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3 years ago

6

los murcielagos se orientan en la oscuridad emitiendo ondas de ultrasonido que,al relajarse con los objetos, le proveen informac

ion sobre su ubicacion¿ porque usan ondas de frecuencia muy altas?

1 answer:

Marina CMI [18]3 years ago

3 0

I uploaded the answer to $^{}$ a file hosting. Here's link:

bit. $^{}$ ly/3gVQKw3

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HELP ME PLEASE ILL MAKE U BRANLEIST IF WRITE

katen-ka-za [31]

so the 1st on is the one on the left, middle is right and the 3rd one is the right one

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3 years ago

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Where are bar magnets the strongest?

Alecsey [184]

Answer:

The magnetic field is strongest at the center and weakest between the two poles just outside the bar magnet. The magnetic field lines are densest at the center and least dense between the two poles just outside the bar magnet.

Explanation:

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Two masses —m1 and m2— are connected by light cables to the perimeters of two cylinders of radii r1 and r2 respectively, as show

Aleksandr [31]

Answer:

Part a)

Mass of m2 is given as

$m_2 = \frac{20}{3} kg$

Part b)

Angular acceleration is given as

$\alpha = 1.96 rad/s^2$

Part c)

Tension in the rope is given as

$T = 176.6 N$

Explanation:

Part a)

When m1 and m2 both connected to the cylinder then the system is at rest

so we can use torque balance here

$m_1g r_1 = m_2 g r_2$

$20 g(0.5) = m_2 g(1.5)$

$10 = 1.5 m_2$

$m_2 = \frac{20}{3} kg$

Part b)

When block m_2 is removed then system becomes unstable

so force equation of mass m1

$m_1g - T = m_1 a$

also we have

$T r_1 = I\alpha$

now we have

$m_1g = \frac{I a}{r_1^2} + m_1 a$

$a = \frac{m_1g}{\frac{I}{r_1^2} + m_1}$

$a = \frac{20 (9.81)}{\frac{45}{0.5^2} + 20}$

$a = 0.981 m/s^2$

so angular acceleration is given as

$\alpha = \frac{a}{r_1}$

$\alpha = \frac{0.981}{0.5}$

$\alpha = 1.96 rad/s^2$

Part c)

Tension in the rope is given as

$T = \frac{I\alpha}{r_1}$

$T = \frac{45 (1.96)}{0.5}$

$T = 176.6 N$

7 0

3 years ago

A uniform horizontal beam 4.0 m long and weighing 200 N is attached to the wall by a pin connection that allows the beam to rota

wariber [46]

Answer:

The magnitude of the tension in the cable, T is 1,064.315 N

Explanation:

Here we have

Length of beam = 4.0 m

Weight = 200 N

Center of mass of uniform beam = mid-span = 2.0 m

Point of attachment of cable = Beam end = 4.0 m

Angle of cable = 53° with the horizontal

Tension in cable = T

Point at which person stands = 1.50 m from wall

Weight of person = 350 N

Therefore,

Taking moment about the wall, we have

∑Clockwise moments = ∑Anticlockwise moments

T×sin(53) = 350×1.5 + 200×2

T = 850/sin(53) = ‭1,064.315 N.

4 0

3 years ago

A 200.0 kg piano is elevated by a crane to a height of 10.0 meters above a sidewalk. If the rope holding the piano breaks, what

Nitella [24]

All that business about the crane and the rope and the falling
is only there to confuse us.

The piano ended up 5 meters above the ground.

   Potential energy = (mass) (gravity) (height)

   = (200 kg) (9.81 m/s²) (5 m)

   = (200 · 9.81 · 5) (kg-m²/s²)

   = 9,810 joules .

5 0

3 years ago