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3 years ago

Which best compares the gravitational force and the strong force ?

2 answers:

8 0

<span>The heavier the body is, the stronger its gravitational pull. Example, the Milky Way Galaxy has a gravitational pull because of the heavenly bodies such as stars and planets are surrounding it. A strong force is exerted if the mass of another body is bigger than the other body.</span>

Natali [406]3 years ago

3 0

Explanation:

Gravitational force :

Gravitation force is an attractive force which act between two bodies.

This force acts between the objects with mass.

The gravitational force is the proportional to the product of their masses of the bodies and inversely proportional to the square of distance between the bodies.

Gravitational force only attraction force.

The gravitational force is very weak force.

Strong force :

The force which holds nucleus together it called strong force.

This force is attractive and repulsive force.

The strong force like electromagnetic force it has short range.

The range of strong force is 10⁻¹⁵ m.

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Two forces act on a 1250 kg sailboat as it moves through the water with an initial velocity of 11 m/s. The forward force of the

Naddik [55]

Answer:

The velocity of the boat 15 seconds later is 5.6 meters per second.

Explanation:

We assume that sailboat can be modelled as particle, so that we use solely translations equations. It is noticed that the sailboat is move by action of the wind and drag force of the water is opposed to such force, but the last force has a greater magnitude than the first one, meaning that net force is less than zero.

From Newton's Laws we have the following equation of equilibrium for the sailboat:

$\Sigma F = F - f = m\cdot a$ (Eq. 1)

Where:

$F$ - Force from the wind exerted on the sailboat, measured in newtons.

$f$ - Drag force of the water, measured in newtons.

$m$ - Mass of the sailboat, measured in kilograms.

$a$ - Net acceleration of the sailboat, measured in meters per square second.

If we know that $F = 3.90\times 10^{3}\,N$ , $f = 4.35\times 10^{3}\,N$ and $m = 1250\,kg$ , then the net acceleration of the sailboat is:

$a = \frac{F-f}{m}$

$a = \frac{3.90\times 10^{3}\,N-4.35\times 10^{3}\,N}{1250\,kg}$

$a = -\frac{9}{25}\,\frac{m}{s^{2}}$

If sailboat decelerates uniformly, then we can get the final velocity of the boat by using this equation of motion:

$v = v_{o}+a\cdot t$ (Eq. 2)

Where:

$v_{o}$ , $v$ - Initial and final velocities of the sailboat, measured in meters per second.

$t$ - Time, measured in seconds.

If we get that $v_{o} = 11\,\frac{m}{s}$ , $a = -\frac{9}{25}\,\frac{m}{s^{2}}$ and $t = 15\,s$ , then the final velocity of the sailboat is:

$v = 11\,\frac{m}{s} +\left(-\frac{9}{25}\,\frac{m}{s^{2}} \right)\cdot (15\,s)$

$v = 5.6\,\frac{m}{s}$

The velocity of the boat 15 seconds later is 5.6 meters per second.

3 0

3 years ago

The box is pushed to the right with a force of 40 Newtons and it just begins to move what is the maximum static frictional force

Vesnalui [34]

Answer:

The maximum static frictional force is 40N.

Explanation:

When an object of mass M is on a surface with a coefficient of static friction μ, there is a minimum force that you need to apply to the object in order to "break" the coefficient of static friction and be able to move the object (Called the threshold of motion, once the object is moving we have a coefficient of kinetic friction, which is smaller than the one for static friction).

This coefficient defines the maximum static friction force that we can have.

So if we apply a small force and we start to increase it, the static frictional force will be equal to our force until it reaches its maximum, and then we can move the object and now we will have frictional force.

In this case, we know that we apply a force of 40N and the object just starts to move.

Then we can assume that we are just at the point of transition between static frictional force and kinetic frictional force (the threshold of motion), thus, 40 N is the maximum of the static frictional force.

3 0

3 years ago

While online last week, you saw the following advertisement:

Karolina [17]

No scientific testing has been made to check for ion transfer, and the claims are purely empirical. Also, nine out of ten people is hardly a representative sample, and the people can claim whatever they want since "feeling" is subjective. This is most likely a pseudoscientific claim, made to sound legitimate to consumers. The best answer is choice D.

3 0

3 years ago

Read 2 more answers

car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of a. The car

Luba_88 [7]

Answer:

0.572

Explanation:

First examine the force of friction at the slipping point where Ff = µsFN = µsmg.

the mass of the car is unknown,

The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.

First the tangential direction

∑Ft =Fft =mat

And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r

Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2

So going backwards and plugging in Ffc =m2atπr/ 2r =πmat

Ff = √(F2ft +F2fc)= matp √(1+π²)

µs = Ff /mg = at /g √(1+π²)=

1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572

7 0

3 years ago

The force between two charges, q, and 92, is F. If the distance between the

Sholpan [36]

Answer:

Explanation:

F = kQ₁Q₂/d²

F' = k2Q₁2Q₂/d²

F' = 4(kQ₁Q₂/d²)

F' = 4F

4 0

2 years ago