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Bogdan [553]
3 years ago
14

Use the drop-down menus to complete the statements. Outlook allows users to insert symbols and characters not located on the , b

ut they can have shortcut keys. Inserting horizontal lines in the message body breaks up different sections and mostly benefits people using Outlook on .
Physics
2 answers:
Anna11 [10]3 years ago
6 0

Answer:

Outlook allows users to insert symbols and characters not located on the  

keyboard

, but they can have shortcut keys.  

Inserting horizontal lines in the message body breaks up different sections and mostly benefits people using Outlook on  

mobile devices

.

Explanation:

Vera_Pavlovna [14]3 years ago
4 0

Answer:

Keyboard and Mobile devices

Explanation:

got it right on edge 2021

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A client experiences difficulty in performing the prone iso-abs exercise. Which of the following should be suggested as an regre
attashe74 [19]

Answer:

a. Quadruped arm and opposite leg raise

Explanation:

A client experiences difficulty in performing the prone iso-abs exercise, because the regression is Quadruped arm and opposite leg raise.

Iso-abs exercise is a beginner exercise that stabilizes the core muscle. Press the hips together. Slowly lift the pelvis from the floor til the back is straight. Carry  this position for the desired time or til proper form bcomes difficult to maintained.

6 0
3 years ago
A wave has a wavelength of 5mm and a frequency of 10 hertz what is its speed? answer: units:
gayaneshka [121]
Speed=wavelength×frequency
speed= 0.005m × 10 Hz = 0.05m/s
3 0
3 years ago
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5 0
3 years ago
Please help meeeeeeeeeeeeeeee
maxonik [38]

probabilityAnswer:

2/27

Explanation:

The elk can not be eaten so we remove that from the probablity

so we have x/18

songbird = 4/18

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5 0
2 years ago
A block with a mass of 31.8 kg is pushed on a frictionless
OlgaM077 [116]

Answer: Total work done on the block is 3670.5 Joules.

Step by step:

Work done:

W = F.d.\cos\theta

With F the force, d the displacement, and theta the angle of action (which is 0 since the block is pushed along the direction of displacement, and cos 0 = 1)

W = F.d

Given:

F = 75 N

m = 31.8 kg

Final velocity v_f = 15.3 \frac{m}{s}

In order to calculate the Work we need to determine the displacement, or distance the block travels. We can use the information about F and m to first figure out the acceleration:

F = ma\\\implies a=\frac{F}{m}=\frac{75 N }{31.8 kg}\approx 2.36 \frac{m}{s^2}\\

Now we can determine the displacement from the following formula:

d = \frac{1}{2}a^2+v_0t+d_0

Here, the initial displacement is 0 and initial velocity is also 0 (at rest):

d = \frac{1}{2}at^2\\

Now we still have "t" as unknown. But we are given one more bit of information from which this can be determined:

v_f = a\cdot t_f\\\implies t_f = \frac{v_f}{a} = \frac{15.2 \frac{m}{s}}{2.36 \frac{m}{s^2}}\approx 6.44 s

(using vf as final velocity, and tf as final time)

So it takes about 6.44 seconds for the block to move. This allows us to finally calculate the displacement:

d = \frac{1}{2}at^2=\frac{1}{2}2.36 \frac{m}{s^2}\cdot 6.44^2 s^2 \approx 48.94 m

and the corresponding work:

W = F\cdot d=75 N\cdot 48.94 m =3670.5J


7 0
3 years ago
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