Answer:
Explanation:
Ke = 1/2mv2 = 1/2.10.(4)2=5.16= 80 J
Answer:
when an object kept at some height and having some mass then ,when it falls down then shifting of energy takes place
Answer:
a) x = 40 t
, y = 39 t
, z = 6 + 32 t - 16 t
², b) x = 80 feet
, y = 78 feet
, the ball came into the field
Explanation:
a) This is a projectile launch exercise, where in the x and y axes there is no acceleration and in the z axis the acceleration of the acceleration of gravity, let's write the equations of motion in each axis
Since the cast is in the center of the field, let's place the coordinate system
x₀ = 0
y₀ = 0
z₀ = 6 feet
x-axis (towards end zone, GOAL zone)
x = xo + v₀ₓ t
x = 40 t
y-axis (field width)
y = y₀ +
t
y = 39 t
z axis (vertical)
z = z₀ + v_{oz} t - ½ g t²
z = 6 + 32 t - ½ 32 t²
z = 6 + 32 t - 16 t
²
b) The player catches the ball at the same height as it came out, so we can find the time it takes to arrive
z = 6
6 = 6 + 32 t - 16 t²
(t - 2)t = 0
t=0 s
t= 2 s
The ball position
x = 40 2
x = 80 feet
y = 39 2
y = 78 feet
the dimensions of the field from the coordinate system (center of the field) are
x_total = 150 feet
y _total = 80 feet
so we can see that the ball came into the field
Explanation:
After some time t the current does not passing through the circuit
=>so the back emf is zero
=>here the inductor opposes decay of the circuit
- Ldi/dt = Ri
di/dt = - R/Li
di/i = - R/Ldt
now we applying the integration on both sides
log i=-R/Lt+C
here t=0=>i=io
Log io=C
=>Log i=-R/L*t + Log io
logi-Log io=-R/L*t
Log[i/io]=-R/L*t
i/io=e^-Rt/L
i=ioe^-Rt/L
the option D is correct