Answer:
55N
Explanation:
Using Newton's second law of motion:
F=ma
Force=mass × acceleration
F=25×2.2
F=55N
So 55 Newtons are needed
Explanation:
Let
is the mass of proton. It is moving in a circular path perpendicular to a magnetic field of magnitude B.
The magnetic force is balanced by the centripetal force acting on the proton as :

r is the radius of path,

Time period is given by :


Frequency of proton is given by :

The wavelength of radiation is given by :


So, the wavelength of radiation produced by a proton is
. Hence, this is the required solution.
<span>Energy can be transformed from one type to another in any convection. Some of the energy is lost to the environment as
HEAT.</span>
Answer:
Explanation:
radius of aorta = 1.5 cm
cross sectional area = π r²
= 3.14 x 1.5²
= 7.065 cm²
volume of blood flowing out per second out of heart
= a x v , a is cross sectional area , v is velocity of flow
= 7.065 x 11.2
= 79.128 cm³
heart beat per second = 67 / 60
= 1.116666
If V be the volume of heart
1.116666 V = 79.128
V = 70.86 cm³.
The radius of a nucleus of hydrogen is approximately

, while we can use the Borh radius as the distance of an electron from the nucleus in a hydrogen atom:

The radius of a dime is approximately

: if we assume that the radius of the nucleus is exactly this value, then we can find how far is the electron by using the proportion

from which we find

So, if the nucleus had the size of a dime, we would find the electron approximately 500 meters away.