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Maksim231197 [3]
3 years ago
7

Gravitational Force: Two small objects, with masses m and M, are originally a distance r apart, and the magnitude of the gravita

tional force on each one is F. The masses are changed to 2m and 2M, and the distance is changed to 4r. What is the magnitude of the new gravitational force?
Physics
1 answer:
Alona [7]3 years ago
4 0

Law of universal gravitation:

F = GMm/r²

F = gravitational force, G = gravitational constant, M & m = masses of the objects, r = distance between the objects

F is proportional to both M and m:

F ∝ M, F ∝ m

F is proportional to the inverse square of r:

F ∝ 1/r²

Calculate the scaling factor of F due to the change in M:

k₁ = 2M/M = 2

Calculate the scaling factor of F due to the change in m:

k₂ = 2m/m = 2

Calculate the scaling factor of F due to the change in r:

k₃ = 1/(4r/r)² = 1/16

Multiply the original force F by the scaling factors to obtain the new force:

Fk₁k₂k₃

= F(2)(2)(1/16)

= F/4

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A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 35.5 m/s. a) What is the coeff
postnew [5]

Answer:

The coefficient of friction present between the roadway and the wheels of the truck is <u>0.833</u>.

Explanation:

Given:

Radius of the curve (R) = 150 m

Maximum speed of truck (v) = 35.5 m/s

Let the coefficient of friction between the roadway and the wheels of the truck be "μ".

As the truck is moving around a circular curve. So, the force acting on it is centripetal force which acts in the radial inward direction towards the center of the circular curve.

The centripetal force acting on the truck is given as:

F_c=\frac{mv^2}{R}

Now, the friction between the roadway and the wheels of the truck is responsible for providing the necessary centripetal force. So, frictional force is equal to the centripetal force necessary for circular motion.

Frictional force is given as:

f=\mu N

Where, 'N' is the normal force. Since there is no vertical motion, the normal force is equal to weight of truck. So,

N=mg

Therefore, frictional force, f=\mu mg

Now, frictional force = centripetal force

f=F_c\\\\\mu mg=\frac{mv^2}{R}\\\\\mu = \frac{v^2}{Rg}

Plug in the given values and solve for 'μ'. This gives,

\mu=\frac{(35\ m/s)^2}{(150\ m)(9.8\ m/s^2)}\\\\\mu=\frac{1225\ m^2/s^2}{1470\ m^2/s^2}\\\\\mu=0.833

Therefore, the coefficient of friction present between the roadway and the wheels of the truck is 0.833

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3 years ago
Based on its location on the periodic table, which element would be most
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Answer: Sodium Na

Hydrogen H

8 0
3 years ago
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The system below has a friction force of 25 N acting on the cart which 8 kg. The mass hanging off the edge has a mass of 6 kg. F
photoshop1234 [79]

The cart will be pulled to the right by the hanging mass, so by Newton's second law, the net force on the cart is

<em>T</em> - 25 N = (8 kg) <em>a</em>

where <em>T</em> is the tension in the rope and <em>a</em> is the acceleration.

The hanging mass has a net force of

(6 kg) <em>g</em> - <em>T</em> = (6 kg) <em>a</em>

where <em>g</em> = 9.8 m/s².

Adding these equations together eliminates <em>T</em>, and we can solve for <em>a</em> :

(<em>T</em> - 25 N) + ((6 kg) <em>g</em> - <em>T </em>) = (14 kg) <em>a</em>

33.8 N = (14 kg) <em>a</em>

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5 0
3 years ago
What is the formula for Ba+2 and F-?
Alinara [238K]

Answer:

BaF2

Explanation:

since you got the valence numbers just do the scissors move where you:

give the F the 2 and give the Ba the 1 so it be like

BaF2 here is the chemical compound

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3 years ago
Which describes acceleration involving only a change in direction?
Viefleur [7K]
If we consider any system moving with u<span>niform circular motion we can notice that the MAGNITUDE of the accelaration remains constant. However,  there is a change in the direction of the acceleration at every instant of time .

 As the object moves through the circle the acceleration changes its direction always pointing to the center of the circle.</span>
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3 years ago
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