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2 years ago

A possum is trying to cross the street. Its velocity v as a function of time t is given in the graph below

Physics

1 answer:

AlladinOne [14]2 years ago

4 0

Answer:

-10

Explanation:

The rectangular area is the base bbb times height hhh.

\begin{aligned}\Delta x&= A \\\\ &=(2\,\text s)\left (-5\,\dfrac{\text m}{\text s}\right ) \\\\ &= -10\,\text m\end{aligned}

Δx

=(2s)(−5

)

=−10m

The possum displaces 10\,\text m10m10, start text, m, end text to the left.

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A uniform metal meter-stick is balanced with a 1.0 kg rock attached to the left end of the stick. If the support is located 0.25

Dafna1 [17]

Answer:

c) 1.0 kg

Explanation:

The mass of the stick will be located at the centre of the metre rule. Since the rock is located 0.25m from the pivot, the mass of the meter rule is also 0.25m to the Right of the support

According to law of moment

Sum of clockwise moment = sum of anti clockwise moments

Clockwise moment = M×0.25(mass of metre rule is M)

CW moment = 0.25M

Anti clockwise moment = 0.25×1

ACW moments = 0.25kgm

Equate;

0.25M = 0.25

M = 0.25/0.25

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Hence the mass of the metre rule is 1.0kg

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A 150 kg bike takes a roundabout with a radius of 53.0 m. The roundabout it’s only ¾ of a circle. the time taken was 0.35 minute

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All 2023 ariya ac synchronous drive motors produce ____% torque at 0 mph for impressive off-the-line acceleration and smooth cru

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All 2023 Ariya ac synchronous drive motors produce 100% torque at 0 mph for impressive off-the-line acceleration and smooth cruising.

<h3>How to calculate the torque?</h3>

Mathematically, the torque of an automobile vehicle can be calculated by using this formula:

Torque = Fd

<u>Where:</u>

F is the force.

d is the distance.

Generally, torque is a rotational force which is developed by the crankshaft of an automobile vehicle and its capacity to move at a specific acceleration.

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2 years ago

An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

solong [7]

Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

8 0

2 years ago