Arena is accelerating her car at a rate of 46 mis2 for 10 seconds. Her starting velocity was 0 mis

Cu ce viteza s-a deplasat un vehicul daca la ora 16:45 se afla in dreptul bornei kilometrice 169 iar la ora 17:50 se afla la bor

jolli1 [7]

Answer:

7

Explanation:

the answer is 7

3 0

3 years ago

In uniform circular motion, how does the acceleration change when the speed is increased by a factor of 3? When the radius is de

crimeas [40]

Answer:

The acceleration will become 9/2 times.

a' =9/2 a

Explanation:

We know that acceleration of a particle when it is moving in the circular path is given as

$a=\omega^2\ r$

r=radius

ω= angular speed

If the speed ω '= 3 ω

If the radius , $r'=\dfrac{r}{2}$

The final acceleration =a'

$a'=\omega^2'\ r'$

$a'=(3\omega)^2\times \dfrac{r}{2}$

$a'=9\omega^2\times \dfrac{r}{2}$

$a'=\omega^2\times \dfrac{9r}{2}$

$a'= \dfrac{9r}{2}\times \omega^2\times r$

$a'=\dfrac{9}{2}a$

Therefore the acceleration will become 9/2 times.

5 0

3 years ago

In what region of the em spectrum is each?

Alisiya [41]

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3 0

4 years ago

An individual experiences a deep cut that severs the radial artery near the elbow. Ignoring air resistance, approximately how hi

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4 0

4 years ago

To understand and apply the formula τ=Iα to rigid objects rotating about a fixed axis. To find the acceleration a of a particle

Alenkinab [10]

Answer:

Hello your question is incomplete attached below is the missing part of the question

In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Assume that the pivot is attached tot he center of the bar.

You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m1>m2.I

answer : part A = 2*[(M1 - M2)/(M1 + M2)]*g/L

part A = attached below

Explanation:

Part A :

Assuming that mass of swing is negligible

α = T/I

where ; T = torque, I = inertia,

hence T = L/2*9*(M1 - M2)

also; I = $M1*(L/2)^2 + M2*(L/2)^2$ = ( M1 + M2) * (L/2)^2

Finally the magnitude of the angular acceleration α

α = 2*[(M1 - M2)/(M1 + M2)]*g/L

Part B attached below

7 0

3 years ago