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2 years ago

6

A 43-kg child sits in a massless swing. With what horizontal force must the seat be pulled so that the ropes form an angle of 35

o with respect to the vertical

1 answer:

GenaCL600 [577]2 years ago

7 0

Answer:

F_x = 295.4N

Explanation:

180°-90°-35°= 55°

F= m(a)

F_y = 43kg(9.81m/s²)

F_y = 421.83N

tan(55°) = 421.83N/F_x

F_x = 421.83N/tan(55°)

F_x = 295.3685458N

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A car traveling west in a straight line on a highway decreases its speed from 30 meters per second to 23 meters per second in 2

garik1379 [7]

U1= 30m/s
u2= 23m/s
t=2s
a=(u2-u1)/t
a=-7/2=-3.5 m/s^2

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Which equation represents mass-energy equivalence? E = m2c E = mc2 E = (mc)2 E = mc

motikmotik

Einstein's energy mass equivalence relation say that if the whole given mass is converted to energy then it would be

$E = mc^2$

where

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c = speed of light in m/s

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3 years ago

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Dafna11 [192]

Falling from an airplane.

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Write the answer:<br>physics ... i need help

Mariana [72]

Answer:

6 gallons

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3 years ago

Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic

Nuetrik [128]

Answer:

(a) A = $3.90 \AA$

(b) $A = 4.50 \AA$

(c) $A = 5.51 \AA$

(d) $A = 9.02 \AA$

Solution:

As per the question:

Radius of atom, r = 1.95 $\AA = 1.95\times 10^{- 10} m$

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = $2\times 1.95 = 3.90 \AA$

(b) For body centered cubic lattice:

$A = \frac{4}{\sqrt{3}}r$

$A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA$

(c) For face centered cubic lattice:

$A = 2{\sqrt{2}}r$

$A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA$

(d) For diamond lattice:

$A = 2\times \frac{4}{\sqrt{3}}r$

$A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA$

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2 years ago