Answer:
310.5 J
Explanation:
The total work done by Martha is equal to the increase in gravitational potential energy of the package, which is equal to

where
(mg) = 45 N is the weight of the package
is the increase in height of the package
The package is carried up 3 flights of stairs, each one with a height of 2.3 m, so the total increase in heigth is

And so, the work done by Martha is

The correct option is this: SCIENTISTS HAVING DIFFERENT INTERESTS ARRIVE AT DIFFERENT CONCLUSIONS.
There are many fields in science and the scientists working in these fields have varying interests. The interests that a scientist has in a certain research will determines his views and conclusions about such a research.<span />
Answer:
The Resultant Induced Emf in coil is 4∈.
Explanation:
Given that,
A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.
To find :-
find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).
So,
Emf induced in the coil represented by formula
∈ =
...................(1)
Where:
.
{ B is magnetic field }
{A is cross-sectional area}
.
No. of turns in coil.
.
Rate change of induced Emf.
Here,
Considering the case :-
&
Putting these value in the equation (1) and finding the new emf induced (∈1)
∈1 =
∈1 =
∈1 =![4 [-N\times\frac{d\phi}{dt}]](https://tex.z-dn.net/?f=4%20%5B-N%5Ctimes%5Cfrac%7Bd%5Cphi%7D%7Bdt%7D%5D)
∈1 = 4∈ ...............{from Equation (1)}
Hence,
The Resultant Induced Emf in coil is 4∈.
Answer:
Work out = 28.27 kJ/kg
Explanation:
For R-134a, from the saturated tables at 800 kPa, we get
= 171.82 kJ/kg
Therefore, at saturation pressure 140 kPa, saturation temperature is
= -18.77°C = 254.23 K
At saturation pressure 800 kPa, the saturation temperature is
= 31.31°C = 304.31 K
Now heat rejected will be same as enthalpy during vaporization since heat is rejected from saturated vapour state to saturated liquid state.
Thus,
=
= 171.82 kJ/kg
We know COP of heat pump
COP = 
= 
= 6.076
Therefore, Work out put, W = 
= 171.82 / 6.076
= 28.27 kJ/kg
Given data
Power (P) = 50 hp,
= 50 × 746, we know that 1 hp = 746 W.
= 37300 Watts (Watt = J/s)
Work = 6.40 ×10⁴ J
Power is defined as rate of doing work and the unit of power is<em> Watt.</em>
Mathematically,
Power = (Work / time) Watts
= 6.40 ×10⁴ / time
37300 W = 6.40 ×10⁴ J /time (Where time in seconds)
=> time = Work/Power
= 6.40 ×10⁴/37300
= <em>1.74 seconds </em>
<em> </em><em>Therefore , the engine need 1.74 seconds to do 6.40 6.40 ×10⁴ J of work </em>
<em> </em>