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erastovalidia [21]
3 years ago
8

In deep space, sphere A of mass MA is located at the origin of an x axis and sphere B of mass MB is located on the axis at a poi

nt x1. 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
Physics
1 answer:
Nookie1986 [14]3 years ago
6 0

Answer:

The gravitational potential energy of the two-sphere system just as B is released is

U = -[(G)(MA)(MB)/x₁]

where G = Gravitational constant

G = (6.7 × 10⁻¹¹) Nm²/kg²

Explanation:

The gravitational potential energy of two masses (m and M), separated by a distance, d, is given as

U = -(GMm/d)

For our question,

Mass of object 1 = MA

Mass of object 2 = MB

Distance between them = x₁

U = -[(G)(MA)(MB)/x₁]

where G = Gravitational constant

G = (6.7 × 10⁻¹¹) Nm²/kg²

Hope this Helps!!!

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4

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This force can either push the block upward at a constant velocity or allow it to slide downward at a constant velocity. The mag
sweet-ann [11.9K]

Answer:

Explanation:

Given

coefficient of kinetic friction(\mu _k)=0.34

inclination \theta =44

weight of block=51 N

(a) When block is moving upward friction force acts downward

thus

Fsin\theta -W-f_r=0

as block is moving with constant velocity thus F_{net} is zero

f_r=\mu _kN=0.34\times Fcos\theta

F\left ( \sin \theta -\mu \cos \theta \right )=W

F=\frac{51}{0.45}=113.31 N

(b)When Block slides down the wall friction changes its direction to oppose the block

Fsin\theta -W+f_r=0

F\left ( \sin \theta +\mu \cos \theta \right )=W

F=\frac{W}{\left ( \sin \theta +\mu \cos \theta \right )}

F=\frac{51}{0.939}=54.299 N

3 0
3 years ago
"A 3 kg crate slides down a ramp. The ramp is 1 m in length and inclined at an angle of 30 degrees. The crate starts from rest a
OlgaM077 [116]

Answer:

v = 2.57 m /sec

Explanation: See Annex Free Body Diagram

From free body diagram and Newton´s second law we have

There is not movements in the y axis direction

cos 30°  =  √3/2       sin 30°  =  1/2

We have  P  = mg   =  3 Kg  *  9.8 m/sec²

P  =   29.4  Kg*m/ sec²      P  =  29.4 [N]

Py  =  P * cos 30°    Py =  29.4 [N] * √3/2   ⇒    Py = 25.43 [N]

Px  =   P * sin 30°    Px =  29.4 [N] * 1/2      ⇒     Px = 14.7  [N]

∑ F   =  m* a         ⇒    ∑ Fy   =  0       ∑ Fx  =  m *a

∑ Fy   =  Fn  -  Py   =  0         Py   = P*cos30°       Py = 25.43 [N]

Fn  =  25.43 [N]

Fr  =  μk * Fn      ⇒   Fr  =  0.19 * 25.43   ⇒ Fr  =  4.83 [N]    

Now

∑ Fx  =  m *a       mg sin30° - Fr =  m*a    ⇒   Px  - Fr  = m*a

14.7 [N]   -  4.83 [N]   =  3 [kg] * a       ⇒   9.87 /3   = a [m /sec²]

a = 3.29 [m/sec²]

From uniformly accelerated movement

distance  =  x₀  + V₀*t ± at²/2     but  x₀   and  V₀    =  0

Then

d = ( 1/2 )*a*t²     ⇒  1 [m]  * 2  =  3.29 [m/sec²] * t²

t  =  0.78 sec

And finally

v =  a*t      ⇒   v  =  3.29 *(.78)     ⇒   v = 2.57 m /sec

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<h3>I hope it helps ❤❤</h3>
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