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LiRa [457]
3 years ago
12

Three identical 50-kg balls are held at the corners of an equilateral triangle, 30 cm on each side. if one of the balls is relea

sed, what is the magnitude of its initial acceleration if the only forces acting on it are the gravitational forces due to the other two masses? (g = 6.67 × 10-11 n • m2/kg2)
Physics
2 answers:
Ivahew [28]3 years ago
6 0

Answer:

a= 6.42×10⁻⁸m/s²

Explanation:

Given Data

mass=m=50 kg

angle=α=30°

Gravitational Constant=G=6.67 × 10-11 Nm²/kg²)

To find

acceleration=a=?

Solution

From Gravitational Law " the force of attraction between all masses in the universe; especially the attraction of the earth's mass for bodies near its surface "

F₁ = G×m²/D²

F = 2*F₁*cos30° = 2*G*50²*cos30°/0.3² = 3.21×10⁻⁶N  

a = F/m = 3.21×10⁻⁶/50

a= 6.42×10⁻⁸m/s²

natima [27]3 years ago
3 0
F1 = G*m²/D² 
<span>F = 2*F1*cos30° = 2*G*50²*cos30°/0.3² = 3.21E-6 </span>

<span>a = F/m = 3.21E-6/50 = 6.42E-8

Hope this helped!
STSN</span>
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The diffusion rate for a solute is 4.0 x 10^-11 kg/s in a solvent- filled channel that has a cross-sectional area of 0.50 cm^2 a
zlopas [31]

Answer:

s = 9.6\times 10^{-12}kg/s

Explanation:

Given:

Solute Diffusion rate  = 4.0 × 10⁻¹¹ kg/s

Area of cross-section = 0.50 cm²

Length of channel  =0.25 cm

Now for the new channel

Area of cross-section = 0.30 cm²

Length of channel  =0.10 cm

let the Solute Diffusion rate  of new channel = s

now equating the diffusion rate per unit volume for both the channels

\frac{4\times 10^{-11}}{0.50\times 0.25}=\frac{s}{0.30\times 0.10}

thus,

s = 9.6\times 10^{-12}kg/s

7 0
4 years ago
This force will cause the path of the particle to curve. Therefore, at a later time, the direction of the force will ___________
melisa1 [442]

Answer:

have a component along the direction of motion that remains perpendicular to the direction of motion

Explanation:

In this exercise you are asked to enter which sentence is correct, let's start by writing Newton's second law.

circular movement

          F = m a

          a = v² / r

          F = m v²/R

where the force is perpendicular to the velocity, all the force is used to change the direction of the velocity

in linear motion

         F = m a

where the force is parallel to the acceleration of the body, the total force is used to change the modulus of the velocity

the correct answer is: have a component along the direction of motion that remains perpendicular to the direction of motion

8 0
3 years ago
Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y
Gnesinka [82]
The representation of this problem is shown in Figure 1. So our goal is to find the vector \overrightarrow{R}. From the figure we know that:

\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}

From geometry, we know that:

\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}

Then using vector decomposition into components:

For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85

Therefore:

R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m

So if you want to find out <span>how far are you from your starting point you need to know the magnitude of the vector \overrightarrow{R}, that is:
</span>
\left | \overrightarrow{R} \right |=&#10;\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}

Finally, let's find the <span>compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

</span>\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}


6 0
3 years ago
5 problem associated with PopulationCensus​
dolphi86 [110]

Answer:

1. High Illiteracy Rate in a Nation has a Negative Impact on the Conducting of Census

2.Corruption Interferes with Census

3.Traditional and Religious Beliefs can Interfere with the Census Exercise

4.insufficient and Ineffective Census Educational Campaign

5.Insufficient Census Experts

Explanation:

5 0
4 years ago
A rubber ball is dropped from a height of 8m. After strikingthe floor, the ball bounces to a height of 5m. a. If the ball had bo
kifflom [539]

Answer:

a) This means the collision between the ball and the floor is elastic.

b) This points to a perfectly inelastic collision between the ball and the floor as they stick together after collision

c) Check Explanation.

Explanation:

Collision of bodies are analysed according to whether both momentum and kinetic energy of the system is conserved, that is, if these two quantities before collision are equal to their values after collision.

In all types of collisions, momentum is usually conserved, but kinetic energy is conserved only in an elastic collision.

A ball dropped from a height of 8 m bounces up back to a height of 5 m.

a. If the ball had bounced to a height of 8m, how would you describe the collision between the ball and the floor?

The ball not bouncing back to a height of 8 m shows energy loss at some point in the total motion of the ball (most likely at the collision). If kinetic energy was conserved, the ball would bounce back up to the height at which it fell from (8 m) after the collision with the floor.

b. If the ball had not bounced at all, how would you describe the collision between the ball and the floor?

If the ball had not bounced at all, this means it lost all of its kinetic energy to the floor, and this points to a perfectly inelastic collision between the ball and the floor as they stick together after collision.

c. What happened to the energy lost by the ball during thecollision?

The energy lost during the collision is converted to another form, most likely responsible for some deformation on the ball & a minute deformation on the floor, converted to some form of heat as a result of the collision or into sound energy, usually, it's a combination of all This!

Hope this Helps!!!

5 0
4 years ago
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