1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Leni [432]
3 years ago
13

A particle moving in the x-y plane has a position vector given by r = 1.89t2i + 1.17t3j, where r is in inches and t is in second

s. Calculate the radius of curvature ? of the path for the position of the particle when t = 2.1 sec.
Engineering
2 answers:
ehidna [41]3 years ago
6 0

Answer:

r=89.970 m

Explanation:

We need to calculate radius of curvature, using first and second derivatives of the functions x(t) and y(t).

From the given equation, we can get, that:

x(t)=1.89t^2 and y(t)=1.17t^3

Then, the first derivative:

x'(t)=2*1.89*t=3.78t and y'(t)=3*1.17*t^2=3.51t^2

The second derivatives are:

x''(t)=3.78 and y''(t)=3.51*2*t=7.02t

Equation for the radius of curvature, can be found as:

r=(x'^2+y'^2)^(3/2)/(x''y'-x'y'')

Note, that the denominator should be taken by its absolute value.

For the given time, we should calculate numerical values for the derivatives. For t=2.1 s:

x'(2.1)=7.938 m/s; x''(2.1)=3.78 m/s^2; y'(2.1)=15.4791 m/s and y''(2.1)=14.742 m/s^2

Using equation of the curvature radius and the values, we can get:

r=89.970 m

schepotkina [342]3 years ago
6 0

Answer:

ρ = 0.0657

Explanation:

We can apply the equation

ρ = 1/║r"(t)║

we have to get r"(t):

r'(t) = 3.78*t i + 3.51*t² j

r"(t) = 3.78 i + 7.02*t j

⇒   ║r"(t)║ = √(3.78²+(7.02*t)²)

If t = 2.1 s we have

║r"(2.1)║ = √(3.78²+(7.02*2.1)²) = 15.219

⇒   ρ = 1 / 15.219 = 0.0657 m

You might be interested in
If a vacuum gau ge reads 9.62 psi, it means that: a. the very highest column of mercury it could support would be 19.58 inches.
scZoUnD [109]

Answer:All of the above

Explanation:

9.62 psi means 497.49 mm of Hg pressure

for (a)19.58 inches is equals to 497.49 mm of Hg

(b)atmospheric pressure is 14.69 psi

vaccum gauge is 9.62psi

absolute pressure is=14.69-9.62=5.07

(c)vaccum means air is sucked and there is negative pressure so it tells about below atmospheric pressure.

thus all are correct

8 0
4 years ago
Fill in the blank to correctly complete the statement below.
frutty [35]

Answer:

The invention of the pendulum-driven ___<u>clocks</u>___ in the 1600s paved the way for a new industrial era.

4 0
4 years ago
Why does the ring on saturn spin
spayn [35]

Answer: THERE IS NO GRAVITY IN SPACE SO ROCKS SPIN

Explanation:

8 0
3 years ago
Read 2 more answers
2.
den301095 [7]
Jae pain seems the most off
4 0
3 years ago
If extension done in a stork position, during backward bending in a standing position of a special test for lumbar spine, produc
Klio2033 [76]
The answer is 7 because I just took the test!
5 0
2 years ago
Other questions:
  • The manufacturer of a 1.5 V D flashlight battery says that the battery will deliver 9 {\rm mA} for 37 continuous hours. During t
    10·1 answer
  • Given the unity feedback system
    5·1 answer
  • Which statement is true for the relay logic diagram shown below?
    9·1 answer
  • The emissivity of galvanized steel sheet, a common roofing material, is ε = 0.13 at temperatures around 300 K, while its absorpt
    7·1 answer
  • Fluid originally flows through a tube at a rate of 100 cm^3/s. To illustrate the sensitivity of the Poiseuille flow rate to vari
    7·1 answer
  • Nguyên lý hoạt động của kim phun
    7·1 answer
  • The toughness of a material does what, when it's been heated?​
    7·1 answer
  • 1. You are asked to write a report about one of the structures that Transportation Engineers
    9·1 answer
  • Air modeled as an ideal gas enters a combustion chamber at 20 lbf/in.2
    10·1 answer
  • which of the following tools is used for measuring small diameter holes which a telescoping gauge cannot fit into? A. telescopin
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!