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3 years ago

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A 55 kg person moves at the constant speed of 7 m/s along a straight stretch of track for 20s. How far does he travel in this ti

me ?

1 answer:

crimeas [40]3 years ago

6 0

By definition,
Distance = Speed * Time

Therefore the distance traveled is
(7 m/s)*(20 s) = 140 m

Answer: 140 m

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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.

Learn more here:brainly.com/question/15345295

6 0

2 years ago

An electric iron connected to a 110 V source draws 9 A of current. How much heat (in joules) does it generate in a minute?

meriva

Q = 110V x 9A x 60s = J

7 0

3 years ago

A thin uniform rod of mass M and length L is bent at its center so that the two segments are perpendicular to each other. Find i

serg [7]

Answer:

$\frac{1}{12}ML^2$

Explanation:

The moments of the whole object is the sum of the moments of the 2 segments of rod at their ends of which length is L/2 and mass M/2:

$I = 2I_{end} = 2\frac{1}{3}\frac{M}{2}\left(\frac{L}{2}\right)^2$

$I = \frac{1}{3}M\frac{L^2}{4}$

$I = \frac{1}{12}ML^2$

5 0

2 years ago

What is the difference between renewable and nonrenewable resources

ololo11 [35]

Non renewable resources are resources that we can only use for so long until they’re gone and they can not be brought back. So like coal would be a non renewable resource. Renewable resources are resources that we can continue to use and we will always have them. An example of a renewable resource would be oxygen. We can breathe as much oxygen as we want but we will never run out of it.

8 0

2 years ago

Read 2 more answers

Students use a simple pendulum with a length of 36.9cm to determine the value of "g". If it takes 14.2s to complete 10 oscillati

wel

Explanation:

It is given that,

Length of the simple pendulum, l = 36.9 cm = 0.369 m

If it takes 14.2 s to complete 10 oscillations, $T=\dfrac{14.2}{10}=1.42\ s$

(a) The time period of the simple pendulum is given by :

$T=2\pi\sqrt{\dfrac{l}{g}}$

$g=\dfrac{4\pi^2l}{T^2}$

$g=\dfrac{4\pi^2\times 0.369}{(1.42)^2}$

$g=7.22\ m/s^2$

(b) On the surface of moon, $g'=\dfrac{g}{6}$

At earth, $g=9.8\ m/s^2$

$g'=1.63\ m/s^2$

As the value of g is less on the moon, so the time period on the moon increases.

(c) The time period on the earth, T = 3 s

On earth, $T=2\pi\sqrt{\dfrac{l}{g}}$

$3=2\pi\sqrt{\dfrac{l}{9.8}}$ ..............(1)

On moon, $T'=2\pi\sqrt{\dfrac{l}{g'}}$

$T'=2\pi\sqrt{\dfrac{l}{1.63}}$ ..............(2)

On solving equation (1) and (2),

T' = 18.03 s

Hence, this is the required solution.

7 0

3 years ago