PLEASE HELP ME! IM TIMED

How can tectonic activity change the climate of the earth?

soldi70 [24.7K]

If you really want to find out the answer just pray about it and it will come to you..

3 0

3 years ago

The "lead" in pencils is a graphite composition with a Young’s modulus of about 1×1010N/m21×1010⁢N/m2. Calculate the change in l

Tanya [424]

Answer:

b) 0.1 mm

Explanation:

Given that

E= 1 x 10¹⁰ N/m²

F= 4 N

d= 0.5 mm

L = 60 mm

We know that elongation due to force F given as

$\Delta L=\dfrac{FL}{AE}$

$\Delta L=\dfrac{FL}{\dfrac{\pi d^2}{4}\times E}$

$\Delta L=\dfrac{4\times 60}{\dfrac{\pi \times 0.5^2}{4}\times 10^4}$

ΔL = 0.12 mm

Therefore the answer is -

b) 0.1 mm

6 0

2 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

brainly.com/question/691705

#LearnwithBrainly

7 0

3 years ago

When a driver presses the brake pedal, his car can stop with an acceleration of 5.4 m/s2. How far will the car travel while comi

notsponge [240]

$Given:\\a=5.4 \frac{m}{s^2} \\v_0= 25\frac{m}{s} \\\\Find:\\s=?\\\\Solution:\\\\s=v_0t- \frac{at^2}{2} \\\\a= \frac{\Delta v}{t} \\\\v_e=0\Rightarrow \Delta v=v_0\\\\t= \frac{v_0}{a} \\\\s= \frac{v_0^2}{a} - \frac{v_0^2}{2a} =\frac{v_0^2}{2a} \\\\s= \frac{ (25\frac{m}{s})2 }{2\cdot 5.4 \frac{m}{s^2} } \approx 57.9m\\\\C\rightarrow Correct\;answer$