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4 years ago

Helpppppppppp!!!!!!!!!!...

Physics

2 answers:

OverLord2011 [107]4 years ago

8 0

Answer:

i can't see the picture pls can you send a link or no

Explanation:

jok3333 [9.3K]4 years ago

5 0

Explanation:

Hey there!

Height(h)= 1.8m

Mass(m)= 14kg

Potential energy= m.h.g

Where "m" means mass, "h" means vertical height and "g" means gravity.

Now, put all values.

P.e = 14kg * 1.8m * 10m/s^2

= 252kg.m^2/s^2

Therefore, The potential energy was 252kg.m^2/s^2.

Hope it helps...

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A student at the top of building of height h throws one ball upward with the initial speed V and then throws a second ball downw

Vlad1618 [11]

Answer:

They are the same (assuming there is no air friction)

Explanation:

Take a look at the picture.

When the first ball (the one thrown upward) gets to the point marked as A, the speed will has the exact same value V but the velocity will now point downward (just like the second ball).

So if you think about it, the first ball, from point A to the ground, will behave exactly like the second ball (same initial speed, same height).

That is why the speeds will be the same when they reach the ground.

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3 years ago

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if a metal object was cut on half how much heat will be required to raise both half to a specific temperature

il63 [147K]

suppose the mass of a metal object be m and its specific heat capacity be s, and, H joules of heat is required to raise its temperature by t degrees Celsius

Than H is given by relation, $H = mst$

Now if this object is cut in two half's, than mass of half part will also be half i.e. M/2

So heat required to heat the half part will be $= (\frac{m}{2}) st =\frac{mst}{2} = H/2$

Hence, the heat required to raise the half object to a specific temperature will also become half.

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4 years ago

Derive the equation of motion of the block of mass m1 in terms of its displacement x. The friction between the block and the sur

Alenkasestr [34]

Answer:

the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$

the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be $\phi$ and $\beta$ respectively.

The velocity of the block to be = v

The equivalent mass of the system = $m_e$

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

$K.E = \frac{1}{2} m_ev^2$ --------------- equation (1)

The angular velocity of the cylinder = $\omega$ : &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

$v = \omega R$

$\omega =\frac{v}{R}$

The kinetic energy of the system in terms of individual mass can be written as:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2$

By replacing $\omega$ with $\frac{v}{R}$ ; we have:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2$

$K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2$ ------------------ equation (2)

Equating both equation (1) and (2); we have:

$m_e = m_1+m_2+\frac{I}{R^2}$

Therefore, the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$ which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

$f_{block }=m_1gsin \beta$

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

$f_{cylinder} = m_2gsin \phi$

Equating the above both equations; we have the equation of motion of the equivalent system to be

$m_e \bar x = f_{cylinder}-f_{block}$

which can be written as follows from the previous derivations

$(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

Finally; the equation of the motion of the block of mass $m_1$ in terms of its displacement is = $(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

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3 years ago

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lorasvet [3.4K]

Answer:

I attached an image that should help.

Explanation:

Check it out.

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3 years ago

Helpppppppppp!!!!!!!!!!...​

Helpppppppppp!!!!!!!!!!...