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xz_007 [3.2K]
3 years ago
9

Block B has mass 6.00 kg and sits at rest on a horizontal, frictionless surface. Block A has mass 2.50 kg and sits at rest on to

p of block B. The coefficient of static friction between the two blocks is 0.400. A horizontal force P is then applied to block A. What is the largest value P can have and the blocks move together with equal accelerations?
Physics
1 answer:
zzz [600]3 years ago
4 0

Answer:

Explanation:

Block A sits on block B and force is applied on block A . Block A will experience two forces 1) force P and 2 )  friction force in opposite direction of motion . Block B will experience one force that is force of friction in the direction of motion .

Let force on block A be P . friction force on it will be equal to kinetic friction, that is μ mg , where μ is coefficient of friction and m is mass of block A

friction force = .4 x 2.5 x 9.8

= 9.8 N

net force on block A = P - 9.8

acceleration = ( P - 9.8 ) / 2.5

force on block B = 9.8

acceleration = force / mass

= 9.8 / 6

for common acceleration

( P - 9.8 ) / 2.5  = 9.8 / 6

( P - 9.8 ) / 2.5 = 1.63333

P = 13.88 N .

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The answer would be B..

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how does the electric force between two charged particles change if the distance between them is increased by a factor of 3? a.
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Delvig [45]

Answer:

Sound waves are produced when something vibrates.  

Explanation:

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Sorry if this if  wrong

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Read 2 more answers
A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of 39.1 cm. The room is slightl
USPshnik [31]

Given Information:  

Wavelength =  λ = 39.1 cm = 0.391 m

speed of sound = v = 344 m/s

linear density = μ = 0.660 g/m = 0.00066 kg/m

tension = T = 160 N

Required Information:

Length of the vibrating string = L = ?

Answer:

Length of the vibrating string = 0.28 m

Explanation:

The frequency of beautiful note is

f = v/λ

f = 344/0.391

f = 879.79 Hz

As we know, the speed of the wave is

v = √T/μ

v = √160/0.00066

v = 492.36 m/s

The wavelength of the string is

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λ = 492.36/879.79

λ = 0.5596 m

and finally the length of the vibrating string is

λ = 2L

L = λ/2

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L = 0.28 m

Therefore, the vibrating section of the violin string is 0.28 m long.

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