گاما رو نصب کن!

{{ number }}
اعلان ها
اعلان جدیدی وجود ندارد!
کاربر جدید

جستجو

پربازدیدها: #{{ tag.title }}

جستجوهای پرتکرار

میتونی لایو بذاری!

$AB$ and $BC$ are straight roads inclined at ${5^ \circ }$ to the horizontal and ${1^ \circ }$ to the horizontal respectively. $A$ and $C$ are at the same horizontal level and $B$ is $45{\text{ }}m$ above the level of $A$ and $C$ (see diagram, which is not to scale). A car of mass $1200{\text{ }}kg$ travels from $A$ to $C$ passing through $B$.

a) For the motion from $A$ to $B$, the speed of the car is constant and the work done against the resistance to motion is $360{\text{ }}kJ$. Find the work done by the car’s engine from $A$ to $B$.

The resistance to motion is constant throughout the whole journey.

b) For the motion from $B$ to $C$ the work done by the driving force is $1660{\text{ }}kJ$. Given that the speed of the car at $B$ is $15{\text{ }}m{\text{ }}{s^{ - 1}}$, show that its speed at $C$ is $29.9{\text{ }}m{\text{ }}{s^{ - 1}}$, correct to 3 significant figures.

c) The car’s driving force immediately after leaving $B$ is 1.5 times the driving force immediately before reaching $C$. Find, correct to 2 significant figures, the ratio of the power developed by the car’s engine immediately after leaving $B$ to the power developed immediately before reaching $C$.

پاسخ تشریحی :
نمایش پاسخ

a) $PE$ gain $ = 1200g \times 45$

WD $ = 1200g \times 45 + 360000$

Work done is $900000{\text{ }}J$ or $900{\text{ }}kJ$

b) WD against resistance

$ = 360 \times \sin 5/\sin 1{\text{ }}\left( {kJ} \right)$ or

$\left\{ {360000 \div \left( {45/\sin {5^ \circ }} \right)} \right\} \times \left( {45/\sin {1^ \circ }} \right){\text{ }}\left( J \right)$ or

$697.24... \times 2578.44...{\text{ }}\left( J \right)$ or

$1798{\text{ }}\left( {kJ} \right)$

$KE$ gain $ = 1660 + 540 - 1798$

$\left[ {402000 = {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}1200\left( {{v^2} - 225} \right)} \right]$

Speed is $29.9{\text{ }}m{s^{ - 1}}$

c) $\frac{{{P_B}}}{{{P_C}}} = \left( {\frac{{D{F_B}}}{{D{F_C}}}} \right) \times \frac{{{v_B}}}{{{v_C}}} = 1.5 \times 15/29.9$

Ratio is $0.75$

تحلیل ویدئویی تست

تحلیل ویدئویی برای این تست ثبت نشده است!