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viktelen [127]
4 years ago
6

What is the surface area to volume ratio of this cube

Physics
1 answer:
Ipatiy [6.2K]4 years ago
3 0
The edge of the cube is 5 cm.
The volume of the cube is  (5cm) · (5 cm) · (5 cm)  =  (5 cm)³  =  125 cm³

The cube has 6 faces.
The area of each face is  (5 cm) · (5 cm)  =  25 cm²
The area of all 6 faces is  (6) · (25 cm²)  =  150 cm²

The ratio of its volume to its surface area is
the ratio of

                             (125 cm³)  to  (150 cm²)

                       =          5/6  cm⁻¹  .

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How much kinetic energy is required to break through?
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Explanation:

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7 0
3 years ago
as the video shows, the star begins its life from a clump of gas that heats up as it contracts. where does the energy that heats
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5 0
2 years ago
Birdman is flying horizontally at a
zavuch27 [327]

Answer:

X=92.49 m

Explanation:

Given that

u= 21 m/s

h= 97 m

Time taken to cover vertical distance h

h= 1/2 g t²

By putting the values

97 = 1/2 x 10 x  t²          ( g = 10 m/s²)

t= 4.4 s

The horizontal distance

X= u .t

X= 21 x 4.4

X=92.49 m

3 0
4 years ago
In abiding by the code of ethics, forensic scientists must possess what characteristics or ethics?
bixtya [17]
Scientists must have an unbiased type of investigation. These type of science requires three major characteristics for it to be functional and sticked to the truth: they must be impartial, accuarate and truthful. Without these characteristics there is no way forensic science must be treated as science. 
3 0
3 years ago
A gymnast of mass 62.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume t
MrRissso [65]

Answer:

a) T = 608.22 N

b) T = 608.22 N

c) T = 682.62 N

d) T = 533.82 N

Explanation:

Given that the mass of gymnast is m = 62.0 kg

Acceleration due to gravity is g = 9.81 m/s²

Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.

So;

To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;

T = mg

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs the rope at a constant rate tension in the string is

= (62.0 kg)(9.81 m/s²)

= 608.22 N

When the gymnast climbs up the rope with an upward acceleration of magnitude

a = 1.2 m/s²

the tension in the string is  T - mg = ma (Since acceleration a is upwards)

T = ma + mg

= m (a + g )

= (62.0 kg)(9.81 m/s² + 1.2  m/s²)

= (62.0 kg) (11.01 m/s²)

= 682.62 N

When the gymnast climbs up the rope with an downward acceleration of magnitude

a = 1.2 m/s² the tension in the string is  mg - T = ma (Since acceleration a is downwards)

T = mg - ma

= m (g - a )

= (62.0 kg)(9.81 m/s² - 1.2 m/s²)

= (62.0 kg)(8.61 m/s²)

= 533.82 N

5 0
3 years ago
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