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3 years ago

He graph below shows the velocity f(t) of a runner during a certain time interval:

Physics

2 answers:

Annette [7]3 years ago

5 0

The following that describes the intercepts on the graph is "The initial velocity of the runner was 4 m/s, and the runner stopped after 8 seconds." It is because the starting point of the line is at 4 and then the ending point is at 8.

iragen [17]3 years ago

4 0

Answer is "the initial velocity of the runner was 4 m/s, and the runner stopped after 8 seconds".

This is a velocity - time graph.

At t = 0 s, the graph has a velocity as 4 m/s. This means, the runner has an initial velocity as 4 m/s.

At t = 4 s, the runner has reached his maximum velocity as 8 m/s and acceleration is 1 m/s².

After t = 4 s, the velocity has decreased with the time means it is a negative acceleration. 

At t = 8 s, the velocity of the runner has reached to zero means runner has stopped after 8 seconds.

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What is the effect on the force of gravity between two objects if the mass of one object remains unchanged while the distance to

drek231 [11]

Answer:

The force of gravity after you double the mass and the distance is half of the initial force: $F_{2}=\frac{1}{2}F_{1}$

Explanation:

The initial force of gravity is:

$F_{1}=\frac{Gm_{1}m_{2}}{r^2}$

where $G$ is the universal gravitational constant, $m_{1}$ is the mass of the first object, $m_{2}$ is the mass of the second object, and $r$ is the distance between the objects.

If the mass of the second object is doubled, now we have $2m_{2}$ , and if the distance between the objects is also doubled instead of $r$ now we have $2r$ .

So the force of gravity now is:

$F_{2}=\frac{Gm_{1}(2m_{2})}{(2r)^2}\\ F_{2}=\frac{2Gm_{1}m_{2}}{4r^2} \\F_{2}=\frac{1}{2} \frac{Gm_{1}m_{2}}{r^2}$

and we know that $F_{1}=\frac{Gm_{1}m_{2}}{r^2}$

so the new force of gravity is:

$F_{2}=\frac{1}{2}F_{1}$

The force of gravity after you double the mass and the distance is half of the initial force.

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3 years ago

Why is there a universe at all

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Answer:GOD ;)

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3 years ago

Which type of transformer is found outside a residential house

KonstantinChe [14]

The type of transformer that is found outside a residential house is :
step-down transformer
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3 years ago

When a liquid is cooled, the kinetic energy of the particles ?

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3 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:

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