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Aleksandr-060686 [28]
3 years ago
14

Due to a technical malfunction a space explorer had to crash land on Planet-X. She manages to fix her space ship and now she is

preparing for launch. However she needs to know the gravitational acceleration on the surface of the planet in order to take off successfully. She builds a mathematical pendulum out of a piece of string and a left over steel bolt. The bolt has a mass of 35.5 g, and the string is 133 cm long. She attaches the pendulum to a fixed point and she lets it swing. She counts 12 complete oscillations in a time period of 70.7 seconds. What is the gravitational acceleration on the surface of Planet-X
Physics
2 answers:
lapo4ka [179]3 years ago
7 0

Answer:

the gravitational acceleration of the Xplanet is 1.344m/s^2

Explanation:

You can use the formula for the calculation of the frequency of a pendulum, in order to find an expression for the gravitational constant:

f=\frac{1}{2\pi}\sqrt{\frac{g}{l}}\\\\g=4\pi^2 f^2l

Where you can notice that mass ob the object does not influence of the gravitatiolan acceleration. By the information of the question, you have the values of f and l. By replacing these values (with units of meter and seconds) you obtain:

f=\frac{12}{70.7}=0.16s^{-1}\\\\g=4\pi^2(0.16s^{-1})^2(1.33m)=1.344\frac{m}{s^2}

Hence, the gravitational acceleration of the Xplante is 1.344m/s^2

ValentinkaMS [17]3 years ago
5 0

Answer:

The gravitational acceleration on the surface of Planet-X is 1.5 ms^{-2}

Explanation:

Firstly, we are to list out the parameters given:

Mass (m) = 35.5 g, Length of string (L) = 133 cm = 1.33 m, t = 70.7 seconds for 12 oscillations, T = 70.7 ÷ 12 = 5.89 seconds

The formula for calculating the period of a simple pendulum (assuming the angle of deflection is lesser than 15º) is given by:

T=2\pi\sqrt{\frac{L}{g}}\\  

We want to calculate for the gravitational acceleration (g), hence, we have to make g the subject of the formula

g=4\pi^{2}\frac{L}{T^{2}}\\

Substitute the parameters into the equation, we have:

g = 4\pi^{2} * 1.33 ÷ 5.89^{2} = 1.51

g = 1.5 ms^{-2}<u />

The gravitational acceleration on the surface of Planet-X is 1.5 ms^{-2}

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A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When
11111nata11111 [884]

Answer:

The value is R_f =  \frac{4}{5}  R

Explanation:

From the question we are told that

   The  initial velocity of the  proton is v_o

    At a distance R from the nucleus the velocity is  v_1 =  \frac{1}{2}  v_o

    The  velocity considered is  v_2 =  \frac{1}{4}  v_o

Generally considering from initial position to a position of  distance R  from the nucleus

 Generally from the law of energy conservation we have that  

       \Delta  K  =  \Delta P

Here \Delta K is the change in kinetic energy from initial position to a  position of  distance R  from the nucleus , this is mathematically represented as

      \Delta K  =  K__{R}} -  K_i

=>    \Delta K  =  \frac{1}{2}  *  m  *  v_1^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K  =  \frac{1}{2}  *  m  * (\frac{1}{2} * v_o )^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K  =  \frac{1}{2}  *  m  * \frac{1}{4} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2

And  \Delta  P is the change in electric potential energy  from initial position to a  position of  distance R  from the nucleus , this is mathematically represented as

          \Delta P =  P_f - P_i

Here  P_i is zero because the electric potential energy at the initial stage is  zero  so

             \Delta P =  k  *  \frac{q_1 * q_2 }{R}  - 0

So

           \frac{1}{2}  *  m  * \frac{1}{4} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2 =   k  *  \frac{q_1 * q_2 }{R}  - 0

=>        \frac{1}{2}  *  m  *v_0^2 [ \frac{1}{4} -1 ]  =   k  *  \frac{q_1 * q_2 }{R}

=>        - \frac{3}{8}  *  m  *v_0^2  =   k  *  \frac{q_1 * q_2 }{R} ---(1 )

Generally considering from initial position to a position of  distance R_f  from the nucleus

Here R_f represented the distance of the proton from the nucleus where the velocity is  \frac{1}{4} v_o

     Generally from the law of energy conservation we have that  

       \Delta  K_f  =  \Delta P_f

Here \Delta K is the change in kinetic energy from initial position to a  position of  distance R  from the nucleus  , this is mathematically represented as

      \Delta K_f   =  K_f -  K_i

=>    \Delta K_f  =  \frac{1}{2}  *  m  *  v_2^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K_f  =  \frac{1}{2}  *  m  * (\frac{1}{4} * v_o )^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K_f  =  \frac{1}{2}  *  m  * \frac{1}{16} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2

And  \Delta  P is the change in electric potential energy  from initial position to a  position of  distance R_f  from the nucleus , this is mathematically represented as

          \Delta P_f  =  P_f - P_i

Here  P_i is zero because the electric potential energy at the initial stage is  zero  so

             \Delta P_f  =  k  *  \frac{q_1 * q_2 }{R_f }  - 0      

So

          \frac{1}{2}  *  m  * \frac{1}{8} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2 =   k  *  \frac{q_1 * q_2 }{R_f }

=>        \frac{1}{2}  *  m  *v_o^2 [-\frac{15}{16} ]  =   k  *  \frac{q_1 * q_2 }{R_f }

=>        - \frac{15}{32}  *  m  *v_o^2 =   k  *  \frac{q_1 * q_2 }{R_f } ---(2)

Divide equation 2  by equation 1

              \frac{- \frac{15}{32}  *  m  *v_o^2 }{- \frac{3}{8}  *  m  *v_0^2  } }   =  \frac{k  *  \frac{q_1 * q_2 }{R_f } }{k  *  \frac{q_1 * q_2 }{R } }}

=>           -\frac{15}{32 } *  -\frac{8}{3}   =  \frac{R}{R_f}

=>           \frac{5}{4}  =  \frac{R}{R_f}

=>             R_f =  \frac{4}{5}  R

   

7 0
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Starting from rest, a 75.0-kg snowboarder slides straight down a 115-m slope in 6.6 s. If the slope is a 35°
Likurg_2 [28]

The acceleration of the snowboarder is 5.6 m/s^2 down along the incline

Explanation:

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where

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g=9.8 m/s^2 is the acceleration of gravity

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Substituting, we find that the net force along the incline is:

F=(75.0)(9.8)(sin 35^{\circ})=421.6 N

According to Newton's second law, the net force on the snowboard is equal to the product between his mass and his acceleration:

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Answer:

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Explanation:

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time of motion, t = 6.1 s

The angular distance traveled by the edge of the wheel is calculated as;

\theta = (\frac{\omega_f + \omega_i}{2} )t\\\\\theta =  (\frac{39.8 + 25.66}{2} )\times 6.1\\\\\theta = 199.653 \ radian

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