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3 years ago

A motorcycle is capable of accelerating at 4.7 m/s^2, starting from rest, how far can it travel in 3 seconds?

Physics

2 answers:

tankabanditka [31]3 years ago

7 0

21.15m should be the answer ^_^

maw [93]3 years ago

5 0

Answer: 21.15m

Explanation:

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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;

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

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

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

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

The change in heat energy for copper metal;

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

For aluminum metal, the quantity of heat transferred is calculated as;

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

The change in heat energy for aluminum metal ;

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

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

A clay vase on a potter's wheel experiences an angular acceleration of 5.69 rad/s2 due to the application of a 16.0-n m net torq

Digiron [165]

The equivalent of the Newton's second law for rotational motions is:
$\tau = I \alpha$
where
$\tau$ is the net torque acting on the object
$I$ is its moment of inertia
$\alpha$ is the angular acceleration of the object.

Re-arranging the formula, we get
$I= \frac{\tau}{\alpha}$
and since we know the net torque acting on the (vase+potter's wheel) system, $\tau=16.0 Nm$ , and its angular acceleration, $\alpha = 5.69 rad/s^2$ , we can calculate the moment of inertia of the system:
$I= \frac{\tau}{\alpha}= \frac{16.0 Nm}{5.69 rad/s^2} =2.81 kg m^2$

8 0

3 years ago

A 2500 kg mustang car accelerates from rest to 30 m/s in 6 sec.What is acceleration of the mustang car?

alexgriva [62]

A = v / t
= 30 / 6
a = 5 m/s^2

5 0

2 years ago

provides one model for solving this problem. The maximum strength of the earth's magnetic field is about 6.9 x 10-5 T near the s

jeka57 [31]

Answer:

The minimum no. of turns is $3.126 \times 10^{5}$

Explanation:

Given:

Magnetic field $B = 6.9 \times 10^{-5}$ T

Frequency $f = 84.5$ Hz

Area of turn $A = 0.021$ $m^{2}$

Voltage $V_{rms} = 170$ V

From the formula of induced emf,

$V = NBA \omega$

Where $\omega = 2\pi f$ and $V = \sqrt{2} V_{rms}$

So number of turn is,

$N = \frac{\sqrt{2} V_{rms} }{AB2\pi f }$

$N = \frac{\sqrt{2} \times 170 }{0.021 \times 6.9 \times 10^{-5} \times 6.28 \times 84.5 }$

$N = 3.126 \times 10^{5}$

Therefore, the minimum no. of turns is $3.126 \times 10^{5}$

4 0

2 years ago

Any five physics problems

kow [346]

Explanation:

There are still some questions beyond the Standard Model of physics, such as the strong CP problem, neutrino mass, matter–antimatter asymmetry, and the nature of dark matter and dark energy.

7 0

2 years ago