Coulomb's Law
Given:
F = 3.0 x 10^-3 Newton
d = 6.0 x 10^2 meters
Q1 = 3.3x 10^-8 Coulombs
k = 9.0 x 10^9 Newton*m^2/Coulombs^2
Required:
Q2 =?
Formula:
F = k • Q1 • Q2 / d²
Solution:
So, to solve for Q2
Q2 = F • d²/ k • Q1
Q2 = (3.0 x 10^-3 Newton) • (6.0 x 10^2 m)² / (9.0 x 10^9
Newton*m²/Coulombs²) • (3.3x 10^-8 Coulombs)
Q2 = (3.0 x 10^-3 Newton) • (360 000 m²) / (297 Newton*m²/Coulombs)
Q2 = 1080 Newton*m²/ (297 Newton*m²/Coulombs)
Then, take the reciprocal of the denominator and start
multiplying
Q2 = 1080 • 1 Coulombs/297
Q2 = 1080 Coulombs / 297
Q2 = 3.63636363636 Coulombs
Q2 = 3.64 Coulumbs
M = 30 g = 0.03 kg, the mass of the bullet
v = 500 m/s, the velocity of the bullet
By definition, the KE (kinetic energy) of the bullet is
KE = (1/2)*m*v²
= 0.5*(0.03 kg)*(500 m/s)² = 3750 J
Because the bullet comes to rest, the change in mechanical energy is 3750 J.
The work done by the wall to stop the bullet in 12 cm is
W = (1/2)*(F N)*(0.12 m) = 0.06F J
If energy losses in the form of heat or sound waves are ignored, then
W = KE.
That is,
0.06F = 3750
F = 62500 N = 62.5 kN
Answer:
(a) 3750 J
(b) 62.5 kN
Answer: 55 ohms
Explanation:
Given that,
Voltage of heater (v) = 110-volt
Current drawn by heater (I) = 2.0 amperes
resistance of the heater (r) = ?
Since voltage, current and resistance are involved, apply the formula for ohms law.
Voltage = current x resistance
i.e v = ir
where r = v / i
r = 110 volts / 2.0 A
r = 55 ohms
Thus, the resistance of the heater is 55 ohms
Answer:
If she stands on the North side of a river flowing to the East at 5 mph,
she must head towards the SouthWest to arrive on the South side of the river directly across from her starting point and we have
x^2 + 5^2 = 10^2 where x is her speed directly across the river
x = (75)^1/2 = 8.66 mph towards the South
sin theta = 5 / 10 = 1/2
She must angle the boat at 30 deg from straight South
Answer:
The angular frequency of the block is ω = 5.64 rad/s
Explanation:
The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.
Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.
The angular frequency of the oscillation ω is
ω = v/r
ω = 62 cm/s ÷ 11 cm
ω = 5.64 rad/s
So, the angular frequency of the block is ω = 5.64 rad/s