Cell phones emit radiofrequency radiation ( radio waves ), a form of non-ionizing radiation, from their antennas
Brainiest if correct? I'd appreciate it SM :D
When object reached the terminal speed then its acceleration is zero
So as per Newton's II law we can say
now in that case we can say that net force is zero so here weight of the object is counter balanced by the drag force when it will reach at terminal speed
so we can write
so here we are given that
so terminal speed will be nearly 2 m/s
Explanation:
F = 20N m= m1 a=10m/s²
m=m2 a=5m/s²
F = ma
<u>for the first one</u><u>:</u><u> </u>
f=m1 × a
20 = m1 ×10
20=10m1
m1=20/10
m1=2
<u>for</u><u> </u><u>the</u><u> </u><u>second</u><u> </u><u>one</u><u> </u><u>:</u>
f=m2×a
20=m2×5
m2= 20/5
m2= 4
since F=ma
F=(m1+m2) ×a
F =(4+2)×a
F =6×a
F=20(from the question above )
20=6×a
a=20/6
a=3.33
Given
Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied
force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient
of kinetic friction µk = 0.3. If a force P
is applied to the body, no relative motion will take place until the applied
force is equal to the force of friction Ff, which is acting opposite to the
direction of motion. Magnitude of static force of friction between block A and
block B, Fs = µsN, where N is
reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA +
mB)a,
6 = (20 / 32.2 + 50 / 32.2)a
2.173a = 6
A = 2.76 ft/s^2
To check slipping occurs between block A and block B, consider block A:
P – Ff = mAaA
6 – Ff = 1.71
Ff = 4.29 lb
And also,
N = wA. We know static friction,
Fs = µsN
Fs = 0.4 x 20
Fs = 8lb
Frictional force is less than static friction. Ff < Fs
<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of
block B aB = 2.76 ft/s^2</span>
