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3 years ago

9

Alternate freezing and thawing often leads to ______.

2 answers:

Evgen [1.6K]3 years ago

8 0

Alternate freezing and thawing often leads to :
A. Creep
this occurrence usually occurs when the soils in that specific region is subjected to the temperature below 0 Celcius

hope this helps

WITCHER [35]3 years ago

7 0

The answer from the given options is "A. Creep".

Alternate freezing and thawing often leads to "creep".

Creep refers to the low mass development of soil and soil material down slants, driven fundamentally by gravity yet facilitated by immersion with water and by substitute freezing and thawing. Generally not distinguishable with the exception of through expanded perception.

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4 years ago

Read 2 more answers

A boy pushes a 50 kg wheelbarrow with a velocity of 0.5 m/s. If he uses a force of 10 N, how far does he push

labwork [276]

Answer:

d=1.25 m

Explanation:

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3 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

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