To solve this problem we will use Newton's second law in order to obtain the weight of a person. The second law tells us that
F = ma
Where,
m = mass
a = Acceleration
In this particular case, the acceleration is equal to that exerted by the earth through gravitational acceleration, so if the person's weight is 75Kg and the gravity is 
, the weight of the body will be,
When the elevator is at rest this reads 735.75N and 75Kg (The same mass of the person)
The weightiness of the added
water displaced is equivalent to the joined weight of the two extra people who come
to be into the boat:
<span>m water g = 2 x 690 N</span>
<span> =
1,380 N</span>
<span>
</span>
The mass of the water displace
is then
<span>m water g = 1,380 N</span>
<span> = 1,380 N / 9.8 m/s^2</span>
<span> = 141 kg</span>
<span>
</span>
Compute the calculation for
density for the volume of water displace and practice this outcome for the mass
of the water displace to get the answer:
<span>p water = mass of water / volume of water</span>
<span>
</span>
<span>volume of water = mass of water / p water</span>
<span> = 141 kg / 1000 kg /m^3 eliminate
kilogram</span>
<span> = 0.14 m^3 the additional volume
of water that is displaced</span>
Answer:
Explanation:
The sensor contains an LDR which has a resistance of 10kohlms in daylight and 100kohlms in the dark.
If the resistor in the circuit is 1 megaohlm, the total resistance in daylight and darkness will be 1.01 megaohms and 1.1 megaohlms.
The percentage difference = (1.1-1.01)/1.1*100% = 8.18%
If the resistor in the circuit is 25 kohlm, the total resistance in daylight and darkness will be 35 kohms and 125 kohlms.
The percentage difference = (125-35)/125*100% = 72%
With the input p.d to the sensing circuit fixed at 12 v, the sensing current will change according to the total resistance. A 72% difference is much more detectable. So the 25 kohm resistor is the better choice.
Answer:
Explanation:
A ) The spheres are non conducting , charge will not move on the surface so neutralization of charge by + ve and - ve charge is not possible. Charges will remain intact on them . The electric field inside them will be zero . Electric field outside shell will not be spherically symmetrical . Lines of force will emanate from the surface of positively charged shell outwardly oriented and end at negatively charged shell .
B )
distance between the centres of spherical shell
= 2 a
potential energy of charges
= k q₁ x q₂ / R
= k x - Q x Q / ( 2a )
= - k Q²/ 2a
So work needed to separate them to infinity will be equal to
= k Q²/ 2a
Moving a distal part of arms or legs to follow a circular path is called circumduction. This medical term is sometimes described as circular movement because the movements of hands and arms are tracing circles in the air or space. Circumduction is also described as conical movement.<span> </span>
