The equation that relates distance, velocities, acceleration, and time is,
d = V₀t + 0.5gt²
where d is distance,
V₀ is the initial velocity,
t is time, and
g is the acceleration due to gravity (equal to 9.8 m/s²)
(1) Dropped rock,
(3 x 10² m ) = 0(t) + 0.5(9.8 m/s²)(t²)
The value of t from this equation is 24.73 s
(2) Thrown rock with V₀ = 26 m/s
(3 x 10² m) = (26)(t) + 0.5(9.8 m/s²)(t²)
The value of t from the equation is 5.61 s
The difference between the tim,
difference = 24.73 s - 5.61 s
difference = 19.12 s
<em>ANSWER: 19.12 s</em>
Answer:
Explanation:
= Velocity of one lump = 
= Velocity of the other lump = 
m = Mass of each lump = 
The collision is perfectly inelastic as the lumps stick to each other so we have the relation
The velocity of the stuck-together lump just after the collision is .
Answer:
a) The current density ,J = 2.05×10^-5
b) The drift velocity Vd= 1.51×10^-15
Explanation:
The equation for the current density and drift velocity is given by:
J = i/A = (ne)×Vd
Where i= current
A = Are
Vd = drift velocity
e = charge ,q= 1.602 ×10^-19C
n = volume
Given: i = 5.8×10^-10A
Raduis,r = 3mm= 3.0×10^-3m
n = 8.49×10^28m^3
a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]
J = (5.8×10^-10) /(2.83×10^-5)
J = 2.05 ×10^-5
b) Drift velocity, Vd = J/ (ne)
Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)
Vd = (2.05×10^-5)/(1.36 ×10^10)
Vd = 1.51× 10^-5
According to the Work-Energy Theorem, the work done on an object is equal to the change in the kinetic energy of the object:
Since the car ends with a kinetic energy of 0J (because it stops), then the work needed to stop the car is equal to the initial kinetic energy of the car:
Replace m=1100kg and v=112km/h. Write the speed in m/s. Remember that 1m/s = 3.6km/h:
Therefore, the answer is: 532,346 J.
