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2 years ago

Use the information from the graph to answer the question. What is the acceleration of the object

Physics

1 answer:

Art [367]2 years ago

4 0

Answer:

-2.5 m/s²

Explanation:

The acceleration of a body is the change in it's velocity with time.

The change in velocity with time can be obtained as the slope of a velocity time graph ;

Acceleration = (change in velocity / change in time)

Taking the slope :

Change in Velocity = △y = y2 - y1

Change in time = △x = x2 - x1

(10, 15) ; (0, 40)

△y / △x = y2 - y1 / x2 - x1 = (40 - 15) / (0 - 10)

△y / △x = 25 / - 10 = - 2.5 m/s²

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A sled is slowing down at the bottom of a snowy hill.

Kitty [74]

Answer:

A sled and its rider are moving at a speed of along a horizontal stretch of snow, as Figure 4.24a illustrates. The snow exerts a kinetic frictional force on the runners of the sled, so the sled slows down and eventually comes to a stop. The coefficient of kinetic friction is 0.050. What is the displacement x of the sled?

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2 years ago

Betty is six years old and often asks her parents many questions. Her parents respond to her questions however trivial they may

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This behavior helps Betty in intellectual development.

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2 years ago

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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:

Given:

$T_1$ = 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 = $1.01325\times 10^5\ Pa$

Assume:

$P_2$ = 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

What is the frequency of an x- ray if the wavelength is 4.5 E - 10m?

AnnZ [28]

V (speed) = F (frequency) x Wavelength
If we rearrange the formula, making frequency the subject;
F (frequency) = Speed ÷ Wavelength
F = 300,000 m\s x 4.5 e -10m
F = 0.08810409956 Hz

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2 years ago

Select the major problems associated with the production of nuclear energy.

liberstina [14]

Hazardous solid wastes

3 0

3 years ago

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