<span>If there isn't any force then the normal contact force will be
N=m*g=7.5*9.81=73.58N
which is 73.58-23=50.58N less
so, there the person must pull at 23 degree upward
break down the tension in two components, vertical and horizontal.
vertical tension= 50.58=T*sin23
T=50.58/sin23=129.45N</span>
= 27.777
Explanation:
A kilometer has 1,000 meters, and an hour has 3,600 seconds, so 100 kilometers per hour is: 100 x 1,000 / 3,600 = 27.777... m/s.
Answer:
physical health
Explanation:
Physical health has many components including: exercise, nutrition, sleep, alcohol & drugs, and weight management.
Answer:
41.74 m/s
Explanation:
The energy used to draw the bowstring = the kinetic energy of the arrow.
Fd = 1/2mv²................................ Equation 1
Where F = force, d = distance move string, m = mass of the arrow, v = speed of the arrow.
make v the subject of the equation
v = √(2Fd/m)...................... Equation 2
Given: F = 201 N, m = 0.3 kg, d = 1.3 m.
Substitute into equation 2
v = √(2×201×1.3/0.3)
v = √(1742)
v = 41.74 m/s.
Hence the arrow leave the bow with a speed of 41.74 m/s
To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,
Here,
= Angle between areal vector and magnetic field direction.
According to Faraday's law, induced emf in the loop is,
At time 
, Induced emf is,
Therefore the magnitude of the induced emf is 10.9V
