let the length of the beam be "L"
from the diagram
AD = length of beam = L
AC = CD = AD/2 = L/2
BC = AC - AB = (L/2) - 1.10
BD = AD - AB = L - 1.10
m = mass of beam = 20 kg
m₁ = mass of child on left end = 30 kg
m₂ = mass of child on right end = 40 kg
using equilibrium of torque about B
(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)
30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)
L = 1.98 m
Answer:
100 meters, 54.5 East of North or 125.5 North of East.
Explanation:
Try drawing it out to get a better visual. Make sure that when you draw the arrows that you make a scale (for example: 1 cm = 10 meters). After drawing it out, draw a line from the origin/starting point and connect it to the end point from the "75 m west" arrow. Then, measure the line you drew and convert it back into meters. Lastly, measure the angle.
Answer:
a. be sure to hold expansion cards by the edge connectors
Explanation:
Removal of loose jewelry is a good safety practice. Also not touching a microchip with a magnetized screwdriver is also a good practice.
But holding expansion cards by the edge connectors is not a good practice, so it is the odd one in the question. Therefore answer option a provides the correct and best answer to the question
Answer:
The range is maximum when the angle of projection is 45 degree.
Explanation:
The formula for the horizontal range of the projectile is given by

The range should be maximum if the value of Sin2θ is maximum.
The maximum value of Sin2θ is 1.
It means 2θ = 90
θ = 45
Thus, the range is maximum when the angle of projection is 45 degree.
If the angle of projection is 0 degree
R = 0
It means the horizontal distance covered by the projectile is zero, it can move in vertical direction.
If the angle of projection is 30 degree.

R = 0.088u^2
If the angle of projection is 45 degree.

R = u^2 / g
Answer:
25.82 m/s
Explanation:
We are given;
Force exerted by baseball player; F = 100 N
Distance covered by ball; d = 0.5 m
Mass of ball; m = 0.15 kg
Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.
We should note that work done is a measure of the energy exerted by the baseball player.
Thus;
F × d = ½mv²
100 × 0.5 = ½ × 0.15 × v²
v² = (2 × 100 × 0.5)/0.15
v² = 666.67
v = √666.67
v = 25.82 m/s