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3 years ago

The period of a pendulum T depends on a constant g and the length of the pendulum ????in a relation given by

Physics

1 answer:

adell [148]3 years ago

4 0

Answer:

The relation is $T = 2\p \sqrt{\frac{l}{g} }$

Explanation:

Generally the mathematically relation showing the relationship between the period(T), the constant (g) and the length(l) is

$T = 2\p \sqrt{\frac{l}{g} }$

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As you go farther down the periodic table, the atoms get _______ and more ________.

VMariaS [17]

Answer:

As we navigate down a group the atoms get bigger and bigger with more and more electrons. This means the outermost electrons get further and further away from the positively charged nucleus.

3 0

3 years ago

Read 2 more answers

A mass of 148g stretches a spring 6cm. The mass is set in motion from its equlibrium position with a downward velocity of 10cm/s

zalisa [80]

Answer:

$u(t)=0.78sin12.78t$

Explanation:

We are given that

Mass,m=148 g

Length,L=6 cm

Velocity,u'(0)=10 cm/s

We have to find the position u of the mass at any time t

We know that

$\omega_0=\sqrt{\frac{g}{L}}=\sqrt{\frac{980}{6}}=12.78 rad/s$

Where g= $980 cm/s^2$

$u(t)=Acos12.78 t+Bsin 12.78t$

u(0)=0

Substitute the value

$A=0$

$u'(t)=-12.78Asin12.78t+12.78 Bcos12.78 t$

Substitute u'(0)=10

$12.78B=10$

$B=\frac{10}{12.78}=0.78$

Substitute the values

$u(t)=0.78sin12.78t$

7 0

3 years ago

A subatomic particle X spontaneously decays into two particles, A and B, each of rest energy 1.40 × 10^2 MeV. The particles fly

adoni [48]

Answer:

E = 389 MeV

Explanation:

The total energy of particle A, will be equal to the sum of rest mass energy and relative energy of particle A. Therefore,

Total Energy of A = E = Rest Mass Energy + Relative Energy

Using Einstein's Equation: E = mc²

E = m₀c² + mc²

From Einstein's Special Theory of Relativity, we know that:

m = m₀/[√(1-v²/c²)]

Therefore,

E = m₀c² + m₀c²/[√(1-v²/c²)]

E = m₀c²[1 + 1/√(1-v²/c²)]

where,

m₀c² = rest mass energy = 140 MeV

v = relative speed = 0.827 c

Therefore,

E = (140 MeV)[1 + 1/√(1 - (0.827c)²/c²)]

E = (140 MeV)(2.78)

6 0

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

Rasek [7]

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

Kinetic energy is equal to

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

$E = \dfrac{1}{2}\times 6.64 \times 10^{-27}\times (7.5\times 10^6)^2$

E = 2.5 x 10⁻¹⁴ J

8 0

3 years ago

What is the energy per photon absorbed during the transition from n = 2 to n = 3 in the hydrogen atom?

adelina 88 [10]

Answer : The energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

Explanation :

First we have to calculate the wavelength of hydrogen atom.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = 10973731.6 m⁻¹

$n_f$ = Higher energy level = 3

$n_i$ = Lower energy level = 2

Putting the values, in above equation, we get:

$\frac{1}{\lambda}=(10973731.6)\left(\frac{1}{2^2}-\frac{1}{3^2} \right )$

$\lambda=6.56\times 10^{-7}m$

Now we have to calculate the energy.

$E=\frac{hc}{\lambda}$

where,

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = $6.56\times 10^{-7}m$

Putting the values, in this formula, we get:

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{6.56\times 10^{-7}m}$

$E=3.03\times 10^{-19}J$

Therefore, the energy of one photon of hydrogen atom is, $3.03\times 10^{-19}J$

3 0

3 years ago