Two 0.006 Kg bullets are fired with speeds of 20.0 m/s and 50.0 m/s respectively. What are their kinetic energies? Which bullet
1 answer:
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The frequency of middle C on a string is
f = 261.6 Hz.
The given linear density is
ρ = 0.02 g/cm = (0.02 x 10⁻³ kg)/(10⁻² m)
= 0.002 kg/m
The length of the string is L = 1 m.
Let T = the tension in the string (N).
The velocity of the standing wave is
In the fundamental mode, the wavelength, λ, is equal to the length, L.
That is
Because v = fλ, therefore
From given information, obtain
T = (0.002 kg/m)*(261.6 1/s)²*(1 m)²
= 136.87 N
Answer: 136.9 N (nearest tenth)
f = 261.6 Hz.
The given linear density is
ρ = 0.02 g/cm = (0.02 x 10⁻³ kg)/(10⁻² m)
= 0.002 kg/m
The length of the string is L = 1 m.
Let T = the tension in the string (N).
The velocity of the standing wave is
In the fundamental mode, the wavelength, λ, is equal to the length, L.
That is
Because v = fλ, therefore
From given information, obtain
T = (0.002 kg/m)*(261.6 1/s)²*(1 m)²
= 136.87 N
Answer: 136.9 N (nearest tenth)
Answer:
a) R₁ = 14.1 Ω, b) R₂ = 19.9 Ω
Explanation:
For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances
all resistors connected
V = i (R₁ + R₂)
with R₁ connected
V = (i + 0.5) R₁
with R₂ connected
V = (i + 0.25) R₂
We have a system of three equations with three unknowns for which we can solve it
We substitute the last two equations in the first
V = i ( )
1 = i ( )
1 = i ( ) =
i² + 0.75 i + 0.125 = 2i² + 0.75 i
i² - 0.125 = 0
i = √0.125
i = 0.35355 A
with the second equation we look for R1
R₁ =
R₁ = 12 /( 0.35355 +0.5)
R₁ = 14.1 Ω
with the third equation we look for R2
R₂ =
R₂ =
R₂ = 19.9 Ω
Answer:
13 sec
Explanation:
Hello
by definition the acceleration is a vector derived magnitude that indicates the speed variation per unit of time
Let
Vi=4.0 m/s
a=1.0 m/s2
Vf=17.0 m/s
ti= (0)sec
t2= unknown
Answer: 13 sec
I hope it helps