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3 years ago

12

When the height of the cylinder increased by a factor of 5, from 100 m to 500 m, what happened to the cylinder’s gravitational p

otential energy?

2 answers:

Grace [21]3 years ago

8 0

PE = m*g*h ⇒ 5* PE = 5*m*g*h

storchak [24]3 years ago

4 0

It increased by a factor of 5

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Can someone tell me the answer and the reason

spin [16.1K]

Answer:

(a) the blocks all had different masses.

Explanation:

The surface is smooth, therefore coefficient of friction is tending to zero.

Forces for each blocks varied from 6N to 8N to 7N to 5N

The blocks were made of different materials and different materials are going to have varying weight for the same size of block.

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3 years ago

Exercise that strengthens your heart, increases your lung capacity, and helps you to burn fate and lose weight is..

olga2289 [7]

Answer : Cardio training

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Cardio helps decrease your heart rate and blood pressure and improve your breathing.

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3 years ago

A train car with mass m1 = 515 kg is moving to the right with a speed of v1 = 7.5 m/s and collides with a second train car. The

Anna [14]

Answer:

For the first situation, we first need to find the mass of the second train car.

In order to do that, we apply the conservation of the amount of movement:

$515*7.5+m2*0=(m1+m2)*4.8$

and we have as a result:

m2 = 289.6875

For the second situation, also we will apply the conservation of the amount of movement:

$515*7.5-289.6875*6 = (515+289.6875)*V$

and we have as a result:

V = 2.64 (it is moving to the right)

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3 years ago

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par

Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

<em>initial temperature of metals, = </em> $T_m$ <em />
<em>initial temperature of water, = </em> $T_i$ <em> </em>
<em>specific heat capacity of copper, </em> $C_p$ <em> = 0.385 J/g.K</em>
<em>specific heat capacity of aluminum, </em> $C_A$ = 0.9 J/g.K
<em>both metals have equal mass = m</em>

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

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2 years ago

A perfectly flexible cable has length L, and initially it is at rest with a length Xo of it hanging over the table edge. Neglect

zaharov [31]

Answer:

$X=X_o+\dfrac{1}{2}gt^2$

Explanation:

Given that

Length = L

At initial over hanging length = Xo

Lets take the length =X after time t

The velocity of length will become V

Now by energy conservation

$\dfrac{1}{2}mV^2=mg(X-X_o)$

So

$V=\sqrt{2g(X-X_o)}$

We know that

$\dfrac{dX}{dt}=V$

$\dfrac{dX}{dt}=\sqrt{2g(X-X_o)}$

$\sqrt{2g}\ dt=(X-X_o)^{-\frac{1}{2}}dX$

At t= 0 ,X=Xo

So we can say that

$X=X_o+\dfrac{1}{2}gt^2$

So the length of cable after time t

$X=X_o+\dfrac{1}{2}gt^2$

6 0

3 years ago