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2 years ago

The critical path is the sequence of activities that determine the latest date by which the project can be evaluated.

Physics

1 answer:

Leviafan [203]2 years ago

3 0

The statement given about the critical path is true.

To find the answer, we have to know more about the critical path.

<h3>What is critical path?</h3>

The longest series of tasks in a project plan that must be finished on time for it to be finished by the deadline is known as the critical path.
The critical route is the set of actions that, when carried out in a timely manner while taking into account all the aforementioned factors, will take the longest to accomplish the project.
Because it is the "red alert" of paths, the critical path is referred to as "critical".
Your project won't be successfully completed before the deadline if it doesn't follow the critical path's timing and order.

Thus, we can conclude that, the statement given about the critical path is true.

Learn more about the critical path here:

brainly.com/question/15091786

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Which device converts chemical energy directly into electrical energy?

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Answer:

Explanation:

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3 years ago

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A rectangular coil with 50 turns of conducting wire and a total resistance of 10.0 Ω initially lies in the yz-plane at time t =

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Answer:

a) 43.20V

b) 2.71W/s

c) 40.25s

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Explanation:

(a) The emf of a rotating coil with N turns is given by:

$emf=NBA\omega sin(\omega t)$

N: turns

B: magnitude of the magnetic field

A: area

w: angular velocity

the emf max is given by:

$emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V$

(b) the maximum rate of change of the magnetic flux is given by:

$\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}$

(c) $emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V$

(d) The torque is given by:

$\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm$

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3 years ago

A glass slab of density 2500kg/m3 measures 20cmx10cmx50cm what is maximum pressure

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2 years ago

Two objects A and B (both masses are m) are connected by a spring of elastic constant k and placed on a frictionless surface. At

umka2103 [35]

Answer:

a) Em_f = ½ m vₐ², b) v = vₐ /√2

Explanation:

a) For this part we use conservation of energy

starting point .. block A moving, spring unstretched

Em₀ = K = ½ m vₐ²

end point. Stretched spring

Em_f = K_e = ½ k x²

energy is conserved

Em₀ = Em_f

Em_f = ½ m vₐ²

b) Let's analyze the movement a little, block A began to move at a speed va, I stretch the spring an amount Dx, it exerts a force on block b that begins to move and the elongation of the spring decreases.

In all this process there is no friction force, therefore the energy is conserved, therefore the maximum energy stored in the spring must be distributed among the bodies.

Em₀ = K_e = E₀

where E₀ is the initial energy of block a

E₀ = ½ m vₐ²

At the point where we are in equilibrium

Em_f = Kₐ + K_b = ½ m vₐ² +1/2 m $v_{b}^2$

so that the spring does not stretch or shrink, the two bodies must go at the same speed.

Em_f = m v²

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therefore both blocks must go at this speed

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3 years ago

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