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pickupchik [31]
3 years ago
5

How much force to move 18 kg at a rate of 3m/s​

Physics
1 answer:
Stolb23 [73]3 years ago
4 0

Answer:

54N

Explanation:

F = ma

F = (18kg)(3m/s)

F = 54N

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Direct tension indicators are sometimes used instead of torque wrenches to ensure that a bolt has a prescribed tension when used
natali 33 [55]

Complete Question

The complete question is shown on the uploaded image

Answer:

The tension on the shank is  T =8391.6 N

Explanation:

From the question we are told that

       The strain on the strain on the head is \Delta l = 0.1 mm/mm = \frac{0.1}{1000} = 0.1 *10^{-3} m/m

         The contact area is  A = 2.8 mm^2 = 2.8* (\frac{1}{1000} )^2 = 2.8*10^{-6} m^2  

Looking at the first diagram

           At  600 MPa of stress

               The strain is  0.3mm/mm

          At   450 MPa of stress

                 The strain is   0.0015 mm/mm

 To find the stress at  \Delta l we use the interpolation method

            \frac{\sigma_{\Delta l} -  \sigma_{0.0015} }{ \sigma _ {0.3} - \sigma_{0.0015} } = \frac{e_{\Delta l }  - e_{0.0015}}{e_{0.3} - e_{ 0.0015}}

Substituting values

              \frac{\sigma _{\Delta l} - 450}{600 - 450} = \frac{0.1 -0.0015}{0.3 - 0.0015}

            \sigma _{\Delta l} -450 = 49.50

             \sigma _{\Delta l} = 499.50 MPa

Generally the force on each head is mathematically represented as

              F = \sigma_{\Delta l} * A

Substituting values

             F = 499.50*10^{6} * 2.8*10^{-6}

                =1398.6N

Now the tension on the bolt shank is as a result of the force on the 6 head which is mathematically evaluated as

              T = 6 * F

                  = 6* 1398.6

              T =8391.6 N

                 

     

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2b2t hope that helps
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An astronaut weighs 104 newtons on the moon, where the strength of gravity is 1.6N/kg. What is her mass?
Leona [35]
Her mass is 65 kilograms
5 0
3 years ago
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Kayla draws the image shown as part of her physical science homework.
AnnZ [28]
What the question for this assessment
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When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal speed. Once he has reache
Masteriza [31]

Answer:

b) True.    the force of air drag on him is equal to his weight.

Explanation:

Let us propose the solution of the problem in order to analyze the given statements.

The problem must be solved with Newton's second law.

When he jumps off the plane

     fr - w = ma

Where the friction force has some form of type.

     fr = G v + H v²

Let's replace

     (G v + H v²) - mg = m dv / dt

We can see that the friction force increases as the speed increases

At the equilibrium point

      fr - w = 0

      fr = mg

      (G v + H v2) = mg

For low speeds the quadratic depended is not important, so we can reduce the equation to

     G v = mg

     v = mg / G

This is the terminal speed.

Now let's analyze the claims

a) False is g between the friction force constant

b) True.

c) False. It is equal to the weight

d) False. In the terminal speed the acceleration is zero

e) False. The friction force is equal to the weight

3 0
3 years ago
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