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4 years ago

At depths of greater than 60 m, ice moves by plastic deformation. What causes this transition from brittle to ductile behavior?

Physics

1 answer:

mafiozo [28]4 years ago

3 0

Answer:

Increase in the intensity of pressure allows very less deformation

Explanation:

For the water the variation of pressure at any depth is directly proportional to the depth of the water above that level.

Thus, at depth greater than 60 m the intensity of pressure on ice is very high and therefore very little deformation is possible.

Hence, the transition from brittle to ductile behavior is observed.

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What is 12 and 1 half % expressed as a fraction

kiruha [24]

12 1/2 i would assume

6 0

3 years ago

Read 2 more answers

A rectangular piece of cardboard, whose area is 352 square centimeters, is made into an open box by cutting a 2-centimeter sq

Usimov [2.4K]

Answer:

Dimension of cardboard is 22 m by 16 m

Explanation:

Given that,

Area = 352 cm²

Side of each square cutting from corner = 2 cm

Volume of box = 432 cm³

Let the two sides are x and y.

The area of the rectangular piece is

$xy=352$

$y=\dfrac{352}{x}$ -------- (1)

The volume of the rectangular piece

$2(x-4)(\dfrac{352}{x}-4)=432$

$x^2-38x+352=0$

$(x-16)(x-22)=0$

x=16,22

Put the value of x in the equation (I)

For x = 16

$y=\dfrac{352}{16}=22$

For x = 22

$y=\dfrac{352}{22}=16$

Dimension of cardboard is 22 m by 16 m

8 0

4 years ago

A ball is thrown straight upward and rises to a maximum

Leviafan [203]

Answer:

Explanation:

As we know that the ball is projected upwards so that it will reach to maximum height of 16 m

so we have

$v_f^2 - v_i^2 = 2 a d$

here we know that

$v_f = 0$

also we have

$a = -9.81 m/s^2$

so we have

$0 - v_i^2 = 2(-9.81)(16)$

$v_i = 17.72 m/s$

Now we need to find the height where its speed becomes half of initial value

so we have

$v_f = 0.5 v_i$

now we have

$v_f^2 - v_i^2 = 2 a d$

$(0.5v_i)^2 - v_i^2 = 2(-9.81)h$

$-0.75v_i^2 = -19.62 h$

$0.75(17.72)^2 = 19.62 h$

$h = 12 m$

3 0

3 years ago

A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.2 rad/s in 3.07 s. (a) f

Hitman42 [59]

(a) The angular acceleration of the wheel is given by
$\alpha = \frac{\omega_f - \omega_i }{t}$
where $\omega_i$ and $\omega_f$ are the initial and final angular speed of the wheel, and t the time.

In our problem, the initial angular speed is zero (the wheel starts from rest), so the angular acceleration is
$\alpha = \frac{(11.2 rad/s) - 0}{3.07 s} =3.65 rad/s^2$

(b) The wheel is moving by uniformly rotational accelerated motion, so the angle it covered after a time t is given by
$\theta (t) = \omega_i t + \frac{1}{2} \alpha t^2$
where $\omega_i = 0$ is the initial angular speed. So, the angle covered after a time t=3.07 s is
$\theta= \frac{1}{2} \alpha t^2 = \frac{1}{2}(3.65 rad/s^2)(3.07 s)^2 = 17.2 rad$

6 0

3 years ago

one of the ways, covered in the article and in class, that we charged an object was rubbing it against carpet on the floor true

astraxan [27]

Answer:

It is possible to statically charge objects by rubbing it against carpet fibers, but I'm not sure if that was in the article that you read.

Explanation:

Static charge can build up via carpet fibers.

5 0

3 years ago