As we know f is inversely proportion to time so its frequency is 1/4 Hz
We will use the Cosine Law:
m D² = A² + B² - 2 AB cos D
∠D = 65° + 58° = 123°
m D² = 220² + 140² - 2 * 220 * 140 * cos 123°
m D² = 48,000 + 19,600 + 33,854.72
m D = 319.15 km
After that we will use the Sine Law:
340 / sin 123° = 220 / sin (theta)
sin (theta) = 0.578
∠(theta) = sin^(-1)0.578 = 35.3°
Answer:
The magnitude of the vector D is 319.15 km and points 35.3° south to east.
The diameter of the semicircular portion of the track is 16 m.
Therefore its radius is
r = 8 m.
The tangential velocity of the skater is 13 m/s.
Therefore the angular speed is
ω = v/r = (13 m/s)/(8 m) = 1.625 rad/s
The horizontal force on the skater is due to centripetal acceleration of
a = r*ω² = (8 m)*(1.625 rad/s)² = 21.125 m/s².
The force acting on the 64-kg skater is
F = m*a = (64 kg)*(21.125 m/s²) = 1352 N
Answer: 1352 N
Taking care of hygiene, washing hands with soap and water
Answer: 8.01×10^-17 J
Explanation: potential difference (v)= 500v, distance between plates (d) = 2cm = 0.02m
q = magnitude of an electronic charge = 1.602×10^-19c
E = strength of electric field = ?
Relationship between potential difference, disntahge between plates and electric field strength is given by the formule below
v =Ed
500 = E × 0.02
E = 500/ 0.02
E =25,000 v/m
But F=Eq
F = 25,000 × 1.602×10^-19
F = 4.005×10^-15 N
Using the work-energy theorem, the work done in moving the electron between the plates equals the kinetic energy of the electron.
Kinetic energy = work done = f × d
Kinetic energy = 4.005×10^-15 × 0.02
Kinetic energy = 8.01×10^-17 J
