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4 years ago

11

Who is the leader of the party's national committee

2 answers:

ad-work [718]4 years ago

6 0

national chairperson is the answer.

cestrela7 [59]4 years ago

3 0

That would be <span>the national chairperson

-I hope this helped.</span>

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Which of the following is an example of a SOLUTION?

Mila [183]

Answer:

Salt water

Explanation:

This is the answer

7 0

2 years ago

The length and width of a rectangular room are measured to be 3.92 ± 0.0035 m and 3.15 ± 0.0055 m. In this problem you can appro

Pavel [41]

Answer:

A) $A=12.2480\ m^2$

B) $12.2480\pm 0.1029\ m^2$

Explanation:

<u>Given:</u>

Length of the room $l= 3.92 ± 0.0035$

Width of the room $w= 3.15 ± 0.0055$

A) Let A be the area of the room

$A=l\times w\\A=3.92\times3.15\\A=12.2480\ \rm m^2$

B)We will calculate uncertainty in each dimension

%uncertainty in length $=\dfrac{0.0035}{3.92}\times 100=0.0892\ %$

%uncertainty in width = $\dfrac{0.0055}{3.15}\times 100=0.0174%$

The uncertainty in area will be sum of uncertainty in length and width

%uncertainty in Area= %uncertainty in length + %uncertainty in width

%uncertainty in Area $=0.0892\ % + 0.0174\ %$

%uncertainty in Area=0.0106

Uncertainty in Area $=0.0106\times 12.2480=0.1029\ \rm m^2$

There Area is $12.2480 ± 0.1029\ \rm m^2$

7 0

3 years ago

A small object with mass 1.30 kg is mounted on one end of arod

Julli [10]

Answer:

(a) $I_{system} = 1.014\ kg.m^{2}$

(b) $\tau = 0.0179\ N-m$

Solution:

As per the question:

Mass of the object, m = 1.30 kg

Length of the rod, L = 0.780 m

Angular speed, $\omega = 5010\ rev/min$

Now,

(a) To calculate the rotational inertia of the system about the axis of rotation:

Since, the rod is mass less, the moment of inertia of the rotating system and that of the object about the rotation axis will be equal:

$I_{system} = ML^{2} = 1.30\times (0.780)^{2} = 0.791\ kg.m^{2}$

(b) To calculate the applied torque required for the system to rotate at constant speed:

Drag Force, F = $2.30\times 10^{- 2}\ N$

$\tau = FLsin\theta 90 = 2.30\times 10^{- 2}\times 0.780\times 1 = 0.0179\ N-m$

3 0

3 years ago

I can’t figure this out!!! Answer what you can , please.

tigry1 [53]

a/b. The ball has velocity vector at time $t$

$\vec v=(v_x,v_y)=(v_0\cos63^\circ,v_0\sin63^\circ-gt)$

where $v_0=16\dfrac{\rm m}{\rm s}$ is the ball's initial speed and $g=9.8\dfrac{\rm m}{\mathrm s^2}$ .

c. At its highest point, the ball has 0 vertical speed. This occurs when

$v_0\sin63^\circ-gt=0\implies t=1.5\,\mathrm s$

d. Recall that

${v_y}^2-{v_{0y}}^2=-2g\Delta y$

so that at its highest point,

$0^2-(v_0\sin63^\circ)^2=-2g\Delta y\implies\Delta y=10\,\mathrm m$

e. This is just twice the time it takes for the ball to reach its maximum height, $t=2.9\,\mathrm s$ .

f. The ball's horizontal position after time $t$ is

$v_0\cos63^\circ\,t$

so that after the time found in part (f), the ball has traveled

$v_0\cos63^\circ(2.9\,\mathrm s)=11\,\mathrm m$

4 0

4 years ago

How would you set up SMART goals for your clients?

aleksklad [387]

Answer: The client must be advised while preparing a goal either for a treatment or for a fitness class.

Explanation:

The client goal must be set in a way that it should be measurable, attainable, time-bound.

The measurable goal is the in the sense that the client can measure own success.

The attainable is in the sense the client can achieve and perform all set plans in the goal.

Time-bound is in the sense that the client can devote sufficient time to obtain success.

6 0

3 years ago