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prisoha [69]
3 years ago
6

A crate is given a big push, and after it is released, it slides up an inclined plane which makes an angle 0.52 radians with the

horizontal. The frictional coefficients between the crate and plane are (\muμs = 0.84, \muμk = 0.48 ). What is the magnitude of the acceleration (in meters/second2) of this crate as it slides up the incline?
Physics
1 answer:
vovangra [49]3 years ago
8 0

Answer:

a = 8.951 m/s²

Explanation:

given,

angle = 0.52 radians

μ_s = 0.84

μ_k = 0.48

acceleration = ?

using

F +  f = m a

mg sin θ +  μk mg cos θ = m a

a = g sin θ  + μk g cos θ

a = 9.8 x  sin 0.52 + 0.48 x 9.8 x  cos 0.52

a = 4.869 + 4.082

a = 8.951 m/s²

the magnitude of acceleration is a = 8.951 m/s²

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a specimen of oil having an initial volume of 5000cm³ is subjected to a pressure of 10⁴N/m² and the volume decreases by 0.20cm³.
inessss [21]

Answer:

 B = 2.5 10⁸ Pa

Explanation:

The volume modulus is defined by

           B = - \frac{P}{ \frac{\Delta V}{V} }

           

The negative fate is for the module to be positive since the volume change is negative

       

It is not necessary to reduce the volumes to the SI system, since they are both in the same units

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             B = 2.5 10⁸ Pa

4 0
2 years ago
Which combination of temperature and pressure correctly describes standard temperature and pressure
raketka [301]

Answer: A combination 0 degrees Celsius and 101.3 kPa or 1 atm correctly describes standard temperature and pressure.

Explanation:

The term standard temperature and pressure is also known as STP and it is most commonly used when we want to calculate the density of a gas.

The term standard temperature means 32^{o} Fahrenheit or 0^{o}C or 273 Kelvin. On the other hand, term standard pressure means 1 atmosheric pressure of a gas.

Thus, we can conclude that a combination 0 degrees Celsius and 101.3 kPa or 1 atm correctly describes standard temperature and pressure.

4 0
3 years ago
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3 years ago
Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.3 m/s parallel to the gro
pashok25 [27]

Answer:

Explanation:

cSep 20, 2010

well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.

Anonymous

Sep 20, 2010

First you need to solve for time by using

d=(1/2)(a)(t^2)+(vi)t

1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s

t=.45 sec

Then you find the horizontal distance traveled by using

v=d/t

1.3m/s=d/.54sec

d=.585m

Then you need to find the time of player B by using

d=(1/2)(a)(t^2)+(vi)t

1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0

t=.61 sec

Finally to find player Bs initial horizontal velocity you use the horizontal equation

v=d/t

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5 0
3 years ago
Read 2 more answers
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ValentinkaMS [17]

Answer:

East component is: 18.64 m/s

Explanation:

If the resultant is 32.5 m/s directed 35 degrees east of north, then we use the sin(35) projection to find the east component of the velocity:

East component = 32.5 m/s * sin(35) = 18.64 m/s

4 0
2 years ago
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