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kiruha [24]
3 years ago
10

You will get the most accurate resting heart rate if you take your pulse for ___ consecutive mornings and average the number

Physics
1 answer:
Setler79 [48]3 years ago
3 0
The answer will be 3
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Three forces are applied to an object, as shown in the diagram below. F1 = 30 N, F2 = 34 N, F3 = 20 N, d1 = 25 mm, d2 = 20 mm, d
lianna [129]
Torque of F1 = 30*0.025 = 0.75Nm anticlockwise
Torque of F2 = 34 * 0.020 = 0.68Nm clockwise
Horizontal component of F3:
cos 50 = x/20
x = 12.856 N

Torque of F3 horizontal component = 12.856 * 0.01 = 0.129Nm anticlockwise

0.129 + 0.75 - 0.68 = 0.199Nm anticlockwise

sorry I'm not sure is it correct
7 0
3 years ago
Cations have fewer _____ than _____.
FrozenT [24]
Cations are positively charged ions. And for positive charged ions, it means the positive charges, protons, are more than the negative charges, the electrons.

Therefore Cations have fewer electrons than protons.

So the answer is:    c. electrons; protons. 
3 0
3 years ago
Read 2 more answers
1.Convert 340 cm into m *(answer=0.34m)
Nataly [62]

Answer:

<em>1</em><em>.</em><em>for </em><em>the </em><em>first </em><em>one </em><em>100c</em><em>e</em><em>n</em><em>t</em><em>i</em><em>m</em><em>e</em><em>t</em><em>e</em><em>r</em><em>s</em><em> </em><em>make </em><em>1</em><em> </em><em>meter </em><em>therefore</em>

<em>100c</em><em>m</em><em>-</em><em>1</em><em>m</em>

<em>3</em><em>4</em><em>0</em><em>c</em><em>m</em><em>-</em><em>x</em>

<em>3</em><em>4</em><em>0</em><em>/</em><em>100</em>

<em>=</em><em>3</em><em>.</em><em>4</em>

<em>the </em><em>answer </em><em>is </em><em>supposed</em><em> to</em><em> be</em><em> </em><em>3</em><em>.</em><em>4</em><em>,</em><em> maybe</em><em> </em><em>there's</em><em> </em><em>a </em><em>mistake</em><em> </em><em>with </em><em>the </em><em>question</em><em> </em><em>or </em><em>the </em><em>answer</em>

<em>2</em><em>.</em><em>t</em><em>h</em><em>e</em><em> </em><em>weight</em><em> </em><em>of </em><em>a </em><em>body </em><em>is </em><em>given </em><em>by </em><em>the </em><em>formula</em>

<em>mass×</em><em>g</em><em>r</em><em>a</em><em>v</em><em>i</em><em>t</em><em>y</em><em>,</em><em>in </em><em>this </em><em>case </em><em>the </em><em>mass </em><em>is </em><em>7</em><em>5</em><em>k</em><em>g</em><em> </em><em>and </em><em>the </em><em>gravity </em><em>is </em><em>9</em><em>.</em><em>8</em>

<em>weight</em><em>=</em><em>7</em><em>5</em><em>×</em><em>9</em><em>.</em><em>8</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>7</em><em>3</em><em>5</em><em>N</em>

<em>3</em><em>.</em><em>f</em><em>o</em><em>r</em><em> </em><em>this </em><em>one </em><em>the </em><em>mass </em><em>of </em><em>a </em><em>body </em><em>is </em><em>given</em><em> by</em><em> the</em><em> formula</em>

<em>mass=</em><em>weight/</em><em>gravity</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>4</em><em>2</em><em>0</em><em>/</em><em>9</em><em>.</em><em>8</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em>4</em><em>2</em><em>.</em><em>8</em><em>k</em><em>g</em>

<em>I </em><em>hope</em><em> this</em><em> helps</em>

4 0
3 years ago
Write an expression for the magnitude of charge moved, Q, in terms of N and the fundamental charge e
NeTakaya

We have that for the Question "Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be said its equation is

Q=\frac{E}{Nr^2}

       

From the question we are told

Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e

<h3>An Expression for the <em>magnitude </em>of charge moved</h3>

Generally the equation for the  <em>magnitude </em>of charge moved, Q   is mathematically given as

Q=\frac{E}{Nr^2}

Therefore

An expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be

 Q=\frac{E}{Nr^2}

 

For more information on this visit

brainly.com/question/16517842

3 0
2 years ago
A 3250 N car is pushed a distance of 35 m the power was 11375 J, how long did it take?
irakobra [83]

Answer:

10.8s

Explanation:

Given parameters:

Force on the car  = 3250N

Distance  = 35m

Power  = 11375W

Unknown:

Time taken = ?

Solution:

To solve this problem;

 Power is the rate at which work is done

         Power = \frac{work done }{time}  

  Work done  = force x distance  = 3250 x 35  = 123200J

Now;

          11375  = \frac{123200}{t}  

           11375t  = 123200  

                   t  = 10.8s

5 0
3 years ago
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