I believe the answer is 40,000 Hz to 100,000 Hz
Answer: hello below is the missing part of your question
A mass m = 10 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5029 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.49. The mass leaves the spring at a speed v = 3.4 m/s.
answer
x = 0.0962 m
Explanation:
<em>First step :</em>
Determine the length of the rough patch/spot
F = Uₓ (mg)
and w = F.d = Uₓ (mg) * d
hence;
d( length of rough patch) = w / Uₓ (mg) = 46.55 / (0.49 * 10 * 9.8) = 0.9694 m
<em>next : </em>
work done on unstretched spring length
Given that block travels halfway i.e. d = 0.9694 / 2 = 0.4847 m
w' = Uₓ (mg) * d
= 0.49 * 10 * 9.81 * 0.4847 = 23.27 J
also given that the Elastic energy of spring = work done ( w')
1/2 * kx^2 = 23.27 J
x =
= 0.0962 m
The electric current is measured in coulombs per second.<u> A flow of one coulomb per second is called one ampere</u>
Answer:
B. Air
Explanation:
I'm taking the exam rn ^^
Answer:
image is vertical at distance -203.62 cm
magnification is 2.110
Explanation:
given data
n = 1.51
distance u = 96.5 cm
concave radius r1 = 24 cm
convex radius r2 = 19.1 cm
to find out
final image distance and magnification
solution
we will apply here lens formula to find focal length f
1/f = n-1 ( 1/r1 - 1/r2) .......................1
put here all value
1/f = 1.51 -1 ( -1/24 + 1/19.1)
f = 183.43
so from lens formula
1/f = 1/v + 1/u .............................2
put here all value and find v
1/183.43 = 1/v + 1/96.5
so
v = −203.62 cm
so here image is vertical at distance -203.62 cm
and
magnification are = -v /u
magnification = 203.62 / 96.5
magnification is 2.110