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denis23 [38]
3 years ago
14

Two boulders have masses of 7.5 x 106 kg and 9.2 x 105 kg. After rolling apart, their centers are separated by a distance of 124

m. Find the gravitational attraction FG between them.
Physics
1 answer:
aleksley [76]3 years ago
4 0

Answer:

0.0299 N

Explanation:

Applying,

F = Gmm'/r²................... Equation 1

Where F = gravitational force between the boulders, m = mass of the first boulder, m' = mass of the second boulder, r = distance between the boulders, G = Universal constant.

From the question,

Given: m = 7.5×10⁶ kg, m' = 9.2×10⁵ kg, r = 124 m

Constant: G = 6.67×10⁻¹¹ Nm²/kg²

Substitute these values into equation 1

F = (6.67×10⁻¹¹× 7.5×10⁶ × 9.2×10⁵ )/124²

F = (460.23)/15376

F = 0.0299 N

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Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2
Bess [88]

Answer:

The acceleration of M_2 is  a =  0.7156 m/s^2

Explanation:

From the question we are told that

    The mass of first block is  M_1 =  2.25 \ kg

    The angle of inclination of first block is  \theta _1 =  43.5^o

    The coefficient of kinetic friction of the first block is  \mu_1  = 0.205

      The mass of the second block is  M_2 = 5.45 \ kg

     The angle of inclination of the second block is  \theta _2 =  32.5^o

      The coefficient of kinetic friction of the second block is \mu _2 = 0.105

The acceleration of M_1 \ and\  M_2 are same

The force acting on the mass M_1 is mathematically represented as

     F_1 = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

=> M_1 a = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

Where T is the tension on the rope

The force acting on the mass M_2 is mathematically represented as    

  F_2 =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

   M_2 a =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

At equilibrium

  F_1 =  F_2

So

 T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

making a the subject of the formula

    a =  \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}

substituting values a =  \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}

    => a =  0.7156 m/s^2

     

3 0
3 years ago
Positive feedback interactions in earth’s systems are always a result of human action. T/F
VLD [36.1K]

Positive feedback interactions in earth’s systems are always a result of human action is a FALSE statement.

<u>Explanation:</u>

Earth is a unstable equilibrium which tends to move out of equilibrium, but several other factors try to bring it back in equilibrium again. Earth has a different actions going on both on its surface and also inside it.

Human can alter, or can modify a very small part of the events that occur on earth’s surface. But they don’t have any control on what’s going inside earth’s core. So positive feedback interactions are not only the results of human interaction, but also different other factors.  

4 0
3 years ago
The maximum speed with which you can throw a stone is about 20 m/s. Can you hit a window 45 m away horizontally and 10 m up from
Allushta [10]

Answer:

 y = 17 m

Explanation:

For this projectile launch exercise, let's write the equation of position

          x = v₀ₓ t

          y = v_{oy} t - ½ g t²

let's substitute

          45 = v₀ cos θ t

          10 = v₀ sin θ t - ½ 9.8 t²

the maximum height the ball can reach where the vertical velocity is zero

 

           v_{y} = v_{oy} - gt

           0 = v₀ sin θ - gt

           0 = v₀ sin θ - 9.8 t

Let's write our system of equations

         45 = v₀ cos θ  t

         10 = v₀ sin θ t - ½ 9.8 t²

         0 = v₀ sin θ - 9.8 t

We have a system of three equations with three unknowns for which it can be solved.

Let's use the last two

        v₀ sin θ = 9.8 t

we substitute

        10 = (9.8 t) t - ½ 9.8 t2

        10 = ½ 9.8 t2

        10 = 4.9 t2

        t = √ (10 / 4.9)

        t = 1,429 s

Now let's use the first equation and the last one

         45 = v₀ cos θ t

         0  = v₀ sin θ - 9.8 t

         9.8 t = v₀  sin θ

         45 / t = v₀ cos θ

we divide

         9.8t / (45 / t) = tan θ

          tan θ = 9.8 t² / 45

          θ = tan⁻¹ ( 9.8 t² / 45 )

          θ = tan⁻¹ (0.4447)

          θ = 24º

Now we can calculate the maximum height

         v_y² = v_{oy}^2 - 2 g y

         vy = 0

          y = v_{oy}^2 / 2g

          y = (20 sin 24)²/2 9.8

          y = 3,376 m

the other angle that gives the same result is

       θ‘= 90 - θ

       θ' = 90 -24

       θ'= 66'

for this angle the maximum height is

 

          y = v_{oy}^2 / 2g

          y = (20 sin 66)²/2 9.8

           y = 17 m

thisis the correct

6 0
3 years ago
What is the motion of the particles in this kind of wave? A hand holds the left end of a set of waves. The waves themselves make
Sonja [21]

Answer:

The transverse wave

Explanation:

4 0
3 years ago
The voltage entering a transformer’s primary winding is 120 volts. The primary winding is wrapped around the iron core 10 times.
loris [4]
<u>Answer</u>

 48 Volts  

<u>Explanation</u>
The question can be solve using the turn rule of a transformer that states;

Np/Ns = Vp/Vs
Where Np ⇒ number of turns in the primary coil.
            Ns ⇒number of turns in the seconndary coil
            Vp ⇒ primary voltage
             Vs ⇒secondary voltage

Np/Ns = Vp/Vs

10/4 = 120/Vp

Vp = (120 × 4)/10

      = 480/10
      = 48 Volts  

 

5 0
3 years ago
Read 2 more answers
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