The answer for <span>electromagnetic radiation released during radioactive decay i</span>s C. He
Answer:
the mark of the broken end is 2.6 cm so, we use the scale from the next full mark i.e. 3cm
Explanation:
<em>we </em><em>now </em><em>measure</em><em> </em><em>the </em><em>length</em><em> </em><em>of </em><em>the </em><em>pencil</em><em> </em><em>by </em><em>keeping </em><em>the </em><em>3</em><em> </em><em>c</em><em>m</em><em> </em><em>mark </em><em>of </em><em>the </em><em>scale</em><em> </em><em>at </em><em>it's</em><em> </em><em>left </em><em>end.</em>
<em>The </em><em>3</em><em> </em><em>cm </em><em>value </em><em>is </em><em>then </em><em>subtracted</em><em> </em><em>from </em><em>the </em><em>scale</em><em> </em><em>reading</em><em> </em><em>at </em><em>the </em><em>right</em><em> </em><em>side </em><em>end </em><em>of </em><em>the </em><em>pencil</em><em> </em><em>to </em><em>obtain </em><em>the </em><em>correct</em><em> </em><em>length</em><em> </em><em>of </em><em>the </em><em>pencil.</em><em> </em><em>✏️</em>
<em>(</em><em>i </em><em>i </em><em>)</em><em> </em>place the scale in the contact with object along it's length
(2) Your eyes must be exactly in front of the point where the measurements to be taken.
Hope_it_helps_mga_ka_joiners_mwehehe
Answer:
The sum of all forces for the two objects with force of friction F and tension T are:
(i) m₁a₁ = F
(ii) m₂a₂ = T - F
1) no sliding infers: a₁ = a₂= a
The two equations become:
m₂a = T - m₁a
Solving for a:
a = T / (m₁+m₂) = 2.1 m/s²
2) Using equation(i):
F = m₁a = 51.1 N
3) The maximum friction is given by:
F = μsm₁g
Using equation(i) to find a₁ = a₂ = a:
a₁ = μs*g
Using equation(ii)
T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N
4) The kinetic friction is given by: F = μkm₁g
Using equation (i) and the kinetic friction:
a₁ = μkg = 6.1 m/s²
5) Using equation(ii) and the kinetic friction:
m₂a₂ = T - μkm₁g
a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²
Answer:
1.25 m
Explanation:
From the question given above, the following data were obtained:
Force ratio = 2.5
Distance of load from the fulcrum = 0.5 m
Distance of effort =.?
The distance of the effort from the fulcrum can be obtained as illustrated below:
Force ratio = Distance of effort / Distance of load
2.5 = Distance of effort / 0.5
Cross multiply
Distance of effort = 2.5 × 0.5
Distance of effort = 1.25 m
Therefore, the distance of the effort from the fulcrum is 1.25 m
