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pashok25 [27]
3 years ago
8

An object of mass 100 kg is initially at rest on a horizontal frictionless surface. At time t = 0, a horizontal force of 10 N is

applied to the object for 1 s and then removed. Which of the following is true of the object at time t = 2s if it is still on the surface?
(A) It is at the same position it had at t 0, since a force of 10 N is not large enough to move such a massive object.
(B) It is moving with constant nonzero acceleration.
(C) It is moving with decreasing acceleration.
(D) It is moving at a constant speed
(E) It has come to rest some distance away from the position it had att 0.
Physics
1 answer:
satela [25.4K]3 years ago
8 0

Answer:

(D) It is moving at a constant speed

Explanation:

Before t = 1s. Due to the force, albeit small, acting on the object, since there's no static friction stopping the object from moving, this mass object would have a constant acceleration and it's velocity would be increasing.

According to Newton's 1st law, an object will stay at a constant speed if the net force acting on it is 0. After t = 1s, horizontally speaking there's no other force exerting on the mass object. There is no friction force at play here as the surface is frictionless.

Therefore the correct statement is (D) It is moving at a constant speed

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In order to calculate friction force you need Limiting friction first .

\\ \sf\longmapsto F_L=\mu sN

u s is coefficient of static friction and N is normal reaction

Or

\\ \sf\longmapsto F_L=\mu smg

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a liquid reactant is pumped through a horizontal, cylindrical, catalytic bed. The catalyst particles are spherical, 2mm in diame
natulia [17]

Answer:

The upper limit on the flow rate = 39.46 ft³/hr

Explanation:

Using Ergun Equation to calculate the pressure drop across packed bed;

we have:

\frac{\delta P}{L}= \frac{150 \mu_oU(1- \epsilon )^2}{d^2p \epsilon^3} + \frac{1.75 \rho U^2(1-\epsilon)}{dp \epsilon^3}

where;

L = length of the bed

\mu = viscosity

U = superficial velocity

\epsilon = void fraction

dp = equivalent spherical diameter of bed material (m)

\rho = liquid density (kg/m³)

However, since U ∝ Q and all parameters are constant ; we can write our equation to be :

ΔP = AQ + BQ²

where;

ΔP = pressure drop

Q = flow rate

Given that:

9.6 = A12 + B12²

Then

12A + 144B = 9.6       --------------   equation (1)

24A + 576B = 24.1    ---------------  equation (2)

Using elimination methos; from equation (1); we first multiply it by 2 and then subtract it from equation 2 afterwards ; So

288 B = 4.9

       B = 0.017014

From equation (1)

12A + 144B  = 9.6

12A + 144(0.017014) = 9.6

12 A = 9.6 - 144(0.017014)

A = \frac{9.6 -144(0.017014}{12}

A = 0.5958

Thus;

ΔP = AQ + BQ²

Given that ΔP = 50 psi

Then

50 = 0.5958 Q + 0.017014 Q²

Dividing by the smallest value and then rearranging to a form of quadratic equation; we have;

Q² + 35.02Q - 2938.8 = 0

Solving the quadratic equation and taking consideration of the positive value for the upper limit of the flow rate ;

Q = 39.46 ft³/hr

3 0
3 years ago
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