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g100num [7]
2 years ago
8

A 50kg mass is sitting on a frictionless surface. An unknown constant force pushes the mass for 2 seconds until the mass reaches

a velocity of 3m/s. What was the force acting on the mass?​
Physics
1 answer:
Marina CMI [18]2 years ago
3 0

Answer:

75N

Explanation:

a = v/t = 3/2

F = ma = 50(3/2) = 75

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Which of the following four circuit diagrams best represents the experiment described in this problem?
Valentin [98]

We don't see any circuit diagrams.  

This worries us for a few seconds, until we realize that we don't know anything about the experiment described in the problem either, so we don't have to worry about it at all.

6 0
2 years ago
The boom hoisting sheave must have pitch diameters of no less than _______times the nominal diameter of the rope used.
alexira [117]

Answer:

18 times

Explanation:

According to the security purposes which is set under the rules and regulation OSHA, which describes all the rights to the worker.

In the boom hoist receiving system all the sheaves which are used should have a pitch diameter of rope not less than 18 times the diameter of the nominal rope which is used.

7 0
2 years ago
A material that has a fracture toughness of 33 MPa.m0.5 is to be made into a large panel that is 2000 mm long by 250 mm wide and
scoray [572]

Answer:

F_{allow} = 208.15kN

Explanation:

The word 'nun' for thickness, I will interpret in international units, that is, mm.

We will begin by defining the intensity factor for the steel through the relationship between the safety factor and the fracture resistance of the panel.

The equation is,

K_{allow} =\frac{K_c}{N}

We know that K_c is 33Mpa*m^{0.5} and our Safety factor is 2,

K_{allow} = \frac{33Mpa*m^{0.5}}{2} = 16.5MPa.m^{0.5}

Now we will need to find the average width of both the crack and the panel, these values are found by multiplying the measured values given by 1/2

<em>For the crack;</em>

\alpha = 0.5*L_c = 0.5*4mm = 2mm

<em>For the panel</em>

\gamma = 0.5*W = 0.5*250mm = 125mm

To find now the goemetry factor we need to use this equation

\beta = \sqrt{sec(\frac{\pi\alpha}{2\gamma})}\\\beta = \sqrt{sec(\frac{2\pi}{2*125mm})}\\\beta = 1

That allow us to determine the allowable nominal stress,

\sigma_{allow} = \frac{K_{allow}}{\beta \sqrt{\pi\alpha}}

\sigma_{allow} = \frac{16.5}{1*\sqrt{2*10^{-3} \pi}}

\sigma_{allow} = 208.15Mpa

So to get the force we need only to apply the equation of Force, where

F_{allow}=\sigma_{allow}*L_c*W

F_{allow} = 208.15*250*4

F_{allow} = 208.15kN

That is the maximum tensile load before a catastrophic failure.

4 0
2 years ago
The 20-lb cabinet is subjected to the force F = (3 + 2t) lb, where t is in seconds. If the cabinet is initially moving down the
Andre45 [30]

Answer:

t₁ = 0.95 s

Explanation:

In this chaos we must use the definition of Newton's second law

      F = m a = m dv / dt

      dv = F dt / m

Let's replace and integrate, let's take the upward direction of the plane as positive, the force is positive

       dv = ∫ (3 + 2t) dt / m

       v = (3 t + 2 t²/ 2) /m

Let's evaluate between the lower limit t = 0 v = -6 ft / s (going down) to the upper limit   t = t and v = 0

       0 - (-6) = (3 (t- 0) + (t² -0)) / m

       t² + 3t -6m = 0

Let's look for the mass

      W = mg

      m = W / g

      m = 20/32

      m = 0.625 slug

Let's solve the second degree equation

     t² + 3t -3.75 = 0

     t = (-3 ± √ (32 + 4 1 3.75)) / 2

     t = (-3 ± 4,899) / 2

     t₁ = 0.95 s

     t₂ = -3.95 s

We take the positive time

6 0
2 years ago
The diagram shows two vectors that point west and north. What is the magnitude of the resultant vector? 13 miles 17 miles 60 mil
Kipish [7]
Using the formula A squared plus B squared equals C squared, we can find the solution by substituting 5 for A and 12 for B.

By squaring 5, we get 25, and by squaring 12, we get 144. Adding these, we get 169. The square root of this is 13.
6 0
2 years ago
Read 2 more answers
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