Given that force is applied at an angle of 30 degree below the horizontal
So let say force applied if F
now its two components are given as


Now the normal force on the block is given as



now the friction force on the cart is given as



now if cart moves with constant speed then net force on cart must be zero
so now we have




so the force must be 199.2 N
Answer:
Mass of the climber = 69.38 kg
Explanation:
Change in length

Load, P = m x 9.81 = 9.81m
Young's modulus, Y = 0.37 x 10¹⁰ N/m²
Area

Length, L = 15 m
ΔL = 5.1 cm = 0.051 m
Substituting
Mass of the climber = 69.38 kg
Answer:

Explanation:
give data:
inside diameter = 5.0 cm
charge q = 0.25 nC
Outside diameter = 15 cm
potential V at inside sphere is = 
potential V at outside sphere is = 
k is constant whose value is = 
then potential difference between two point is
![\Delta V = kq \left [\frac{1}{R}-\frac{1}{r} \right ]](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20kq%20%5Cleft%20%5B%5Cfrac%7B1%7D%7BR%7D-%5Cfrac%7B1%7D%7Br%7D%20%20%5Cright%20%5D)
![\Delta V = 9*10^{9}*0.25*10^{-9} \left [\frac{1}{0.05}-\frac{1}{0.15} \right ]](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%209%2A10%5E%7B9%7D%2A0.25%2A10%5E%7B-9%7D%20%5Cleft%20%5B%5Cfrac%7B1%7D%7B0.05%7D-%5Cfrac%7B1%7D%7B0.15%7D%20%5Cright%20%5D)

Answer:
92.397amu
Explanation: The exact amu of the mystery element is obtained by multiplying the relative abundance of each individual isotope by its respective amu and then summing the results.
The sum of the total relative abundance for all the isotopes should be 100%.
However, the relative abundance of the isotope with 95.502amu is not given; therefore to obtain it we subtract the sum of the known relative abundances from 100% as follows:
Relative abundance of isotope with 95.502amu = 100-(23.63+30.53) = 42.84%