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Marat540 [252]
3 years ago
11

A 10.2-kg mass is located at the origin, and a 4.6-kg mass is located at x = 8.1 cm. Assuming g is constant, what is the locatio

n of the center of mass xcom and the location of the center of gravity xcogof the two masses? Are the locations the same? Why or why not?
Physics
1 answer:
goldfiish [28.3K]3 years ago
3 0

Answer:

center of mass of the two masses will lie at x = 2.52 cm

center of gravity of the two masses will lie at x = 2.52 cm

So center of mass is same as center of gravity because value of gravity is constant here

Explanation:

Position of centre of mass is given as

r_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}

here we have

m_1 = 10.2 kg

m_2 = 4.6 kg

r_1 = (0, 0)

r_2 = (8.1cm, 0)

now we have

r_{cm} = \frac{10.2 (0,0) + 4.6 (8.1 , 0)}{10.2 + 4.6}

r_{cm} = {(37.26, 0)}{14.8}

r_{cm} = (2.52 cm, 0)

so center of mass of the two masses will lie at x = 2.52 cm

now for center of gravity we can use

r_g_{cm} = \frac{m_1gr_1 + m_2gr_2}{m_1g + m_2g}

here we have

m_1 = 10.2 kg

m_2 = 4.6 kg

r_1 = (0, 0)

r_2 = (8.1cm, 0)

now we have

r_g_{cm} = \frac{10.2(9.8) (0,0) + 4.6(9.8) (8.1 , 0)}{10.2(9.8) + 4.6(9.8)}

r_g_{cm} = {(37.26, 0)}{14.8}

r_g_{cm} = (2.52 cm, 0)

So center of mass is same as center of gravity because value of gravity is constant here

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Monochromatic light with a wavelength of 6.4 E -7 meter passes through two narrow slits, producing an interference pattern on a
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Answer:

The distance between the slits is given by  1.3 × 10^{-4} m

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To find:

distance between slits, d = ?

Formula used:

y = \frac{m \times \lambda \times D}{d}

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y = distance of first bright band from central maxima

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d = \frac{m \times \lambda \times D}{y}

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3 years ago
During a rockslide, a 670 kg rock slides from rest down a hillside that is 740 m along the slope and 240 m high. The coefficient
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a) 1.58\cdot 10^6 J

b) 1.15\cdot 10^6 J

c) 0.43\cdot 10^6 J

d) 35.8 m/s

Explanation:

a)

The gravitational potential energy of an object is the energy possessed by the object due to its location with respect to the ground.

It is given by:

U=mgh

where

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object, relative to a reference level

Here, the reference level is taken at the bottom of the hill (where the potential energy is zero).

So, we have:

m = 670 kg is the mass of the rock

g=9.8 m/s^2

h = 240 m is the initial height of the rock

So, the potential energy of the rock just before the slide is

U=(670)(9.8)(240)=1.58\cdot 10^6 J

b)

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W=F_f d

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F_f is the force of friction

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The force of friction is given by:

F_f=-\mu mg cos \theta

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c)

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Therefore, we have:

K_f = U_i -E_t

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d)

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