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KIM [24]
4 years ago
5

The burning rates of two different solid-fuel propellants used in aircrew escape systems are being studied. It is known that bot

h propellants have the same standard deviation of burning rate; σ1 = σ2 = 3 centimetres per second. Two random samples of n1 = n2 = 20 are tested; the sample mean burning rates are 23.7 and 24 centimetres per second respectively. If you want to test the hypothesis that the burning rates of the 2 propellants are different, what is the value of zcalc? In computing zcalc, use the hypothesis test with respect to LaTeX: \mu1-\mu2 μ 1 − μ 2 . Please report your answer in 2 decimal places.
Physics
1 answer:
Katena32 [7]4 years ago
5 0

Answer: z_{\text{calc}}=-0.32

Explanation:

The test statistic for difference of two population means :-

z=\dfrac{\mu_1-\mu_2}{\sigma\sqrt{\dfrac{1}{n_1}+\dfrac{1}{n_2}}}

Given : \sigma=\sigma_1=\sigma_2=3

n_1 = n_2 = 20

\mu_1=23.7

\mu_2=24

Then , z=\dfrac{23.7-24}{(3)\sqrt{\dfrac{1}{20}+\dfrac{1}{20}}}

\Rightarrow\ z=-0.316227766017\approx-0.32

Hence, the value of z_{\text{calc}}=-0.32

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Black holes are the final stage of what type of star?
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A car is a compound machine true or false
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A cue ball, moving with 9.0 N·s of momentum strikes the nine-ball at rest. The nine-ball moves off with 2.0 N·s in the original
lisov135 [29]

Answer:

P = 7.28 N.s

Explanation:

given,

initial momentum of cue ball in x- direction,P₁ = 9 N.s

momentum of nine ball in  x-  direction, P₂ = 2 N.s

momentum in perpendicular direction i.e. y - direction,P'₂ = 2 N.s

momentum of the cue after collision = ?

using conservation of momentum

in x- direction

P₁ + p = x  + P₂

p is the initial momentum of the nine balls which is equal to zero.

9 + 0  = x  + 2

x = 7 N.s

momentum in x-direction.

equating along y-direction

P'₁ + p = y + P'₂

0 + 0 = y + 2

y = -2 N.s

the momentum of the cue ball after collision is equal to resultant of the momentum .

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7 0
3 years ago
Objects with masses of 255 kg and a 555 kg are separated by 0.390 m.
eimsori [14]

Answer:

(a). The net gravitational force is 4.20\times10^{-6}\ N

(b). The position is at 0.232 m.

Explanation:

Given that,

Mass of one object M = 255 kg

Mass of another object M'= 555 kg

Separation = 0.390 m

(a). We need to calculate the net gravitational force

Using formula of force

F_{net}=\dfrac{GM'm}{r^2}-\dfrac{GMm}{r^2}

F_{net}=\dfrac{Gm(M'-M)}{r^2}

Put the value into the formula

F_{net}=\dfrac{6.67\times10^{-11}\times32.0(555-255)}{(0.390)^2}

F_{net}=4.20\times10^{-6}\ N

The net gravitational force is 4.20\times10^{-6}\ N

(b). We need to calculate the position

Force from 555 mass = Force from 255 mass

\dfrac{Gm\times555}{x^2}=\dfrac{Gm\times255}{(0.390-x)^2}

555(0.390-x)^2=255x^2

300x^2-0.78x+84.4155=0

x=0.232\ m

The position is at 0.232 m.

Hence, (a). The net gravitational force is 4.20\times10^{-6}\ N

(b). The position is at 0.232 m.

5 0
3 years ago
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