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givi [52]
2 years ago
10

4. A disobedient student dropped his Physics textbook (mass 0.1kg) from the window (15m above the ground). How fast was it going

when it hit the ground?
Physics
1 answer:
vaieri [72.5K]2 years ago
8 0

Answer:

v= 17.15 m/s

Explanation:

mass of the book=0.1 Kg

height above ground, h= 15 m

Using conservation of energy

Potential energy is converted into kinetic energy

mgh = \frac{1}{2}mv^2

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 15}

v=\sqrt{294}

v= 17.15 m/s

Hence, the book will hit the ground at the speed of 17.15 m/s.

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y en el número racional (Q) se puede escribir como conciente de dos números enteros (z)o como un desimal periódico.Espresa el 1/
umka21 [38]

Answer:

Sorry I don't understand this language I'm sorry

6 0
2 years ago
Gulls are often observed dropping clams and other shellfish from a height to the rocks below, as a means of opening the shells.
kumpel [21]

Answer:

v = 17.71 m / s

Explanation:

We can work this exercise with the kinematics equations. In general the body is released so that its initial velocity is zero, the acceleration of the acceleration of gravity

                v² = v₀² - 2 g (y -y₀)

                v² = 0 - 2g (y -y₀)

when it hits the stone the height is zero and part of the height of the seagull I

              v² = 2g y₀

              v = Ra (2g i)

let's calculate

              v =√ (2 9.8 16)

              v = 17.71 m / s

8 0
3 years ago
In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these s
riadik2000 [5.3K]

Answer:

Part a)

\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}

Part b)

\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

\tau = I\alpha

here we will have

\tau = (m_1g - m_2g)(\frac{L}{2})

now we will have inertia of two masses given as

I = (m_1 + m_2)(\frac{L}{2})^2

now we have

I = (m_1 + m_2)\frac{L^2}{4}

now the angular acceleration is given as

\alpha = \frac{\tau}{I}

so we have

\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}

Part b)

Now if the rod is not massles then we will have total inertia given as

I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}

so we will have

I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}

now the acceleration is given as

\alpha = \frac{\tau}{I}

\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}

7 0
3 years ago
Determine the gain in the potential energy when a 8.0 kg box is raised 17.2 m.
Marysya12 [62]

Answer:

<h2>The answer is 1376 J</h2>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 8 × 10 × 17.2

We have the final answer as

<h3>1376 J</h3>

Hope this helps you

6 0
3 years ago
Read 2 more answers
5) The motion that repeats itself in equal intervals of time is called<br>olaration is equal to​
kotykmax [81]

Answer:

the motion that repeat itself in equal interval of time is called periodic motion and it is equal to harmonic motion. for example pendulum

6 0
3 years ago
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