Amass transfer experiment is run to determine the mass transfer coefficient of water into air. air is flowed across a pan of water, which is on a balance. as water evaporates, mass is transferred into the air and the balance measures a loss in water mass. let a = water (�h = 18 �⁄���) and b = air (�p = 29 �⁄���). air at 50 cm/s and 30ºc, with 10% relative humidity, flows in a chamber pressurized to 1.1 bar. the air flows across a pan of length l = 30 cm and width w = 10 cm. after 5 minutes, the balance reads that approximately 1 gram of water has been lost. relevant information: �h _`a(30º�) = 4.247 ���. determine the mass transfer coefficient, in cm/s

when cloud cover increases then the suns ray's which provides the earth with light and warmth gets covered due to the due to the suns ray's being blocked, the temperature in that region

therefore to conclude i would like to say that,

=more the cloud cover, the lesser the

happy to

pls mark as brainliest.

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases . this process does not change the mass number.

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as -decay. in this the mass number remains same.

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.