Answer:
a = - 1.987 × 10⁶ ft/s²
t = 6.84 × 10⁻⁴ s
Explanation:
v₀ = 910 ft/s
x = 5 in.
relation v = v₀ - k x
v = 0 as body comes to rest
0 = 900 - 5k/12
k = 2184 s⁻¹
acceleration
where
(A) a = -k × v
at v= 910 ft/s
a = - 1.987 × 10⁶ ft/s²
(B) at x = 3.9 in.
v = 910 - 3.9(2184)/12
v = 200.2 m/s
t = 6.84 × 10⁻⁴ s
Answer:
Explanation:
Given an LC circuit
Frequency of oscillation
f = 299 kHz = 299,000 Hz
AT t = 0 , the plate A has maximum positive charge
A. At t > 0, the plate again positive charge, the required time is
t =
t = 1 / f
t = 1 / 299,000
t = 0.00000334448 seconds
t = 3.34 × 10^-6 seconds
t = 3.34 μs
it will be maximum after integral cycle t' = 3.34•n μs
Where n = 1,2,3,4....
B. After every odd multiples of n, other plate will be maximum positive charge, at time equals
t" = ½(2n—1)•t
t'' = ½(2n—1) 3.34 μs
t" = (2n —1) 1.67 μs
where n = 1,2,3...
C. After every half of t,inductor have maximum magnetic field at time
t'' = ½ × t'
t''' = ½(2n—1) 1.67μs
t"' = (2n —1) 0.836 μs
where n = 1,2,3...
Answer:
"the" (and any subsequent words) was ignored because we limit queries to 32 words
Answer:
force required to push the block = 219.714 lb
Explanation:
GIVEN DATA:
weight W of block = 250 lb
coefficient of friction = 0.2
consider equilibrium condition in x direction
.........................(1)
consider equilibrium condition in Y direction
.....................(2)
SOLVING 1 and 2 equation we get N value
N = 326.36 lb
putting N value in either equation we get force required to push the block = 219.714 lb
To solve the problem, we can use the equivalent of Newton's second law for rotational motions:
where
is the net torque acting on the object
I is the moment of inertia of the body
is the angular acceleration of the object.
Using the data of the problem:
and
, we find the net torque acting on the object:
