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lesya692 [45]
1 year ago
14

Technician A says that the definition of torque is how far the crankshaft twists in degrees.Technician B says that torque can re

late to tightness of bolts.Who is correct?
Engineering
1 answer:
leonid [27]1 year ago
6 0
Technician B is correct because torque is a force of an object.
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A photovoltaic panel of dimension 2m×4m is installed on the
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Answer:

Explanation:

Simply put, a solar panel works by allowing photons, or particles of light, to knock electrons free from atoms, generating a flow of electricity. Solar panels actually comprise many, smaller units called photovoltaic cells. (Photovoltaic simply means they convert sunlight into electricity. The attached diagram give an ilustsration of the photovotaic pannel mounted on a roof top.

Solution

To Determine the electric power generated for

a) A still summer day.

E = A * r * H * PR

E = Total Amount of Energy in kilowatt

A = Total Surface Area

r = efficiency Rating

H = global radiation value

PR = Performance Ratio

kwh = watt * Time/1000

kwh = 100 * 35/1000

3.5

b)

kwh = watt * Time/1000

kwh = 30 *15/1000

4.5

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3 years ago
How to calculate maximum shear stress in the cross-section of a beam if the neutral axis is not on the thinnest part of the cros
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—By the maximum shear stress is then calculated by: where b = 2 (ro − ri) is the effective width of the cross section, Ic = π (ro4 − ri4) / 4 is the centroidal moment of inertia, and A = π (ro2 − ri2) is the area of the cross section.

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3 years ago
Choose the true statement from those shown below: A Merchant Account allows you to use SSL on your web site. Disadvantages of us
Slav-nsk [51]

Answer:

The answer is A. that is, a merchant account allows you to use SSL on your website.

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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
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