Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance
<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,
Solving Eq.1 for acceleration,
</span></span> v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span>
a = 1058 m/s</span>²
<span>Now applying Newton's 2nd law of motion,
</span>
<span>F = ma
= 0.145</span>×<span>1058
F = 153.4 N</span>
Answer:
a. Wavelength = λ = 20 cm
b. Next distance of maximum intensity will be 40 cm
Explanation:
a. The distance between the two speakers is 20cm. SInce the intensity is maximum which refers that we have constructive interference and the phase difference must be an even multiple of π and equivalent path difference is nλ.
Now when distance increases upto 30 cm between the speakers, the sound intensity becomes zero which means that there is destructive interference and equivalent path is now increased from nλ to nλ + λ/2.
This we get the equation:
(nλ + λ/2) - nλ = 30-20
λ/2 = 10
λ = 20 cm
b. at what distance, sound intensity will be maximum again.
For next point calculation for maximum sound intensity, the path difference must be increased (n+1) λ. The distance must increase by λ/2 from the point of zero intensity.
= 30 + λ/2
= 30 + 20/2
=30+10
=40 cm
Answer:

t'=1.1897 μs
Explanation:
First we will calculate the velocity of micrometeorite relative to spaceship.
Formula:

where:
v is the velocity of spaceship relative to certain frame of reference = -0.82c (Negative sign is due to antiparallel track).
u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)
u' is the relative velocity of micrometeorite with respect to spaceship.
In order to find u' , we can rewrite the above expression as:


u'=0.9806c
Time for micrometeorite to pass spaceship can be calculated as:

(c = 3*10^8 m/s)


t'=1.1897 μs
D. to be structural material
Answer:
Explanation:
This is a recoil problem, which is just another application of the Law of Momentum Conservation. The equation for us is:
which, in words, is
The momentum of the astronaut plus the momentum of the piece of equipment before the equipment is thrown has to be equal to the momentum of all that same stuff after the equipment is thrown. Filling in:
![[(90.0)(0)+(.50)(0)]_b=[(90.0)(v)+(.50)(-4.0)]_a](https://tex.z-dn.net/?f=%5B%2890.0%29%280%29%2B%28.50%29%280%29%5D_b%3D%5B%2890.0%29%28v%29%2B%28.50%29%28-4.0%29%5D_a)
Obviously, on the left side of the equation, nothing is moving so the whole left side equals 0. Doing the math on the right and paying specific attention to the sig fig's here (notice, I added a 0 after the 4 in the velocity value so our sig fig's are 2 instead of just 1. 1 is useless in most applications).
0 = 90.0v - 2.0 and
2.0 = 90.0v so
v = .022 m/s This is the rate at which he is moving TOWARDS the ship (negative was moving away from the ship, as indicated by the - in the problem). Now we can use the d = rt equation to find out how long this process will take him if he wants to reach his ship before he dies.
12 = .022t and
t = 550 seconds, which is the same thing as 9.2 minutes