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3 years ago

A ball dropping from 70 cm took 0.6 seconds to travel a horizontal distance of 34 cm.

Physics

1 answer:

Alenkasestr [34]3 years ago

4 0

Answer:

vₓ = 0.566 m / s, W_total = 9.1 J

Explanation:

This exercise is a parabolic type movement, for the x axis where there is no acceleration

x = v t

vₓ = x / t

vₓ = 0.34 / 0.6

vₓ = 0.566 m / s

the work done is

X axis

In this axis there is no acceleration, therefore the sum of the forces is zero and since the work is the force times the distance, we conclude that the lock in this axis is zero.

W₁ = 0

Y axis

in this axis the force that exists is the force of gravity, that is, the weight of the body

W₂ = Fy y

W₂ = mg and

W₂ = m 9.8 0.70

W₂ = m 9.1

the work is a scalar for which we have to add the quantities obtained

W_total = W₁ + W₂

W_total = 0 + 9.1 m

W_total = 9.1 m

In order to complete the calculation, the mass of the body is needed if we assume that the mass is m = 1

W_total = 9.1 J

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Carbon is allowed to diffuse through a steel plate 11 mm thick. The concentrations of carbon at the two faces are 0.88 and 0.41

Serhud [2]

Answer:

The temperature is 2584.5 K

Explanation:

Given:

Activation energy $Q = 80000$ $\frac{J}{mol}$

Preexponential $D= 6.2 \times 10^{-7}$ $\frac{m^{2} }{s}$

Diffusion flux $J = 6.4 \times 10^{-10}$ $\frac{kg}{m^{2} s}$

Thickness of plate $\Delta x = 11 \times 10^{-3}$ m

Concentration of carbon at two faces $\Delta C = (0.88 - 0.41 ) = 0.47$ $\frac{kg}{m^{3} }$

From the formula of temperature in terms of diffusion flux,

$T = (\frac{Q}{R} ) \frac{1}{\ln (\frac{D\Delta C}{J\Delta x} )}$

Where $R =$ 8.314 $\frac{J}{mol.K}$ ( gas constant )

Put the values and find the temperature,

$T = (\frac{80000}{8.314} ) \frac{1}{\ln (\frac{6.2 \times 10^{-7} \times 0.47 }{6.4 \times 10^{-10}\times 11 \times 10^{-3} } )}$

$T = 2584.5$ K

Therefore, the temperature is 2584.5 K

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3 years ago

In a Rutherford scattering experiment a target nucleus has a diameter of 1.34×10-14 m. The incoming α particle has a mass of 6.6

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Answer:

E = 2.5 x 10⁻¹⁴ J

Explanation:

given,

diameter = 1.33 x 10⁻¹⁴ m

mass = 6.64 x 10⁻²⁷ kg

wavelength is equal to diameter

de broglie wavelength equal to diameter

$\lambda = \dfrac{h}{mv}$

$1.33 \times 10^{-14}= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times v}$

$v= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times 1.33 \times 10^{-14}}$

v = 7.5 x 10⁶ m/s

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$E = \dfrac{1}{2}mv^2$

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E = 2.5 x 10⁻¹⁴ J

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Answer:

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