Answer:
2.04m/s²
Explanation:
Complete Question
<em>A stationary 10 kg object is located on a table near the surface of the earth. The coefficient of kinetic friction between the surfaces is 0.2. A horizontal force of 40 N is applied to the object. Find the acceleration of the object.</em>
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According to Newtons second law;
\sum F_x = ma_x
F_m - F_f = ma_x
Fm is the applied force
Ff is the frictional force
m is the mass
a is the acceleration
Substitute the given values
40N - nmg = 10a
40 - 0.2(10)(9.8) = 10a
40 - 19.6 = 10a
20.4 = 10a
a = 20.4/10
a = 2.04m/s²
<em>Hence the acceleration of the object is 2.04m/s²</em>
Answer: Can I get a picture???
There must be a equal opposite reaction.
E=mv2
The most possible cause of 35% increase in time measured is due to the acceleration due to gravity which is 70% less than 9.80 m/s² at this location.
Explanation:
We know that for a linear motion
S=ut+(1/2)*a*t²
Where S= Distance covered
u= initial velocity
t=time period
a= acceleration due to gravity
Since the body is falling vertically downwards "a" is to be replaced by "g"
Moreover, The body is falling from rest "u"=0
Hence the equation reduces to S=1/2*g*t²
t²=2S/g
Since the "g" is in denominator thus a decreased value of "g" would mean increased period. Thus an increase of 35% in the period would translate to a 70% decrease in "g".