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3 years ago

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África tiene más de 1000 lenguas, pero solo 50 son habladas por más de 500 mil personas, explica si esto es un aspecto positivo

o negativo y porque

1 answer:

Misha Larkins [42]3 years ago

7 0

Answer:

put it in English please

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Which best explains why the diagram shows refraction but not reflection?

Aleonysh [2.5K]

It shows the ray passing through the boundary.

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3 years ago

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Block with mass m =7.6 kg is hung from a vertical spring. when the mass hangs in equilibrium, the spring stretches x = 0.29 m. w

PSYCHO15rus [73]

1) 256.9 N/m

The force applied to the spring is equal to the weight of the block hanging on the spring:

$F=mg=(7.6 kg)(9.8 m/s^2)=74.5 N$

And the spring constant can be found by using Hook's law, because we know that the displacement caused by this force is x = 0.29 m:

$F=kx\\k=\frac{F}{x}=\frac{74.5 N}{0.29 m}=256.9 N/m$

2) 1.08 Hz

The angular frequency of oscillation of the spring is given by the formula:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{256.9 N/m}{7.6 kg}}=5.81 rad/s$

And the frequency of oscillation is given by:

$\omega=2\pi f\\f=\frac{2 \pi}{\omega}=\frac{2\pi}{5.81 rad/s}=1.08 Hz$

3) 2.19 m/s

The velocity at time t of the block is given by:

$v=v_0 cos (\omega t)$

where

$v_0 = 4.4 m/s$ is the initial velocity of the block

$\omega=5.81 rad/s$ is the angular frequency

t is the time

Substituting t=0.36 s, we find the speed of the block at that time:

$v(0.36 s)=(4.4 m/s) ( cos ((5.81 rad/s)(0.36 s)) = -2.18 m/s$

And the negative sign means that the direction of the velocity is upward (because the initial velocity was downward)

4) 25.6 m/s^2

The maximum acceleration is given by:

$a_0 = \omega^2 A$

where A is the amplitude of the oscillation.

We can find the amplitude by using the law of conservation of energy: in fact, the kinetic energy at the equilibrium point must be equal to the elastic potential energy at the point of maximum displacement:

$K=U\\\frac{1}{2}mv_0^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{mv_0^2}{k}}=\sqrt{\frac{(7.6 kg)(4.4 m/s)^2}{256.9 N/m}}=0.76 m$

So, the maximum acceleration is

$a_0 = \omega^2 A=(5.81 rad/s)^2 (0.76 m)=25.6 m/s^2$

5) 95.1 N

The magnitude of the net force acting on the block is given by the difference between the weight and the restoring force of the spring:

$F=mg-kx$

First, we need to find the position x at t=0.36 s, which is given by

$x(t)=A sin(\omega t)=(0.76 m)(sin ((5.81 rad/s)(0.36 s))=0.66 m$

And so, the net force is

$F=(7.6 kg)(9.8 m/s^2)-(256.9 N/m)(0.66 m)=-95.1 N$

And the negative sign means the direction of the force is upward.

8 0

3 years ago

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Una caja de 40.0 kg se desliza 4.00 m hacia arriba por un plano inclinado de 26° con respecto a la acción de una fuerza de 340 N

otez555 [7]

Answer:

W = 320.30 J

Explanation:

To calculate the net work done over the block you take into account all implied forces:

$\Sigma F=F+F_gsin\theta-F_fcos\theta$ (1)

The gravitational force and friction force are against the applied force F.

θ = 26°

F: applied force = 340N

Fg: gravitational force = Mg = (40.0kg)(9.8m/s^2) = 392N

Ff: friction force = $\mu N=\mu Mg=(0.250)(392N)=98N$

Next, you replace to obtain the net force:

$F_N=(340N)-(392N)sin(26\°)-(98N)cos(26\°)\\\\F_N=80.07N$

Finally, the net work, for 4 m, is:

$W_N=F_Nd=(80.07N)(4m)=320.30J$

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3 years ago

A 120N box is being pushed across the surface that has a frictional force of 3N. If the box is being pushed with a constant forc

VashaNatasha [74]

Answer:

Net Force = 10 N

acceleration: $a=0.817 \frac{m}{s^2}$

Explanation:

1) The net force on the box is the applied force (F = 13 N) minus the friction force (f = 3 N), Therefore, net force = 10 N

See attached free body diagram.

2) To find the box's acceleration, we need first to find the mass of the box (since they are giving us its weight in Newtons). To do such we use the equation for weight and solve for the box's mass:

$weight=m*g\\120 N = m*9.8\frac{m}{s^2} \\\frac{120}{9.8} kg = m\\m=12.24 kg$

Now we can find the box's acceleration by using the equation for the net force:

$F=m*a\\10N = 12.24 kg * a\\\frac{10}{12.24} \frac{m}{s^2} =a\\a=0.817 \frac{m}{s^2}$

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3 years ago

In a spaceship accelerating at g, far from earth's gravity, how does the motion of a dropped ball compare with dropping one on e

Degger [83]

On earth, the speed at which a ball falls depends on the distance at which you release it. If you release it in the same place, and at the same distance from the ground, the speed will be the same. While in a spacecraft with constant acceleration, if you repeat the process from the same distance from the ground, the ball will have more and more speed due to inertia. But of a spaceship with a constant speed, that can not be differentiated, this is called equivalence principle, but that's another story...

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3 years ago