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3 years ago

What is the motion of the object?

Physics

1 answer:

aleksley [76]3 years ago

4 0

Answer:

<em>Thus, the object is accelerating to the left</em>

Explanation:

<u>The Net Force</u>

The net force is the result of adding all the forces as vectors acting on a body.

$\vec F=\vec F_1+\vec F_2+...+\vec F_n$

Each vector can be expressed in its rectangular components Fx and Fy, and the sum is the sum of the rectangular components separately.

Second Newton's law gives the relation between the net force and the acceleration of the body:

$\vec F = m.\vec a$

We can see the acceleration is a vector with the same direction as the net force.

The diagram shows two vertical forces and two horizontal forces.

The vertical forces are acting in opposite directions and with the same magnitude, thus they cancel out, leaving zero net force in the y-axis.

The horizontal forces are opposite and with different magnitudes. Since the force acting to the left (F3) has a greater magnitude than the force acting to the right (F4), there is a net force directed to the left with a magnitude of 60 N - 20 N = 40 N

Thus, the object is accelerating to the left

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Use dimensional analysis to determine how the linear acceleration a in m/s2 of a particle traveling in a circle depends on some,

astraxan [27]

I have to say, i love this kind of problems.

So, we got the linear acceleration, as we all know, linear the acceleration its, in dimensional units:

$[a]=[\frac{distance}{time^2}]$ .

Now, we got the radius

$[r] = [distance]$

the angular frequency

$[\omega] = \frac{1}{s}$

and the mass

$[m]=[mass]$ .

Now, the acceleration doesn't have units of mass, so it can't depend on the mass of the particle.

The distance in the acceleration has exponent 1, and so does in the radius. As the radius its the only parameter that has units of distance, this means that the radius must appear with exponent 1. Lets write

$a \propto r$ .

The time in the acceleration has exponent -2 As the angular frequency its the only parameter that has units of time, this means that the angular frequency must appear, but, the angular frequency has an exponent of -1, this means it must be squared

$a \propto r \omega^2$ .

We are almost there. If this were any other problem, we would write:

$a = A r \omega^2$

where A its an dimensionless constant. Its common for this constants to appears if we need an conversion factor. If we wanted the acceleration in cm/s^2, for example. Luckily for us, the problem states that there is no dimensionless constant involved, so:

$a = r \omega^2$

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3 years ago

Plate Tectonic Theory

Vitek1552 [10]

Answer:

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3 years ago

Consider a basketball player spinning a ball on the tip of a finger. If a player performs 1.99 J of work to set the ball spinnin

scoundrel [369]

To solve this problem it is necessary to apply the concepts related to rotational kinetic energy, the definition of the moment of inertia for a sphere and the obtaining of the radius through the circumference. Mathematically kinetic energy can be given as:

$KE= I\omega^2$

Where,

I = Moment of inertia

$\omega =$ Angular velocity

According to the information given we have that the radius is

$\Phi= 2\pi r$

$0.749m = 2\pi r$

$r = 0.1192m$

With the radius obtained we can calculate the moment of inertia which is

$I = \frac{2}{3}mr^2$

$I = \frac{2}{3}(0.624)(0.1192)^2$

$I = 5.91*10^{-3} kg \cdot m^2$

Finally, from the energy equation and rearranging the expression to obtain the angular velocity we have to

$\omega = \sqrt{\frac{2KE}{I}}$

$\omega = \sqrt{\frac{2(1.99)}{5.91*10^{-3}}}$

$\omega = 25.95rad/s$

Therefore the angular speed will the ball rotate is 25.95rad/s

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3 years ago

The law of reflection says that the angle of incidence is

nekit [7.7K]

The law of reflection states that the angle of incidence is equal to the angle of reflection. Furthermore, the law of reflection states that the incident ray, the reflected ray and the normal all lie in the same plane.

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3 years ago

An athlet starting from stationary moves with an acceleration 2.2m/s^2 for 5 seconds, then for other 5 seconds

ycow [4]

Answer:

The graph is shown below.

The athlete's speed at the finish line is 11 m/s

Total distance covered is 82.5 m.

Explanation:

The graph of speed versus time is drawn below.

Acceleration in the first 5 seconds, $a=2.2$ m/s²