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AysviL [449]
3 years ago
7

If an object moves in uniform circular motion in a circle of radius R = 2.00 meters, and the object takes 5.00 seconds to comple

te ten revolutions, calculate the centripetal acceleration.
Physics
2 answers:
ollegr [7]3 years ago
7 0
Hi!

The linear velocity is given by:
v = ωR

And ω is given by: ω = 2π/T (where T is the time for 1 revolution)

Now, put these equations together:
v = 2πR/T

If it takes 5 seconds for 10 revolutions, then it takes 5/10 = 0.5 seconds for each revolution.

V = (2π * 2)/0.5
V = 4π/0.5
V = 8π rad/s

Then, we goes to the centripetal acceleration:
a = V²/R
a = 64π²/2
a = 32π² rad/s²

;)

yulyashka [42]3 years ago
3 0

Answer:

a=315.75\ m/s^2.

Explanation:

Time taken to complete 10 revoution, t=5.0 \ s.

Radius, r=2.0\ m.

We know, v=\dfrac{2\times \pi \times r}{T}.

( T is time taken to complete one revolution).

T=\dfrac{5}{10}=0.5 \ s.

v=\dfrac{2\times \pi\times 2}{0.5} \ m/s=25.13\ m/s.

Centripetal acceleration, a=\dfrac{v^2}{r}.

Therefore, a=\dfrac{25.13^2}{2}=315.75 \ m/s^2.

Hence, it is the required solution.

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an airplane accelerates down a runway at 5.20 m/s^2 for 32.8s until it finally lifts off the ground. determine the distance trav
hammer [34]

Answer:

7,272.7m

Explanation:

Our knowns:

a = 5.2m/s^2

t = 32.8s

V_{i} = 0 m/s

Our Unknowns:

Distance (x): ?

We can use displacement formula:

ΔX = V_{i} t + \frac{1}{2} a t^{2}

Plug in knowns:

X = 0 + 1/2 ((5.2)(32.8))^2

Answer:

ΔX = 7272.7m

(I beleive this should be right) let me know if it helped!

6 0
3 years ago
A ladder is slipping down a vertical wall. If the ladder is 10 ft long and the top of it is slipping at the constant rate of 4 ​
Andrei [34K]

Answer:

Speed will be: 1.33ft/s

Explanation:

If ladder is 10 ft long and bottom is 8 ft from the wall then by Pythagoras theorem we can find the height of the wall where ladder touches. (before it started slipping)

10^2 = 8^2 + x^2

thus x^2 = 100 - 64 = 36

Giving x= 6 ft. If the ladder is falling with a speed of 4ft/s it will take 1.5 seconds to cover the 6ft distance.

This shows that the bottom of ladder will travel from 8ft to 10 ft in 1.5 seconds. Thus covering 2 ft in 1.5 seconds making the speed to be:

v = S / t

v = 2 / 1.5

v = 1.33 ft/s

8 0
4 years ago
Discuss 4 way a scienicst can.reduce bias
Yanka [14]
Make sure to thank me

The answers are by:

Investigation
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Blind Studies
7 0
4 years ago
Read 2 more answers
Low-pressure areas tend to ____ before occluding and ____ after occluding
alexgriva [62]

Out of the following given choices;

a. decelerate; accelerate

b. accelerate; decelerate

c. stop; accelerate

d. decelerate; stop

<span>The answer is B.  This is because the cold front usually rotates around the warm front as cold air mass coverage in the low-pressure system. This causes the cold air mass to accelerate to catch up with the warm front. When an occluded front ( that is the boundary that separates the older cool air mass already in place from new incoming cold air mass), that the system decelerates. </span>






4 0
4 years ago
Zero, a hypothetical planet, has a mass of 4.5 x 1023 kg, a radius of 3.2 x 106 m, and no atmosphere. A 10 kg space probe is to
jok3333 [9.3K]

a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J

b.) ​K 2 +GmM( r 11− r 21)=6.9×10 7 J

Applying Law of  Energy conservation :

K 1+U 1

=K 2+U 2

⇒K 1− r 1GmM

=K 2− r 2 GmM

where M=5.0×10 23kg,r1

=> R=3.0×10 6m and m=10kg

(a) If K 1​

=5.0×10 7J and r 2

=4.0×10 6 m, then the above equation leads to

K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J

(b) In this case, we require K 2

=0 and r2

=8.0×10 6m, and solve for K 1:K 1

​=K 2 +GmM (r 11− r 21)=6.9×10 7 J

Learn more about Kinetic energy on:

brainly.com/question/12337396

#SPJ4

7 0
2 years ago
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