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3 years ago

When trying to solve a frame problem it will typically be necessary to draw many free body diagrams. a)-True b)-False

Engineering

1 answer:

klasskru [66]3 years ago

6 0

Answer:

True

Explanation:

When trying to solve a frame problem in Engineering or Physics, it will typically be necessary to draw more than one body diagram.

When we have several parts of the frame or a set of frames, we have the anchor point, as well as the intersections of frames. Besides that, usually, there is a particle or rigid body together with the frame system. In this sense, usually, it is required to analyze a body diagram for the particle or rigid body suspended, as well as the intersections of the frames. So, usually, it will be required a minimum of two body diagrams.

If the system is more complex, or there are many intersections points, it will be required more than two body diagrams.

Finally, indeed, it will typically be necessary to draw many-body diagrams.

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Concerned with the number of maintenance visits the rocket can undergo before being out of service, you have been informed that

Ainat [17]

Answer:

(a) Mn = M₁ + (n-1) (M₂ -M₁) = 1 + (n- 1) 1 = n (b) n > 10 (exceed 10) or n =11 (c) n >50 or n= 51

After making a journey of 51 times, the rocket will be discarded

Explanation:

Solution

(a) Let Mn denotes the number of maintenance visits after the nth journey

Then M₁ = 1 , M₂ = 1 +M₁ = 2, M₃ = 1 +M₂ = 3

We therefore, notice that M follows an arithmetic sequence

So,

Mn = M₁ + (n-1) (M₂ -M₁)

= 1 + (n- 1) 1 = n

or Mn =n

(b) For what value of n we will get fro Mn > 10

Thus,

n > 10 (exceed 10) or n =11

(c)Similarly of Mn is greater than 50 or Mn>50, the rocket will not be used or reused

So,

n >50 or n= 51

After making a journey of 51 times, the rocket will be discarded

7 0

3 years ago

The present worth of income from an investment that follows an arithmetic gradient is projected to be $475,000. The income in ye

Nikitich [7]

Answer:

G = $37,805.65

Explanation:

I found this on another site:

475,000 = 25,000(P/A,10%,6) + G(P/G,10%,6)

475,000 = 25,000(4.3553) + G(9.6842)

9.6842G = 366,117.50

G = $37,805.65

4 0

3 years ago

What is the resistance of a resistor if the current flowing through it is 3mA and the voltage across it is 5.3V?

Flura [38]

Answer: 1766.667 Ω = 1.767kΩ

Explanation:

V=iR

where V is voltage in Volts (V), i is current in Amps (A), and R is resistance in Ohms(Ω).

3mA = 0.003 A

Rearranging the equation, we get

R=V/i

Now we are solving for resistance. Plug in 0.003 A and 5.3 V.

R = 5.3 / 0.003

= 1766.6667 Ω

= 1.7666667 kΩ

The 6s are repeating so round off to whichever value you need for exactness.

6 0

1 year ago

If you are a mechanical engineer answer these questions:

Natasha_Volkova [10]

Answer:

1. Yes, they are all necessary.

2. Both written and verbal communication skills are of the utmost importance in business, especially in engineering. Communication skills boost you or your teams' performance because they provide clear information and expectations to help manage and deliver excellent work.

3 0

3 years ago

Provide an argument justifying the following claim: The average (as defined here) of two Java ints i and j is representable as a

ahrayia [7]

Answer:

public static int average(int j, int k) {

return (int)(( (long)(i) + (long)(j) ) /2 );

}

Explanation:

The above code returns the average of two integer variables

Line 1 of the code declares a method along with 2 variables

Method declared: average of integer data type

Variables: j and k of type integer, respectively

Line 2 calculates the average of the two variables and returns the value of the average.

The first of two integers to average is j

The second of two integers to average is k

The last parameter ensures average using (j+k)/2

3 0

2 years ago