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4 years ago

A golf ball with an initial angle of 34° lands exactly 240 m down the range on a level course.

Physics

2 answers:

Luba_88 [7]4 years ago

6 0

Answer:

a. The initial Velocity is 50 m/s approximately.

b. The maximum height at that velocity is 79.8 m.

Explanation:

We are given with:

initial angle = Ф = 34°

Range at that angle = R = 240 m

a. Initial Velocity = v = ?

We know that the range of a projectile is given as:

R = v²sin2Ф/g where g = 9.8 m/s²

From this equation we can calculate v as well:

v² = R*g/sin2Ф

Putting the values:

v² = 240*9.8/sin (2*34°)

v² = 2352/sin68°

v² = 2352/0.93

v² = 2529

Taking square root:

v = 50 m/s approx

b. Max height reached by the ball if the initial velocity is 50 m/s with initial angle 34° can be calculated as:

H = v²sin²Ф/g

H = 2500*sin²34°/2(9.8)

H = 2500*.3127/19.6

H = 79.77 m

H = 39.9 m

g100num [7]4 years ago

3 0

Answer:50.39 m/s,

40.46 m

Explanation:

Given

launch angle $=34^{\circ}$

Range=240 m

We know that Range $=\frac{u^2sin2\theta }{g}$

$240=\frac{u^2sin(68)}{9.81}$

u=50.39 m/s

(b)maximum height of projectile is given by

$H=\frac{u^2sin^2\theta }{2g}$

$H=\frac{50.39^2(sin34)^2}{2\times 9.81}$

H=40.46 m

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Answer:

(a) 1560 W

(b) 576 W

Explanation:

Voltage, V = 120 V

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current in vacuum cleaner, i' = 4.8 A

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P = V I = 120 x 13 = 1560 W

(b) Power consumed by vacuum cleaner

P' = V I' = 120 x 4.8 = 576 W

E = P x t = 1560 x 15 x 60 = 1404000 J

Energy consumed by the vacuum cleaner

E' = P' x t' = 576 x 40 x 60 = 1382400 J

the ratio of energies is

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Answer:

La intensidad del campo eléctrico es 70312.5 $\frac{N}{C}$ .

Explanation:

La perturbación que crea en torno a ella una carga eléctrica se representa mediante un vector denominado campo eléctrico.

Se dice que un campo eléctrico es uniforme en una región del espacio cuando la intensidad de dicho campo eléctrico es el mismo en todos los puntos de dicha región.

El campo eléctrico E creado por la carga puntual q en un punto cualquiera P se define como:

$E=k*\frac{q}{r^{2} }$

donde q es la carga creadora del campo, k es la constante electrostática y r es la distancia desde la carga fuente al punto P.

En este caso, los datos son:

k= 9*10⁹ $\frac{N*m^{2} }{C^{2} }$
q= 5*10⁻⁶ C
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Reemplazando:

$E=9*10^{9}\frac{N*m^{2} }{C^{2} } *\frac{5*10^{-6} C}{(0.8 m)^{2} }$

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E= 70312.5 $\frac{N}{C}$

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olga55 [171]

Answer:

Fraction = 59049/60025

Explanation:

Let m be the mass of the neutron and M be the mass of the plutonium nucleus (at rest)

Now, formula for kinetic energy before collision is;

K_i = ½mu²

Formula for kinetic energy after collision is;

K_f = ½mv²

Where;

u is the velocity of the neutron before collision

v is the velocity of the neutron after collision.

From collision principle where momentum before collision equals momentum after collision, we can say that;

(m - M)u = (m + M)v

Thus,

v = [(m - M)u]/(m + M)

Putting [(m - M)u]/(m + M) for v in the final kinetic energy equation gives;

K_f = ½m([(m - M)u]/(m + M))²

K_f = ½mu²((m - M)²/(m + M)²)

To get the fraction of the neutron's kinetic energy is transferred to the plutonium nucleus, it is simply;

K_f/K_i = [½mu²((m - M)²/(m + M)²)]/½mu²

This gives;

K_f/K_i = ((m - M)²/(m + M)²)

But mass of plutonium = 244m

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K_f/K_i = ((m - 244m)²/(m + 244m)²)

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Answer:

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n_f = 2 m_l = 0, 1

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There are only two lines plus the central line, so there are three spectral lines

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n_f = 3 ml = -1, 0, 1

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