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uranmaximum [27]
3 years ago
9

A 15 kg mass is lifted to a height of 2m. What is gravitational potential energy at this position

Physics
2 answers:
7nadin3 [17]3 years ago
7 0

Answer: 294J

Explanation:

U_g=mg(y_f-y_i)

U_g=(15)(9.8)(2-0)=294J

JulijaS [17]3 years ago
7 0

Answer:

294.3 J

Explanation:

U = mgh

U = energy

m = mass

g = gravity

h = height

so

U = 15kg * 9.81 m/s^2 * 2m

294.3 J (joules) [kg *m^2/s^2]

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The order of the positive and negative feedback loops are positive, positive, negative, positive, positive, negative.

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A system component known as a feedback loop is one in which all or a portion of the output is used as input for subsequent actions. A minimum of four phases comprise each feedback loop. Input is produced in the initial phase. Input is recorded and stored in the subsequent stage. Input is examined in the third stage, and during the fourth, decisions are made using the knowledge from the examination.

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5 0
2 years ago
What two bodily functions are increased by a warm up
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6 0
3 years ago
A block of mass m=1.2 kg is held at rest against a spring with a force constant k=730N/m. Initially the spring is compressed a d
igor_vitrenko [27]

Answer:

Explanation:

potential energy of compressed spring

= 1/2 k d²

= 1/2 x 730 d²

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This energy will be given to block of mass of 1.2 kg in the form of kinetic energy .

Kinetic energy after crossing the rough patch

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= 3.174 J

Loss of energy

= 365 d² - 3.174  

This loss is due to negative work done by frictional force

work done by friction = friction force x width of patch

= μmg d ,   μ = coefficient of friction , m is mass of block , d is width of patch

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365 d² - 3.174   = .2587

365 d² = 3.4327

d² = 3.4327 / 365

= .0094

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= 9.7 cm

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5 0
3 years ago
Monochromatic light of wavelength λ=136.8μ m is shone at normal incidence through a thin film of thickness t resting atop a full
inn [45]

Answer:

1.8 × 10⁻⁸ Hm

Explanation:

Given that:

The refractive index of the film = 19

The wavelength of the light = 136.8 μ m

The thickness can be calculated by using the formula shown below as:

Thickness=\frac {\lambda}{4\times n}

Where, n is the refractive index of the film

{\lambda} is the wavelength

So, thickness is:

Thickness=\frac {136.8\ \mu\ m}{4\times 19}

Thickness = 1.8 μ m

Since,

1 μ m = 10⁻⁸ Hm

So,

Thickness = 1.8 × 10⁻⁸ Hm

7 0
3 years ago
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8 0
2 years ago
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