Answer:
80
Explanation:
<em>the </em><em>mechanical</em><em> </em><em>advantage</em><em> </em><em>is </em><em>the </em><em>ratio </em><em>of </em><em>the </em><em>load </em><em>to </em><em>the </em><em>effort</em><em> </em><em>so </em><em>it </em><em>doesn't</em><em> </em><em>have </em><em>units.</em><em>t</em><em>o</em><em> </em><em>calculate</em><em> </em><em>it </em><em>you </em><em>use </em><em>the </em><em>formula</em>
<em>mechanical</em><em> advantage</em><em>=</em><em>load/</em><em>effort</em>
<em>in </em><em>this</em><em> case</em><em> </em><em>the </em><em>load </em><em>is </em><em>6</em><em>0</em><em>0</em><em>0</em><em>N</em><em> </em><em>and </em><em>the </em><em>effort</em><em> </em><em>is </em><em>7</em><em>5</em><em>N</em>
<em>Ma=</em><em>6</em><em>0</em><em>0</em><em>0</em><em>/</em><em>7</em><em>5</em>
<em>=</em><em>8</em><em>0</em>
<em>I </em><em>hope</em><em> this</em><em> helps</em>
Question 1:
Looking for: "speed" (velocity: v)
Given: 112.0 meters (distance: d) 4.0 seconds (time: t)
Formula used: v=d/t
since velocity is equal to the distance (112m) divided by the time (4sec) you would have 112m/4s which is 28m/s. m/s is meters traveled per second traveling aka meters per second.
Question 2:
Looking for: average "speed" don't let the word average confuse you it is still velocity.
Given: 60km (distance: d) and 3.5 hours (time: t)
Formula required: v=d/t
Since velocity needs to be in the units of m/s first convert your time, which was provided in hours, to seconds.
3.5 hours 60 minutes per hour: 3.5 x 60= 210 minutes
210 minutes 60 seconds per minute: 210 x 60= 12600 seconds
also, kilometers needs to be converted to meters. The prefix kilo means 1000 so 1 kilometer = 1000 meters.
60 x 1000 = 60,000 meters
Plug the values into the equation required:
v=d/t
v= 60,000m / 12,600sec
v= 4.76 m/s
The answer your looking for is, A. alpha particle.
Answer:
the sheets approach while the object is near
Explanation:
An electroscope is an apparatus that has two metal sheets attached, when these sheets have a charge and distribute evenly between them and the sheets repel.
When I approach an object charged with a counter (negative) charge, part of the charge of the electroscope moves near the charged external object, to neutralize the electric field, so as the charge on one of the plates decreases the electroscope has approached , as the objects are not touched the system remains in this configuration while the object is close. When the object is released, the electric field it creates disappears, so the positive charges repel inside the electroscope and the sheets repel to the initial position.
In short, the sheets approach while the object is near