Acquisition of resources from an external source is called?

1. Given: R= 25 , E = 100 V<br> Solve for I

Lyrx [107]

Answer:

4 amps

Explanation:

V = I*R

I / V/R

I = 100 / 25

I = 4 amps

5 0

3 years ago

Write an application that solicits and inputs three integers from the user and then displays the sum, average, product, smallest

Ganezh [65]

Answer:

3423=6^H

Explanation:

6 0

3 years ago

A minivan starts from rest on the road whose constant radius of curvature is 40 meters and whose bank angel is 10 degrees. the m

zimovet [89]

Based on the calculations, the magnitude (a) of it's total acceleration is equal to 2.71 m/s².

<u>Given the following data:</u>

Angle of inclination = 10°.

Radius of curvature, r = 40 meters.

Acceleration of the minivan, A = 1.8 m/s².

Initial velocity, u = 0 m/s (since it's starting from rest).

Time, t = 5 seconds.

<h3>How to determine the magnitude (a) of it's total acceleration?</h3>

First of all, we would determine the final velocity of the minivan by applying the first equation of motion as follows:

V = u + at

V = 0 + 1.8 × 5

V = 9 m/s.

Next, we would calculate the centripetal acceleration of this minivan:

Ac = V²/r

Ac = 9²/40

Ac = 2.025 m/s².

Now, we can determine the magnitude (a) of it's total acceleration:

a = √(Ac² + A²)

a = √(2.025² + 1.8²)

a = 2.71 m/s².

Read more on acceleration here: brainly.com/question/24728358

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8 0

2 years ago

A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft

alexandr402 [8]

Answer with Explanation:

By the equation or Torque we have

$\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}$

where

T is the torque applied on the shaft

$I_{p}$ is the polar moment of inertia of the shaft

$\tau$ is the shear stress developed at a distance 'r' from the center of the shaft

$\theta$ is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft $I_{p}=\frac{\pi R^4}{2}$

For a hollow shaft $I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}$

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

$\frac{2T}{\pi R^4}\times r=\tau _{solid}$

2) For hollow shaft we have

$\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4}$

Comparing the above 2 relations we see

$\frac{\tau _{solid}}{\tau _{hollow}}=0.76$

Similarly for angle of twist we can see

$\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316$

Part b)

Strength of solid shaft = $\tau _{max}=\frac{T\times R}{I_{solid}}$

Weight of solid shaft = $\rho \times \pi R^2\times L$

Strength per unit weight of solid shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}$

Strength of hollow shaft = $\tau '_{max}=\frac{T\times R}{I_{hollow}}$

Weight of hollow shaft = $\rho \times \pi (R^2-0.7R^2)\times L$

Strength per unit weight of hollow shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}$

Thus $\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16$

3 0

4 years ago

A brittle failure has extensive plastic deformation in the vicinity of the advancing crack. This process proceeds relatively slo

Tomtit [17]

Answer:

False ( b )

Explanation:

In a brittle failure the cracks spreads rapidly without a significant deformation, and the cracks are very unstable with the cracks extending without an increase in the amount of applied stress.

Therefore the above description in the question is false.

3 0

3 years ago