Answer:
coefficient of friction =0.268
magnitude of force P=289.78N
Explanation:
The coefficient of friction is obtained by mgsinФ/mgcosФ=tanФ=tan15=0.268
force P is horizontal as stated in the question, horizontal component of P=mgcosФ=30*10*cos15=289.78
Answer:
Approximately
.
Explanation:
This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.
This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:
,
where
is the final velocity of the moving object,
is the initial velocity of the moving object,
is the (linear) acceleration of the moving object, and
is the (linear) displacement of the object while its velocity changed from
to
.
The angular analogue of that equation will be:
, where
and
are the initial and final angular velocity of the rotating object,
is the angular acceleration of the moving object, and
is the angular displacement of the object while its angular velocity changed from
to
.
For this object:
, whereas
.
The question is asking for an angular acceleration with the unit
. However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:
.
Rearrange the equation
and solve for
:
.
Answer:
Volume of 22 drop will be 0.68 ml
Explanation:
We have given volume of one drop = 0.031 ml
We know that 1 liter = 1000 ml
So 
So 0.031 ml will be equal to 
We have to find the volume of 22 drop
For finding volume of 22 drop we have to multiply volume of one drop by 22
So volume of 22 drop will be 
So volume of 22 drop will be 0.68 ml
- R=ρ×L/A
ρ=R×A/L
A=pie(r²)=pie(0.6²)=pie(0.36)=1.13
ρ=2ohm×1.13mm²/1500mm
ρ=0.0015ohm.mm
2. we were told that the wires were made of the same substance so the resistivity(ρ)is the same for both wires so:
R=ρ×L/A
R=0.0015ohm.mm×500mm/0.09mm²
R=8.3'ohm
so our resistance for the second wire is :
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>R</em><em>=</em><em>8</em><em>.</em><em>3</em><em>'</em><em>ohm</em>