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vekshin1
3 years ago
5

A high school physics student is sitting in a seat reading this question. The magnitude of the force with which the seat is push

ing up on the student to support him is closest to what?
Physics
2 answers:
liberstina [14]3 years ago
7 0

Answer:

Force due to seat must be equal to the Weight of the student

Explanation:

As we know that the student is sitting on the seat at rest

So here we can say that at this position the student must be at equilibrium position

so we can write force equation for this position of student

vertically downward the net force is due to his/ her weight which is given as

W = mg

now this weight of the student is counter balanced by the the normal force due to seat which is in opposite direction

Now for equilibrium position of the object we will have

F_n = mg

so the seat will push the student upwards by force which is equal to the weight.

Dennis_Churaev [7]3 years ago
4 0
The force pushing down  is the force of Gravity. On a chair it is in perfect balance with the force pushing up (the normal force)
 

in terms of magnitude 
FN = FG = mg

the forces are in opposite direction

hope this helps
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A car going at v = 29.7 m/s (67 mph) rounds a curve of radius R = 50.0 m, where the road is banked at an angle of θ = 30.0°. Wha
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Answer:

μ = 0.6

Explanation:

given,

speed of car = 29.7 m/s

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θ = 30.0°

minimum static friction = ?

now,

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N cos θ - f sinθ = mg

N cos θ -μN sinθ = mg

N = \dfrac{m g}{cos\theta-\mu sin \theta}

now, writing the forces acting along x- direction

N sin θ + f cos θ = F_{net}

N cos θ + μN sinθ = F_{net}

\dfrac{m g}{cos\theta-\mu sin \theta}(cos \theta + \mu sin\theta)=F_{net}

taking cos θ from nominator and denominator

F_{net} =\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{mv^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{v^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}g

\mu=\dfrac{v^2 -r g tan\theta}{v^2tan\theta + r g}

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\mu=\dfrac{29.7^2 -50 \times 9.8tan 30^0}{29.7^2\times tan 30^0 +50 \times 9.8}

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7 0
3 years ago
A ball is projected upward at time t = 0.0 s, from a point on a roof 60 m above the ground. The ball rises, then falls until it
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(remember the direction of the ball too (signs))

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7 0
2 years ago
A chamber fitted with a piston contains 1.90 mol of an ideal gas. Part A The piston is slowly moved to decrease the chamber volu
yarga [219]

Answer:

The work done is 5136.88 J.

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Given that,

n = 1.90 mol

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W=−5136.88\ J

The Work done on the system.

Hence, The work done is 5136.88 J.

5 0
3 years ago
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