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Colt1911
3 months ago
8

A set of notations (SSS, SAS, ASA and RHS) is used to describe/prove

Mathematics
1 answer:
Svet_ta [12.7K]3 months ago
5 0

Detailed breakdown:

locate the triangles

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Finally, suppose m1→∞, while m2 remains finite. what value does the the magnitude of the tension approach?
AnnZ [12381]
The tension does not approach infinity.
<span>Let's analyze free body diagrams (FBDs) for each mass, considering the direction of motion of m₁ as positive.

For m₁: m₁*g - T = m₁*a

For m₂: T - m₂*g = m₂*a

Assuming a massless cord and pulley without friction, the accelerations are the same.

From the second equation: a = (T - m₂*g) / m₂

Substitute into the first:
m₁*g - T = m₁ * [(T - m₂*g) / m₂]
Rearranging:
m₁*g - T = (m₁*T)/m₂ - m₁*g
2*m₁*g = T * (1 + m₁/m₂)
2*m₁*m₂*g = T * (m₂ + m₁)
T = (2*m₁*m₂*g) / (m₂ + m₁)
Taking the limit as m₁ approaches infinity:
T = 2*m₂*g

This aligns with intuition since the greatest acceleration m₁ can have is -g. The cord then accelerates m₂ upward at g while gravity acts downward, leading to a maximum upward acceleration of 2*g for m₁.</span>
5 0
3 months ago
Of 50 people, 38 have brown hair, 29 have brown eyes, and 23 have both brown hair and brown eyes. how many have neither brown ha
babunello [11817]
Is the process like this: 38+29=67, then 67-23=44
50-44=6
thus possibly A?
8 0
1 month ago
Read 2 more answers
Problem 8-4 A computer time-sharing system receives teleport inquiries at an average rate of .1 per millisecond. Find the probab
Svet_ta [12734]

Response:  a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

Detailed explanation:

In Problem 8-4, the computer time-sharing system experiences teleport inquiries at an average rate of 0.1 per millisecond. We are tasked with determining the probabilities of the inquiries over a specific period of 50 milliseconds:

Given that

\lambda=0.1\ per\ millisecond=5\ per\ 50\ millisecond=5

Applying the Poisson process, we find that

(a) at most 12

probability=  P(X\leq 12)=\sum _{k=0}^{12}\dfrac{e^{-5}(-5)^k}{k!}=0.9980

(b) exactly 13

probability= P(X=13)=\dfrac{e^{-5}(-5)^{13}}{13!}=0.0013

(c) more than 12

probability= P(X>12)=\sum _{k=13}^{50}\dfrac{e^{-5}.(-5)^k}{k!}=0.0020

(d) exactly 20

probability= P(X=20)=\dfrac{e^{-5}(-5)^{20}}{20!}=0.00000026

(e) within the range of 10 to 15, inclusive

probability=P(10\leq X\leq 15)=\sum _{k=10}^{15}\dfrac{e^{-5}(-5)^k}{k!}=0.0318

Thus, a) 0.9980, b) 0.0013, c) 0.0020, d) 0.00000026, e) 0.0318

6 0
2 months ago
Jamie is the accountant of a logistics company. In the month of December, $5,000 of wages remained unpaid. The accounting period
babunello [11817]
Wages expenses. 5000
To wages payable. 5000
3 0
2 months ago
Read 2 more answers
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