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Olegator [25]
3 years ago
10

The latitude of the city of Arlington is about 32.7357°. Calculate the following: (a) The angular speed of the Earth. (b) The li

near speed and acceleration of city of Arlington, (c) Find the ratio of the linear speed in b) to the linear speed of a point on the equator.
Physics
1 answer:
julia-pushkina [17]3 years ago
6 0

Answer:

Explanation:

a ) The earth rotates by 2π radian in 24 x 60 x 60 s

so angular speed ( w )  = 2π / (24 x 60 x 60)  = 7.268 x  10⁻⁵ rad / s

b ) Linear speed of city of Arlington ( v )  = w r = w R Cosλ where R is radius of the earth and λ is latitude .

v = 7.268 x 10⁻⁵ x  6.371 x 10⁶ cos 32.7357

389.5 m /s

acceleration = w² r = w² R Cos 32.7357

= (7.268 x 10⁻⁵ )² x 6.371 x 10⁶ x cos 32.7357

=283.08 x 10⁻⁴ m/s²

c) velocity ratio =

w r /w R =

R cos 32.73/ R  

= Cos 32.73

= 0.84 .

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A 0.5 kg object, initially at rest, is pulled to the right along a frictionless horizontal surface by a constant horizontal forc
sergij07 [2.7K]

Answer:

A.) 12.5 J

B.) 12.5 J

C.) 7.1 m/s

Explanation:

Given that a 0.5 kg object, initially at rest, is pulled to the right along a frictionless horizontal surface by a constant horizontal force of 25 N for a distance of 0.5m.

a. What is the work done by the force?

Work done = force × distance

Work done = 25 × 0.5

Work done = 12.5 J

b. What is the change in the kinetic energy of the block?

Work done = energy

Change in Kinetic energy = work done

Change in kinetic energy = 12.5 J

c. What is the speed of the block after the force is removed?

Kinetic energy = 1/2mV^2

12.5 = 1/2 × 0.5 × V^2

25 = 0.5V^2

V^2 = 25/0.5

V^2 = 50

V = 7.1 m/s

6 0
2 years ago
Clarisse had three substances. A white substance was waxy and malleable. A red crystal was translucent, and it cracked when she
melomori [17]

Answer:

The red substance was ionic. The white substance was molecular.

Explanation:

B.C.

6 0
2 years ago
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If two objects, like the eggs in the video, experience the same change in momentum but over time periods of
antoniya [11.8K]

Explanation:

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2 years ago
Determine the amount of time for polonium-210 to decay to one fourth its original quantity. The half-life of polonium-210 is 138
ira [324]

Answer: 276 days

Explanation:

This problem can be solved using the Radioactive Half Life Formula:  

A=A_{o}.2^{\frac{-t}{h}} (1)  

Where:  

A=\frac{1}{4}A_{o} is the final amount of the material

A_{o} is the initial amount of the material  

t is the time elapsed  

h=138 days is the half life of polonium-210

Knowing this, let's substitute the values and find t from (1):

\frac{1}{4}A_{o}=A_{o}2^{\frac{-t}{138 days}} (2)  

\frac{A_{o}}{4A_{o}}=2^{\frac{-t}{138 days}} (3)  

\frac{1}{4}=2^{\frac{-t}{138 days}} (4)  

Applying natural logarithm in both sides:

ln(\frac{1}{4})=ln(2^{\frac{-t}{138 days}}) (5)  

-1.386=-\frac{t}{138days}ln(2) (6)  

Clearing t:

t=276days (7)  

7 0
3 years ago
An undamped spring-mass system contains a mass that weighs and a spring with spring constant . It is suddenly set in motion at b
balandron [24]

Answer:

Explanation:

When all other forces acting on the mass in a damped mass-spring system are grouped together into one term denoted by F(t), the differential equation describing

motion is

Mx''+ βx' + kx = F(t).

Note for an undamped system

β=0,

Then, the differential equation becomes

Mx'' + kx = F(t).

The force is in the form

F=Fo•Sinωo•t

Let solved for the homogeneous or complementary solution, I.e f(t) = 0

Using D operator

MD² + k = 0

MD²=-k

D²=-k/M

Then, D= ±√(-k/m)

D=±√(k/m) •i

So we have a complex root

Therefore, the solution is

x= C1•Cos[√(k/m)t] + C2•Sin[√(k/m)]

This is simple harmonic motion that once again we prefer to write in the form

x(t) = A•Sin[ √(k/M)t + φ]

Where A=√(C1²+C2²)

and angle φ is defined by the equations

sin φ = C1/A and cos φ = C2/A.

Quantity √(k/M), often denoted by ω, is called the angular frequency.

This is called the natural frequency (ωn) of the system

ωn=√(k/M)

ωn²= k/M

Now, for particular solution

Xp=DSinωo•t

Xp' = Dωo•Cosωo•t

Xp"=-Dωo²•Sinωo•t

Now substituting this into

Mx'' + kx = F(t).

M(-Dωo²•Sinωo•t) + k(DSinωo•t)=FoSinωo•t

Now, let solve for D

D(-Mωo²•Sinωo•t +kSinωo•t) = FoSinωo•t

D=Fo•Sinωo•t/(-Mωo²•Sinωo•t +kSinωo•t)

D=Fo•Sinωo•t / Sinωo•t(-Mωo²+k)

D=Fo / (-Mωo²+k)

D=Fo / (k-Mωo²)

Divide through by k

D=Fo/k ÷ (1 -Mωo²/k)

Note from above

ωn²= k/M

Therefore,

D=Fo/k ÷ (1-ωo²/ωn²)

D=Fo/k ÷ [1-(ωo/ωn)²]

Then,

Xp=DSinωo•t

Xp=(Fo/k ÷ [1-(ωo/ωn)²]) Sinωo•t

Then the general solution is the sum of the homogeneous solution and particular solution

Xg(t)=(Fo/k ÷ [1-(ωo/ωn)²]) Sinωo•t + A•Sin[ √(k/M)t + φ]

Check attachment for the graph of homogeneous, particular and general solution.

Also, check for better way of writing the equations.

8 0
3 years ago
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