Ask question

Ask question

All categories

3 years ago

14

n deep space, sphere A of mass 47 kg is located at the origin of an x axis and sphere B of mass 110 kg is located on the axis at

x = 3.4 m. Sphere B is released from rest while sphere A is held at the origin. (a) What is the gravitational potential energy of the two-sphere system just as B is released? (b) What is the kinetic energy of B when it has moved 2.6 m toward A?

1 answer:

Volgvan3 years ago

6 0

Answer:

a)-1.014x $10^{-7$ J

b)3.296 x $10^{-7$ J

Explanation:

For Sphere A:

mass 'Ma'= 47kg

xa= 0

For sphere B:

mass 'Mb'= 110kg

xb=3.4m

a)the gravitational potential energy is given by

$U_{i$ = -GMaMb/ d

$U_{i$ = - 6.67 x $10^{-11}$ x 47 x 110/ 3.4 => -1.014x $10^{-7$ J

b) at d= 0.8m (3.4-2.6) and $U_{i$ =-1.014x $10^{-7$ J

The sum of potential and kinetic energies must be conserved as the energy is conserved.

$K_{i$ + $U_{i$ = $K_{f$ + $U_{f$

As sphere starts from rest and sphere A is fixed at its place, therefore $K_{i$ is zero

$U_{i$ = $K_{f$ + $U_{f$

The final potential energy is

$U_{f$ = - GMaMb/d

Solving for ' $K_{f$ '

$K_{f$ = $U_{i$ + GMaMb/d => -1.014x $10^{-7$ + 6.67 x $10^{-11}$ x 47 x 110/ 0.8

$K_{f$ = 3.296 x $10^{-7$ J

You might be interested in

A bowling ball is launched from the top of a building at an angle of 35° above the horizontal with an initial speed of 15 m/s. T

Mamont248 [21]

Let $y_0$ be the height of the building and thus the initial height of the ball. The ball's altitude at time $t$ is given by

$y=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ\,t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity.

The ball reaches the ground when $y=0$ after $t=2.9\,\mathrm s$ . Solve for $y_0$ :

$0=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ(2.9\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(2.9\,\mathrm s)^2$

$\implies y_0\approx16.258\,\mathrm m$

so the building is about 16 m tall (keeping track of significant digits).

3 0

3 years ago

A woman launches a boat from one shore of a straight river and wants to land at the point directly on the opposite shore. If the

I am Lyosha [343]

Answer:

If she stands on the North side of a river flowing to the East at 5 mph,

she must head towards the SouthWest to arrive on the South side of the river directly across from her starting point and we have

x^2 + 5^2 = 10^2 where x is her speed directly across the river

x = (75)^1/2 = 8.66 mph towards the South

sin theta = 5 / 10 = 1/2

She must angle the boat at 30 deg from straight South

4 0

2 years ago

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

7nadin3 [17]

Answer:

(a) r = 1.062·R $_E$ = $\frac{531}{500} R_E$

(b) r = $\frac{33}{25} R_E$

(c) Zero

Explanation:

Here we have escape velocity v $_e$ given by

$v_e =\sqrt{\frac{2GM}{R_E} }$ and the maximum height given by

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$

Therefore, when the initial speed is 0.241v $_e$ we have

v = $0.241\times \sqrt{\frac{2GM}{R_E} }$ so that;

v² = $0.058081\times {\frac{2GM}{R_E} }$

v² = ${\frac{0.116162\times GM}{R_E} }$

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$ is then

$\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}$

Which gives

$-\frac{0.941919}{R_E} = -\frac{1}{r}$ or

r = 1.062·R $_E$

(b) Here we have

$K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}$

Therefore we put $\frac{0.241GM}{R_E}$ in the maximum height equation to get

$\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}$

From which we get

r = 1.32·R $_E$

(c) The we have the least initial mechanical energy, ME given by

ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

3 0

3 years ago

Read 2 more answers

Ask Your Teacher Cam Newton of the Carolina Panthers throws a perfect football spiral at 6.9 rev/s. The radius of a pro football

faltersainse [42]

Answer:

$a=159.32\ m/s^2$

Explanation:

It is given that,

Angular speed of the football spiral, $\omega=6.9\ rev/s=43.35\ rad/s$

Radius of a pro football, r = 8.5 cm = 0.085 m

The velocity is given by :

$v=r\omega$

$v=0.085\times 43.35$

v = 3.68 m/s

The centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(3.68)^2}{0.085}$

$a=159.32\ m/s^2$

So, the centripetal acceleration of the laces on the football is $159.32\ m/s^2$ . Hence, this is the required solution.

6 0

3 years ago

Which country had the largest population in 1997

nignag [31]

China i hope this helped

6 0

3 years ago