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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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if a person can jump maximum along distance of 3m ,on the earth how far could be jump on the moon where acceleration due to grav

allochka39001 [22]

Answer:

The person can jump 48 m on the Moon

Explanation:

The question parameters are;

The maximum long jump distance of a person on Earth, $R_{max}$ = 3 m

The acceleration due to gravity on the Moon = 1 ÷ 16 of that on Earth

The distance the person can jump on the Moon is given as follows;

A person performing a jump across an horizontal distance on Earth (under gravitational force) follows the path of the motion of a projectile

The horizontal range, $R_{max}$ , of a projectile motion is found by using the following formula

$R_{max} = \dfrac{u^2}{g}$

Where;

g = The acceleration due to gravity = 9.8 m/s²

Therefore, we have;

$R_{max} = 3 \, m = \dfrac{u^2}{9.8 \, m/s^2 }$

u² = 3 m × 9.8 m/s² = 29.4 m²/s²

Therefore, on the Moon, we have;

The acceleration due to gravity on the Moon, $g_{Moon}$ = 1/16 × g

∴ $g_{Moon}$ = 1/16 × g = 1/16 × 9.8 m/s² ≈ 0.6125 m/s²

$R_{max \ Moon} = \dfrac{u^2}{g_{Moon}} = \dfrac{29.4 \ m^2/s^2}{0.6125 \, m/s^2 } \approx 48 \, m$

The maximum distance the person can jump on the Moon with the same velocity which was used on Earth is $R_{max \ Moon}$ ≈ 48 m

8 0

3 years ago

A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block

Usimov [2.4K]

Answer:

a) $v'=-0.227\ m.s^{-1}$

b) $v=1.36\ m.s^{-1}$

Explanation:

Given:

mass of the lighter block, $m'=0.02\kg$

velocity of the lighter block, $u'=0.68\ m.s^{-1}$

mass of the heavier block, $m=0.04\ kg$

velocity of the heavier block, $u=0\ m.s^{-1}$

Using conservation of linear momentum:

$m'.u'+m.u=m'.v'+m.v$

where:

$v'=$ final velocity of the lighter block

$v=$ final velocity of the heavier block

$m'.u'=m'.v'+m.v$

$m'(u'-v')=m.v$ ........................(1)

Since kinetic energy is conserved in elastic collision:

$\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2$

$m'(u'^2-v'^2)=m.v^2$

$m'(u'-v')(u'+v')= m.v^2$

divide the above equation by eq. (1)

$v=u'+v'$ .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

$m'(u'-v')=m(u'+v')$

$\frac{m'+m}{m'-m} =\frac{u'}{v'}$

$\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}$

$v'=-0.227\ m.s^{-1}$ (negative sign denotes that the direction is towards left)

now we substitute the value of v' from eq. (2) in eq. (1)

$m'(u'-v+u')=m.v$

$2m'.u'=(m-m')v$

$2\times 0.02\times 0.68=(0.04-0.02)\times v$

$v=1.36\ m.s^{-1}$

6 0

3 years ago

A jet accelerates from rest down a runway at 1.75m/s² for a distance of 1500 m before takeoff.

babunello [35]

A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.

Read more on Brainly.com - brainly.com/question/18743384#readmore

8 0

3 years ago

A tourist drops (from rest) a ping pong ball from the top of the tower, which has a height of 324 meters. Assuming no air resist

Elanso [62]

Answer:

8.13secs

Explanation:

From the question weal are given

Height H =324m

Required

time it takes to drop t

Using the equation of motion

H = ut + 1/2gt²

Substitute the given values

324 = 0(t)+1/2(9.8)t²

324 = 1/2(9.8)t²

324 = 4.9t²

t² =324/4.9

t² = 66.12

t = √66.12

t = 8.13secs

Hence the time taken to drop is 8.13secs

4 0

3 years ago

En un momento dado , la nadadora de una prueba de natación de 100 m espalda está debajo de la cuerda falsa de salida. Indica a)

Virty [35]

Answer:

I only speak English

Explanation:

I'm sorry can you type it in English

7 0

3 years ago