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3 years ago

You are on a frozen pond, and the ice starts to crack. if you lie down on the ice and begin to crawl, this will

1 answer:

6 0

If you lie down on the ice and begin to crawl, this will spread out your weight more evenly and make it less likely for the ice to suffer a critical failure, in which you might fall below the surface of the ice and into the water. You have better weight distribution by being on more than two points - only your two feet. By crawling you disperse your weight to possibly up to eight points or more. This includes hands, elbows, knees, feet, chest. The dispersing of the weight takes significant pressure off of just one or two points, because as you are walking you are also most likely lifting a foot. By crawling, you remain in constant contact with your weight spread more evenly.

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The uniform rods AB and BC weigh 24 ky and kg, respectively,and the small wheel at C is of negligible weight. If the wheel ismov

victus00 [196]

The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.

<h3>What is Potential and Kinetic energy?</h3>

Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.

mass of rod, mab = 2.4kg

mass of rod, mbc = 4kg

conservation of energy

$T_{1} + V_{1} = T_{2} + V_{2}$

$h_{ab} = h_{bc} = 0.18m$

potential energy at position 1,

$V1 = m_{ab} gh_{ab} + m_{bc} gh_{bc}$

V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18

V1 = 11.30112

kinetic energy T1 at position 1 is zero

potential energy at position 2 is zero

K.E at position 2,

$T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} + \frac{1}{2} m_{bc} v^{2}_{G} + \frac{1}{2} lw^{2}_{bc}$

$l_{ab} =\frac{m_{ab} l^{2}_{ab} }{3}$

= 1/3 *4 * (0.36)²

=0.10368kg m²

$l =\frac{m_{bc} l^{2}_{bc} }{12}$

= 1/12 *4 * (0.6)²

=0.12kg m²

on putting the values in above equation we get,

T₂ = 1.0667vb²

0 + 11.30112 = 1.0667vb² + 0

vb = 3.2549 m/s

to learn more about Kinetic and potential energy go to - brainly.com/question/18963960

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5 0

1 year ago

How can submarines use echolocation to tell how close they are to the bottom of the ocean?

Lapatulllka [165]

To locate a specific target or to determine how close submarines are to the seafloor, they use active and passive sound navigation and ranging (or a SONAR, in simple terms.) It emits pulses of sound waves that travel through the water, reflect off the target and relayed back to the ship. By determining how fast the sound wave travels back, the computers on the sub calculate how far they are from the target.

Hope this helps.

3 0

2 years ago

Read 2 more answers

True or False.<br>1. Electromagnetic radiation comes from nature such as deep within<br>the Earth.

Dahasolnce [82]

Answer:

yes

Explanation:

the mantle rotates around the iron core and creates a charge. that charges radiates outward as the magnetosphere

5 0

2 years ago

Over [174]

What Kepler's constant ? ? ! ?

The only constant in Kepler's laws is in the third one, where it says something to the
effect that (square of a body's period) / (cube of its distance from the central body)
is a constant.

That means it's a constant for multiple little ones orbiting the same central body.
But it's not the same constant for other central bodies.

It's one constant for the planets, asteroids, and comets orbiting the sun.

It's a different constant for the moon, TV satellites, weather satellites,
and military satellites orbiting the Earth.

4 0

3 years ago

One billiard ball is shot east at 2.2 m/s. A second, identical billiard ball is shot west at 0.80 m/s. The balls have a glancing

Leona [35]

Answer:

(a). The speed of the first ball after the collision is 1.95 m/s.

(b). The direction of the first ball after the collision is 44.16° due south of east.

Explanation:

Given that,

Velocity of one ball u₁= 2.2i m/s

Velocity of second ball u₂=- 0.80i m/s

Final velocity of the second ball v₂= 1.36j m/s