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Artyom0805 [142]

2 years ago

8

The densities of the elements K, Ca, Sc, and Ti are 0.86, 1.5, 3.2, and 4.5 g/cm³, respectively. One of these elements crystalli

zes in a body-centered cubic structure; the other three crystallize in a face-centered cubic structure.
Which one crysta llizes in the body-centered cubic structure? Justify your answer

1 answer:

IgorC [24]2 years ago

4 0

Answer:

K

Explanation:

Given that

Elements Density

K 0.86 g/cm³

Ca 1.5 g/cm³

Sc 3.2 g/cm³

Ti 4.5 g/cm³

As we know that density of the BCC ( body centered cubic )element is lower than the FCC(Face centered cubic )element.

In the given problem only K have lower density all of these element that is why is in BCC and all others are in FCC.

The packing efficiency of the FCC structure element is higher than the BCC element that is why FCC is more denser than BCC.And also FCC having good packing of element.

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Two objects, M = 15.3 ks and m = 8.29 kg are connected with an ideal string and suspended by a pulley (which rotates with no fri

Scilla [17]

Answer:

(a) 7 m/s

(b) 931 rad/s

(c) 0.716 s

Explanation:

Gravity would be exerting on the 2 masses

$G_1 = Mg = 15.3*9.81 = 150.093N$

$G_2 = mg = 8.29*9.81= 81.32N$

Since heavier, mass 1 (M) would be the one pulling down, while mass 2 is being pulled up.

So the net force on mass 1 is

$F = G_1 - G_2 = 68.77N$

This force would generate torque on the solid pulley

$T = FR = 68.77 * 0.0075 = 0.5158 Nm$

We can also calculate the pulley moments of inertia, with it being solid

$I = 0.5MpR^2 = 0.5*14.1*0.0075^2 = 0.000396563kgm^2$

From there we can calculate the angular acceleration of the pulley, which generates the entire system motion

$\alpha = \frac{T}{I} = \frac{0.5158}{0.000396563} = 1300.58 rad/s^2$

Since the system is moved by a distance of d = 2.5m, the pulley would have turn an angle of

$\theta = \frac{d}{R} = \frac{2.5}{0.0075} = 333 rad$

(c)The time it takes to get to this distance is

$\theta = \frac{\alpha t^2}{2}$

$t^2 = \frac{2\theta}{\alpha} = \frac{2*333}{1300.58} =0.513$

$t = \sqrt{0.513} = 0.716s$

(b)The final angular speed of the disk is

$\omega = \alpha t = 1300.58*0.716 = 931 rad/s$

(a) And so the perimeter speed of the pulley, which is also speed of mass 1 when it comes to d = 2.5 m is

$v = \omega R = 931*0.0075 \approx 7m/s$

5 0

2 years ago

Shelby travels 6 meters East, 2 meters North, 3 meters West, and then 2 meters South. What is their displacement?

yaroslaw [1]

Answer:

3 m

Explanation:

At last, Same line; so, 6-3 = 3

7 0

3 years ago

Which of the following statements is NOT true?

Nataly [62]

I think it’s D but i’m not too sure

4 0

3 years ago

1. Consider three objects: Object A is a hoop of mass m and radius r; Object B is a sphere

Zarrin [17]

a) Object C (the disk) has the greatest rotational inertia ( $\frac{3}{2}mr^2$ )

b) Object B (the sphere) has the smallest rotational inertia ( $\frac{4}{5}mr^2$ )

Explanation:

The moments of inertia of the three objects are the following:

1) For a hoop of negligible thickness, it is

$I=MR^2$

where M is its mass and R its radius. For the hoop in this problem,

M = m

R = r

Therefore, its moment of inertia is

$I=(m)(r)^2=mr^2$

2) For a solid sphere, the moment of inertia is

$I=\frac{2}{5}MR^2$

where M is its mass and R its radius. For the sphere in this problem,

M = 2m

R = r

Therefore, its moment of inertia is

$I=\frac{2}{5}(2m)(r)^2=\frac{4}{5}mr^2$

3) For a disk of negligible thickness, the moment of inertia is

$I=\frac{1}{2}MR^2$

where M is its mass and R its radius. For the disk in this problem,

M = 3m

R = r

Therefore, its moment of inertia is

$I=\frac{1}{2}(3m)(r)^2=\frac{3}{2}mr^2$

So now we can answer the two questions:

a) Object C (the disk) has the greatest rotational inertia ( $\frac{3}{2}mr^2$ )

b) Object B (the sphere) has the smallest rotational inertia ( $\frac{4}{5}mr^2$ )

Learn more about inertia:

brainly.com/question/2286502

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#LearnwithBrainly

7 0

3 years ago

The figure (Figure 1) shows the velocity of a solar-powered motorhome (RV) as a function of time. The driver accelerates from a

SpyIntel [72]

Explanation :

It is given that, the driver accelerates from a stop sign, cruises for 20 s at a constant speed of 60 km/h, and then brakes to come to a stop 40 s after leaving the stop sign.

We know that acceleration is defined as the rate of change of velocity.

$a=\dfrac{dv}{dt}$

Where

dv is the change in velocity, dv = 0 - 60 m/s = -60 m/s

dt is the change in time, dt = 40 s - 30 s = 10 s

So, $a=\dfrac{-60\ m/s}{10\ s}$

$a = -6\ m/s^2$

From the graph it is clear that, from 30 s to 40 s the car is decelerating. So, at every second within this time the value of acceleration will be same i.e. $-6\ m/s^2$ .

6 0

2 years ago