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Roman55 [17]
3 years ago
12

On your first trip to Planet X you happen to take along a 280 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You'r

e curious about the free-fall acceleration on Planet X, where ordinary tasks seem easier than on earth, but you can't find this information in your Visitor's Guide. One night you suspend the spring from the ceiling in your room and hang the mass from it. You find that the mass stretches the spring by 21.7 cm . You then pull the mass down 11.0 cm and release it. With the stopwatch you find that 11 oscillations take 14.0 s .
Can you now satisfy your curiosity?
Physics
1 answer:
arsen [322]3 years ago
7 0

Answer:

5.31143691523 m/s²

Explanation:

m = Mass = 280 g

x = Displacement of spring = 21.7 cm

Time period

T=\dfrac{14}{11}\\\Rightarrow T=1.27\ s

Angular velocity is given by

\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{1.27}\\\Rightarrow \omega=4.94739\ rad/s

\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow k=\omega^2m\\\Rightarrow k=4.94739^2\times 0.28\\\Rightarrow k=6.85346698739\ N/m

From Hooke's law

mg=kx\\\Rightarrow g=\dfrac{kx}{m}\\\Rightarrow g=\dfrac{6.85346698739\times 0.217}{0.28}\\\Rightarrow g=5.31143691523\ m/s^2

The acceleration due to gravity on the planet is 5.31143691523 m/s²

Yes, I have been able to satisfy my curiosity.

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A projectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. The two protons then mo
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Answer:

(a) The speed of the target proton after the collision is:V_{2f} =433(m/s), and (b) the speed of the projectile proton after the collision is: v_{1f}=250(m/s).

Explanation:

We need to apply at the system the conservation of the linear momentum on both directions x and y, and we get for the x axle:m_{1} v_{1i} =m_{1} v_{1f}Cos\beta _{1} +m_{2} v_{2f}Cos\beta _{2}, and y axle:0=m_{1} v_{1f}Sin\beta _{1}+m_{2} v_{2f}Sin\beta _{2}. Now replacing the value given as: v_{1i}=500(m/s), \beta_{1}=+60^{o} for the projectile proton and according to the problem \beta_{1}and\beta_{2} are perpendicular so \beta_{2}=-30^{o}, and assuming that m_{1}=m_{2}, we get for x axle:500=v_{1f}Cos\beta _{1}+ v_{2f}Cos\beta _{2} and y axle: 0=v_{1f}Sin\beta _{1}+v_{2f}Sin\beta _{2}, then solving for v_{2f}, we get:v_{2f}=-v_{1f}\frac{Sin\beta_{1}}{Sin\beta_{2}}= \sqrt{3}v_{1f} and replacing at the first equation we get:500=\frac{1}{2} v_{1f} +\frac{\sqrt{3}}{2} *\sqrt{3}*v_{1f}, now solving for v_{1f}, we can find the speed of the projectile proton after the collision as:v_{1f}=250(m/s) and v_{2f}=\sqrt{3}*v_{1f}=433(m/s), that is the speed of the target proton after the collision.

5 0
3 years ago
If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her
Marysya12 [62]

Answer:

 μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

          fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ  = 0

          fr sin θ  - cos θ  (W / 2 + 0,3 W_painter) = 0

          fr = cotan θ  (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

     

X axis

        F1 - fr = 0

        F1 = fr

the friction force has the expression

       fr = μ N

Y Axis

       N - W - W_painter = 0

       N = W + W_painter

       

we substitute

      fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

     μ (W + W_painter) = cotan θ  (W / 2 + 0,3 W_painter)

      μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values ​​they give

      μ = cotan θ  (12/2 + 0.3 55) / (12 + 55)

      μ = cotan θ  (22.5 / 67)

      μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

       cotan 45 = 1 / tan 45 = 1

the result is

    μ = 0.336

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A bicycle tire is spinning counterclockwise at 2.70 rad/s. During a time period Δt = 1.50 s, the tire is stopped and spun in the
Andru [333]

Answer:

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αav = -3.6 rad/s²

Explanation:

<u>Given</u>:

           Initial angular velocity = ωi = 2.70 rad/s

           Final angular velocity = ωf = -2.70 rad/s (negative sign is  

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           Change in time period = Δt = 1.50 s

<u>Required</u>:

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<u>Solution</u>:

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               Δω = ωf - ωi

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          <u> Average angular acceleration (αav):</u>

               αav = Δω/Δt

               αav = -5.4/1.50

              αav = -3.6 rad/s²

Since, the angular velocity is decreasing from 2.70 rad/s (in counter clockwise direction) to rest and then to -2.70 rad/s (in clockwise direction) so, the change in angular velocity is negative.

7 0
3 years ago
Read 2 more answers
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