amplitude x, we will use the relationship F = kx where F is the force exerted by the baby of mass 13kg

x = F/k

Since F = mg

x = mg/k

Assume g = 9.81m/s²

x = (13×9.81)/788

x = 127.53/788

x = 0.162m

The minimum amplitude that she requires is 0.162m

6 0
3 years ago
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