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Igoryamba
3 years ago
8

A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal

, frictionless surface of a frozen lake. F(x) is the only horizontal force on the box.
if the box is initially at rest x=0, ​what is the speed after it has travelled 17.0m?
Physics
1 answer:
dsp733 years ago
3 0

Answer:

v\approx 8.570\,\frac{m}{s}

Explanation:

The equation of equlibrium for the box is:

\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a

The formula for the acceleration, given in \frac{m}{s^{2}}, is:

a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}

Velocity can be derived from the following definition of acceleration:

a = v\cdot \frac{dv}{dx}

v\, dv = a\, dx

\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx

\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg}  \int\limits^{17\,m}_{0\,m}\, dx  - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx

\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}

v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}]  }

The speed after the box has travelled 17 meters is:

v\approx 8.570\,\frac{m}{s}

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A 1460-kg submarine rises straight up towards the surface. Seawater exerts both an upward buoyant force of 16670 N on the submar
Amiraneli [1.4K]

Answer:

a = 0.63 m/s²

Explanation:

given,

mass of submarine = 1460-kg

upward buoyant force = 16670 N

downward resistive force = 1150 N

submarine acceleration = ?

assuming g = 10 m/s²

now,

B - (R + mg) = ma

16670 - 1150 - 1460 × 10 = 1460 × a

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hence, the acceleration of submarine is equal to a = 0.63 m/s²

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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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So we can read off the graph that our velocity at the end, or our final velocity appears to be near 2.0 m/s

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