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3 years ago

I need help... What is Torque?

Engineering

2 answers:

faust18 [17]3 years ago

8 0

Answer: a force that causes something to rotate

Explanation:

Vitek1552 [10]3 years ago

7 0

Answer:

a twisting force that tends to cause rotation

Explanation:

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The substance called olivine may have any composition between Mg2SiO4 and Fe2SiO4, i.e. the Mg atoms can be replaced by Fe atoms

Ipatiy [6.2K]

Answer:

The answer is "0.147 nm and 99.63 mol %"

Explanation:

In point (a):

$\to nk1 = 062$

$\to \text{Bragg angle}$ $\theta =37.21^{\circ}$

$\to \text{diffraction angle}$ $2 \theta = 74.42^{\circ}$

$\to \lambda = 0.1790 nm$

find:

d(062)=?

formula:

$\to nx = 2d \sin \theta$

$\to d(062) = \frac{1 \times 0.1790^{\circ}}{2 \times \sin 37.21^{\circ}}\\$

$= \frac{0.1790^{\circ}}{2 \times 0.604738126}\\\\= \frac{0.29599589}{2}\\\\= 0.147 \\$

In point (b):

$\to Mg_2SiO_4\longleftrightarrow Fe_2SiO_4$

$d= 0.14774 \ \ \ \ \ olivine = 0.147 \ \ \ \ \ 0.15153$

formula:

$\to d=\frac{a}{\sqrt{n^2+k^2+i^2}}\\$

that's why the composition value equal to 99.63 %

3 0

3 years ago

The flowrate through a rectangular channel is 20 cfs. The upstream width of the channel is 10 ft, and the depth of the water in

Liula [17]

To solve this problem we will use the Froude number that relates the Forces of Inertia with the Forces of Gravity. There will be jump in the downstream only if Froude Number (Fr) is greater than 1 at upstream. Our values are given as,

$Q = 20cfs\\w= 10ft\\D= 1ft$

Then the velocity would be:

$V = \frac{Q}{wD}V = \frac{20}{10*1}V = 2ft/s$

The number of Froude is given as,

$Fr = \frac{V}{gD}^{1/2}$

Where,

V = Velocity

g = Gravity

D = Diameter

Replacing we have that

$Fr = \frac{2}{32.2}^{1/2}\\Fr = 0.352\\Fr$

There will be no Jump, correct answer is B.

5 0

3 years ago

A rocket is launched vertically from rest with a constant thrust until the rocket reaches an altitude of 25 m and the thrust end

hjlf

Answer:

a) $v=19.6 m/s$

b) $H=19.58 m$

c) $v_{f}=29.57 m/s$

Explanation:

a) Let's calculate the work done by the rocket until the thrust ends.

$W=F_{tot}h=(F_{thrust}-mg)h=(35-(2*9.81))*25=384.5 J$

But we know the work is equal to change of kinetic energy, so:

$W=\Delta K=\frac{1}{2}mv^{2}$

$v=\sqrt{\frac{2W}{m}}=19.6 m/s$

b) Here we have a free fall motion, because there is not external forces acting, that is way we can use the free-fall equations.

$v_{f}^{2}=v_{i}^{2}-2gh$

At the maximum height the velocity is 0, so v(f) = 0.

$0=v_{i}^{2}-2gH$

$H=\frac{19.6^{2}}{2*9.81}=19.58 m$

c) Here we can evaluate the motion equation between the rocket at 25 m from the ground and the instant before the rocket touch the ground.

Using the same equation of part b)

$v_{f}^{2}=v_{i}^{2}-2gh$

$v_{f}=\sqrt{19.6^{2}-(2*9.81*(-25))}=29.57 m/s$

The minus sign of 25 means the zero of the reference system is at the pint when the thrust ends.

I hope it helps you!

6 0

3 years ago

Select the correct answer. Which material is commonly preferred for detailed projects such as connectors, jewelry, or small part

Naya [18.7K]

Answer:

the mean time I was wondering if you have any questions or need any further information please contact me at the mean time

4 0

3 years ago

A large steel tower is to be supported by a series of steel wires; it is estimated that the load on each wire will be 19,000N. D

zhuklara [117]

Answer:

11.6 mm

Explanation:

With a factor of safety of 5 and a yield strength of 900 MPa the admissible stress is:

σadm = strength / fos

σadm = 900 / 5 = 180 MPa

The stress is the load divided by the section:

σ = P / A

σ = 4*P / (π*d^2)

Rearranging:

d^2 = 4*P / (π*σ)

$d = \sqrt{4*P / (\pi*\sigma)}$

$d = \sqrt{4*19000 / (\pi*180*10^6)} = 0.0116 m = 11.6 mm$

7 0

3 years ago