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exis [7]
3 years ago
13

What is mercury barometer​

Physics
1 answer:
Allisa [31]3 years ago
3 0
A mercury barometer is a device that is used to measure atmospheric pressure at a location. :)
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2.486 L is equal to:
swat32
Is equal to 2486 milliliters
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II) A 0.40-kg ball, attached to the end of a horizontal ord, is rotated in a circle of radius 1.3 m on a friction- less horizont
Lostsunrise [7]

Hi there!

In this instance, the object spinning in a horizontal circle will experience a net force in the horizontal direction due to tension.

The net force is equivalent to the centripetal force, so:

∑F = T

mv²/r = T

Solve for v:

v = √rT/m

v = 13.96 m/s

3 0
3 years ago
When the k. E of
Ierofanga [76]

\sqrt{2}Answer:

KE2 = 2 KE1

1/2 M V2^2 = 2 * (1/2 M V1^2)

V2^2 = 2 V1^2

V2 = \sqrt{2} V1

Since momentum = M V  the momentum increases by \sqrt{2}

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3 years ago
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A man pushes on piano with mass 170 kg; it slides at constant velocity down a ramp that is inclined at 20.0 ∘ above the horizont
nikdorinn [45]

Answer

given,                            

mass of the piano = 170 kg              

angle of the inclination = 20°                

moves with constant velocity hence acceleration = 0 m/s²    

neglecting friction                                  

so, force required to pull the piano                    

F = m g sin θ                                                      

F = 170 × 9.81 × sin 20°                                        

F = 570.39 N                                                    

so, force required by the man to push the piano is F = 570.39 N

4 0
3 years ago
The 60.0 kg skier shown below is skiing down a 35.0 degree incline where the magnitude of the friction force is 38.5N
Sunny_sXe [5.5K]

Answer:

a) 4.98m/s²

b) 481.66N

Explanation:

a) Using the Newtons second law of motion

\sum F_x = ma_x\\F_m - F_f = ma_x\\Wsin \theta - F_f = ma_x\\mgsin \theta - F_f = ma_x\\

m is the mass of the object

g is the acceleration due to gravity

Fm is the moving force acting along the plane

Ff is the frictional force opposing the moving froce

a is the acceleration of the skier

Given

m = 60kg

g = 9.8m/s²

\theta = 35°

Ff = 38.5N

Required

acceleration of the skier a

Substituting into the formula;

60(9.8)sin 35^0 - 38.5 = 60a\\588sin35^0 - 38.5 = 60a\\337.26 - 38.5 = 60a\\298.76 = 60a\\a = 298.76/60\\a = 4.98m/s^2\\

Hence the acceleration of the skier is 4.98m/s²

b) The normal force on the skier is expressed as;

N = Wcosθ

N = mgcosθ

N = 60(9.8)cos 35°

N = 588cos 35°

N = 481.66N

Hence the normal force on the skier is 481.66N

5 0
3 years ago
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