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Vedmedyk [2.9K]
3 years ago
13

Look at paths of the projectiles. What effect does air resistance have on the projectile?

Physics
1 answer:
inna [77]3 years ago
6 0

Answer:

Therefore the resistance of the air makes the movement not parabolic but shorter in each direction

Explanation:

The projectile motion is described by the kinematics equations giving a parabolic trajectory, where on the x axis there is no acceleration and on the y axis the acceleration is the acceleration of gravity.

When the air resistance is taken into account it can be approximated as a force that opposes the movement that for low speeds is proportional to the speed of the space.

Consequently, the movement in the axis and the acceleration is less, in some cases it can be so small that the constant handle speed, in this case, is called terminal velocity.

On the x-axis the friction force creates an acceleration in the negative direction of the movement that the projectile has to brake.

Therefore the resistance of the air makes the movement not parabolic but shorter in each direction.

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Onur drops a basketball from a height of 10m on Mars, where the acceleration due to gravity has a magnitude of 3.7m/s2.​ We want
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Explanation:

It is given that, Onur drops a basketball from a height of 10 m on Mars, where the acceleration due to gravity has a magnitude of 3.7 m/s².

The second equation of kinematics gives the relationship between the height reached and time taken by it.

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8 0
3 years ago
A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until t
postnew [5]

Answer:

Explanation:

Given

mass of saturated liquid water m=1.4\ kg

at 200^{\circ} specific volume is \nu =0.001157\ m^3\kg(From Table A-4,Saturated water Temperature table)

V_1=m\nu _1

V_1=1.4\times 0.001157

V_1=1.6198\times 10^{-3}\ m^3

Final Volume V_2=4V_1

V_2=4\times (1.6198\times 10^{-3})

V_2=6.4792\times 10^{-3}\ m^3

Specific volume at this stage

\nu _2=\frac{V_2}{m}

\nu _2=\frac{6.4792\times 10^{-3}}{1.4}

\nu _2=0.004628\ m^3/kg

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

T_2=T_1^{*}+\frac{T_2^{*}-T_1^{*}}{\alpha _2^{*}-\alpha _1^{*}}\times (\alpha _2-\alpha _1^{*})

T_2=370^{\circ}+\frac{373.95-370}{0.003106-0.004953}\times (0.004628-0.004953)

T_2=370.7^{\circ} C

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