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Gnom [1K]
4 years ago
12

P(5)=(0.11 x14)^5/5!e-^0.11-14

Engineering
1 answer:
seraphim [82]4 years ago
7 0

Answer:

The value of P(5)=0.09264

Explanation:

The equation given is;

P(5)=\frac{(0.11*14)^5}{5!e^-0.11-14} \\\\\\P(5)=(1.54)^5/120*2.7183^{-0.11} -14\\\\\\P(5)=8.662/120*0.8958-14\\\\\\P(5)=8.662/107.5-14\\\\P(5)=8.662/93.5\\\\P(5)=0.09264

The value of P(5)=0.09264

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The 5-kg collar has a velocity of 5 m>s to the right when it is at A. It then travels along the smooth guide. Determine its s
Gnoma [55]

Answer:

The speed at point B is 5.33 m/s

The normal force at point B is 694 N

Explanation:

The length of the spring when the collar is in point A is equal to:

lA=\sqrt{0.2^{2}+0.2^{2}  }=0.2\sqrt{2}m

The length in point B is:

lB=0.2+0.2=0.4 m

The equation of conservation of energy is:

(Tc+Ts+Vc+Vs)_{A}=(Tc+Ts+Vc+Vs)_{B} (eq. 1)

Where in point A: Tc = 1/2 mcVA^2, Ts=0, Vc=mcghA, Vs=1/2k(lA-lul)^2

in point B: Ts=0, Vc=0, Tc = 1/2 mcVB^2, Vs=1/2k(lB-lul)^2

Replacing in eq. 1:

\frac{1}{2}m_{c}v_{A}^{2}+0+m_{c}gh_{A}+      \frac{1}{2}k(l_{A}-l_{ul})  ^{2}=\frac{1}{2}m_{c}v_{B}^{2}+0+0+\frac{1}{2}k(l_{B}-l_{ul})  ^{2}

Replacing values and clearing vB:

vB = 5.33 m/s

The balance forces acting in point B is:

Fc-NB-Fs=0

\frac{m_{C}v_{B}^{2}   }{R}-N_{B}-k(l_{B}-l_{ul})=0

Replacing values and clearing NB:

NB = 694 N

6 0
4 years ago
Read 2 more answers
Fixed objects that fix a location are called
Zielflug [23.3K]
Fix3d locationz i thinkz
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3 years ago
Remy noticed that after oiling his skateboard wheels, it was easier to reach the speeds he needed to perform tricks. How did the
Sidana [21]

Answer:

The oil reduced friction between the moving parts of the skateboard. ( A )

Explanation:

The oil reduced Friction between the moving parts of the skateboard and this is because the reduction in friction between moving parts causes an increase in speed.

Remy will oil the moving parts that connects the wheels of the skateboard to the Board, because this is where the most friction is found. the friction between the wheels and the ground cannot be affected by oiling the wheels

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3 years ago
4.A compression spring has a diameter of 0.67 in and a coil diameter of 0.067 in, with 30 total coils. The spring is to be made
Akimi4 [234]

Answer:

Fatigue factor of safety is 2.0267

Explanation:

Solution is attached below.

4 0
4 years ago
A normal shock wave takes place during the flow of air at a Mach number of 1.8. The static pressure and temperature of the air u
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Answer:

The pressure upstream and downstream of a shock wave are related as

\frac{P_{1}}{P_{o}}=\frac{2\gamma M^{2}-(\gamma -1)}{\gamma +1}

where,

\gamma= Specific Heat ratio of air

M = Mach number upstream

We know that \gamma _{air}=1.4

Applying values we get

\frac{P_{1}}{100kPa}=\frac{2\times 1.4\times 1.8^{2}-(1.4 -1)}{1.4 +1}\\\\\frac{P_{1}}{100kPa}=3.61\\\\\therefore P_{1}=361.33kPa(Absloute)

Similarly the temperature downstream is obtained by the relation

\frac{T_{1}}{T_{o}}=\frac{[2\gamma M^{2}-(\gamma -1)][(\gamma -1)M^{2}+2]}{(\gamma +1)^{2}M^{2}}

Applying values we get

\frac{T_{1}}{423}=\frac{[2\times 1.4\times 1.8^{2}-(1.4-1)][(1.4-1)1.8^{2}+2]}{(1.4+1)^{2}\times 1.8^{2}}\\\\\therefore \frac{T_{1}}{423}=1.53\\\\\therefore T_{1}=647.85K=374.85^{o}C

The Mach number downstream is obtained by the relation

M_{d}^{2}=\frac{(\gamma -1)M^{2}+2}{2\gamma M^{2}-(\gamma -1)}\\\\\therefore M_{d}^{2}=\frac{(1.40-1)\times 1.8^{2}+2}{2\times1.4\times 1.8^{2}-(1.4-1)}\\\\\therefore M_{d}^{2}=0.38\\\\M_{d}=0.616

3 0
4 years ago
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