Answer:
3.4 mT
Explanation:
L = 0.53 m
i = 7.5 A
Theta = 19 degree
F = 4.4 × 10^-3 N
Let B be the strength of magnetic field.
Force on a current carrying conductor placed in a magnetic field.
F = i × L × B × Sin theta
4.4 × 10^-3 = 7.5 × 0.53 × B × Sin 19
B = 3.4 × 10^-3 Tesla
B = 3.4 mT
Before answering this question, first we have to understand the effect of ratio of surface area to volume on the rate of diffusion.
The rate of diffusion for a body having larger surface area as compared to the ratio of surface area to volume will be more than a body having less surface area. Mathematically it can written as-
V∝ R [ where v is the rate of diffusion and r is the ratio of surface area to volume]
As per the question,the ratio of surface area to volume for a sphere is given 
The surface area to volume ratio for right circular cylinder is given 
Hence, it is obvious that the ratio is more for right circular cylinder.As the rate diffusion is directly proportional to the surface area to volume ratio,hence rate of diffusion will be more for right circular cylinder.
Hence the correct option is B. The rate of diffusion would be faster for the right cylinder.
Answer:
Earth's crust, called the lithosphere, consists of 15 to 20 moving tectonic plates. The heat from radioactive processes within the planet's interior causes the plates to move, sometimes toward and sometimes away from each other. This movement is called plate motion, or tectonic shift.
beainliest?
Answer:
51.96 m/s^-1
Explanation:
a) see the attachment
b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be
v_ox=v_o*cosФ
=60*cos (30)
= 51.96 m/s^-1
Explanation:
Sucrose is a disaccharide which is composed of fructose and glucose. Sucrose molecule has oxygen atoms bonded to hydrogen atoms (O-H bonds - Polar groups) on all ends of its double 6-Carbon ring. The areas near the oxygen atoms are slightly negative, and the areas near the hydrogen atoms are slightly positive that is, the O-H bonds are polar. They bond with the neighbouring Oxygen and Hydrogen atoms because of their
dipole - dipole attractions and hence hydrogen bonds are formed.
However, the covalent bonds within the molecule aren't broken. But rather, the hydrogen bonds holding the sucrose molecules in the crystalline lattice.