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Andrej [43]
3 years ago
9

The GPE of a 70kg person standing on a chair 1m off the ground is how many joules?

Physics
1 answer:
Artemon [7]3 years ago
8 0
GPE I am assuming is gravitational potential energy. I'll denote it as U for simplicity. 

U = mgy
U = (70kg)(9.81m/s^2)(1m) = 686.7J

U = 686.7J

Hope this helps!
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ID like to say thx everyone who is on brainly u guys helped me get through my quiz so thx
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Answer:

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Explanation:

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A 5.0-kg block of wood is placed on a 2.0-kg aluminum frying pan. How much heat is required to raise the temperature of both the
Shalnov [3]

Heat required to raise the temperature of a given system is

Q = ms\Delta T

here we know that

m = mass

s = specific heat capacity

\Delta T = change in temperature

now as we know that

mass of wood = 5 kg

mass of aluminium pan = 2 kg

change in temperature = 45 - 20 = 25 degree C

specific heat capacity of wood = 1700 J/kg C

specific heat capacity of aluminium = 900 J/kg C

now here we will find the total heat to raise the temperature of both

Q = m_1s_1\Delta T_1 + m_2s_2\Delta T_2

Q = 5 * 1700 * 25 + 2 * 900 * 25

Q = 212500 + 45000

Q = 257500 J

So heat required to raise the temperature of the system is 257500 J

4 0
3 years ago
How will the force of friction affect a wood block being pushed over a table?
olasank [31]

Answer:

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friction acts in the opposite direction of motion but does not affect the motion of the object

5 0
2 years ago
Suppose a uniform electric field of 4 N/C is in the positive x direction. When a charge is placed at and fixed to the origin, th
Yuliya22 [10]

Answer:

E_total = 3 N / A

Explanation:

The electric field is a vector magnitude so when adding we must use vectors, in this case as the initial field E = 4N / c goes towards the axis axis and the field created by the fixed charge (E1) is also on the axis x we can add in scalar form.

               E_total = E + E₁

the expression for the field of a point charge is

                E₁ = k q₁ / r²

for the point x = 2m, they do not say that the total field is zero, so the charge q1 must be negative

                 E_total = E -k q₁ / r₂

we substitute

                   0 = E - k q₁ / r²

                   q₁ = \frac{E r^2}{k}

let's calculate

                   q₁ = \frac{4 \ 2^2}{9 \ 10^{-9}}

                   q₁ = 1.78 10⁻⁹ C

now we can calculate the field for position x = 4 m

                   E_total = 4 - 9 10⁹  1.78 10⁻⁹ / 4²2

                   E_total = 3 N / A

8 0
2 years ago
what is the rotational kinetic energy of the earth? use the moment of inertia you calculated in part a rather than the actual mo
Ivenika [448]

The Earth's rotational kinetic energy is the kinetic Energy that the Earth

has due to rotation.

The rotational kinetic energy of the Earth is approximately <u>3.331 × 10³⁶ J</u>

Reasons:

<em>The parameters required for the question are;  </em>

<em>Mass of the Earth, M = </em><em>5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = </em><em>6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = </em><em>24.0 hrs</em><em>.</em>

The \ moment  \ of \  inertia \  of \  uniform \  sphere \  is \ I =   \mathbf{\dfrac{2}{5} \cdot M \cdot R^2}

Which gives;

\mathbf{I_{Earth}} =   \dfrac{2}{5} \times 5.97 \times 10 ^{24} \cdot \left(6.38 \times 10^6 \right)^2 = 9.7202107 \times 10^{37}

\mathrm{The \ rotational \  kinetic  \ energy \  is} \   E_{rotational} = \mathbf{\dfrac{1}{2} \cdot I \cdot \omega^2}

\mathrm{The \ angular \ speed, \ \omega} = \mathbf{\dfrac{2 \dcdot \pi}{T}}

Therefore;

\omega = \dfrac{2 \cdot \pi}{24}  = \dfrac{\pi}{24}

Which gives;

\mathbf{E_{rotational}} = \dfrac{1}{2} \times  9.7202107 \times 10^{37} \times  \left(  \dfrac{\pi}{12} \right)^2 = 3.331 \times 10^{36}

The rotational kinetic energy of the Earth, E_{rotational} = <u>3.331 × 10³⁶ Joules</u>

Learn more here:

brainly.com/question/13623190

<em>The moment of inertia from part A  of the question (obtained online) is that of the Earth approximated to a perfect sphere</em>.

<em>Mass of the Earth, M = 5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = 6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = 24.0 hrs</em>

3 0
3 years ago
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