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2 years ago

6

A hockey puck with a mass of 0.16 kg is sitting at rest on a frozen pond. Suddenly, the wind begins to blow, accelerating the pu

ck across the ice at 625 m/s2. If friction is ignored, the force of the wind can be calculated to be _______.
A. 625.16 N
B. 100 N
C. 3,906 N
D. 0.000256 N

2 answers:

lions [1.4K]2 years ago

5 0

YES THE ANSWER IS B 100 N THX TO U I GOT IT RIGHT

cricket20 [7]2 years ago

3 0

The answer is B 100n

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You are provided with three polarizers with filters making angles of (A) 90 ∘ , (B) 180 ∘ and (C) −45 ∘ with respect

irinina [24]

Answer:

Order of maximum transmission of the polarizer is A, C and B.

Solution:

As per the question:

For the first polarizer, the angle is quite insignificant:

(A) $90^{\circ}$ :

The light intensity after passing through the first polarizer is $I_{o}$ and this intensity does not depend on the angle of the polarizer.

Consider $90^{\circ}$ with the vertical, the intensity is given by:

$I = I_{o}cos^{2}90^{\circ}$

$I = I_{o}cos(2(45^{\circ})) = I_{o}(\frac{1+cos90^{\circ})}{2} = \frac{I_{o}}{2}$

(B) $180^{\circ}$ :

Suppose the second polarizer is $45^{\circ}$ with the vertical.

Now, intensity through the second polarizer:

$I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(- 45 - 90)$

$I' = \frac{I_{o}}{2}cos^{2}135^{\circ} = \frac{I_{o}}{4}$

Now, if we consider the second polarizer to be $180^{\circ}$ ,

$I' = \frac{I_{o}}{2}cos^{2}180^{\circ} = \frac{I_{o}}{2}cos^{2}(180^{\circ} - 90^{\circ}) = 0$

(C) $- 45^{\circ}$ :

Now,

Intensity through the third polarizer, if it is $180^{\circ}$ with the vertical:

$I' = Icos^{2}(\theta_{2} - \theta_{1}) = \frac{I_{o}}{2}cos^{2}(180 - (- 45))$

$I' = \frac{I_{o}}{8}$

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2 years ago

Your friend says, “chemical changes are caused by an input in energy. In physical changes, there is no transfer of energy” is yo

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Answer:

Ok, let's suppose the simplest of the physical changes:

We have an object that is not moving (so it is not accelerated)

and there is change, now the object moves.

Because there was a change, means that there was an acceleration, and by the second Newton's law.

Force equals mass times acceleration:

F = m*a

There must be a force.

So suppose that you pushed the object, then some energy that you had, you transferred it to the object, that now is moving and now has kinetic energy.

Now, is kinda true that in a closed system the total energy is always constant, but it depends on what is our system.

So if we think in our system as you and the object, then in the whole system the energy does not change because the energy that you lost is now on the object, but again, there was a transfer of energy.

So no, your friend is not correct.

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2 years ago

A 90° elbow in a horizontal pipe is used to direct water flow upward at a rate of 40 kg/s. The diameter of the entire elbow is 1

tekilochka [14]

Answer:

(a) The gauge pressure is 4,885.3 Pa

(b) The anchoring force needed to hold the elbow in place is approximately 296.5 N

The direction of the anchoring force is approximately ≈ 134.8

Explanation:

The given parameters in the fluid dynamics question are;

Pipe elbow angle, θ = 90°

The mass flowrate of the water, $\dot m$ = 40 kg/s

The diameter of the elbow, D = 10 cm

The pressure at the point of discharge of the fluid = Atmospheric pressure

The elevation between the inlet and exit of the elbow, z = 50 cm

The weight of the elbow and the water = Negligible

(a) The velocity of the fluid, v₁ = v₂ = v is given as follows;

ρ·v·A = $\dot m$

v = $\dot m$ /(ρ·A)

Where;

ρ = The density of the fluid (water) = 997 kg/m³

A = The cross-sectional area of the of the elbow = π·D²/4 = π×0.1²/4 ≈ 0.00785398163

A = 0.00785398163 m²

v = 40 kg/s/(0.00785398163 m² × 997 kg/m³) ≈ 5.1083 m/s

Bernoulli's equation for the flow of fluid is presented as follows;

$\dfrac{P_1}{\rho \cdot g} +\dfrac{v_1}{2 \cdot g} + z_1 = \dfrac{P_2}{\rho \cdot g} +\dfrac{v_2}{2 \cdot g} + z_2$

v₁ = v₂ for the elbow of uniform cross section

P₁ - P₂ = ρ·g·(z₂ - z₁)

P₁ - P₂ = The gauge pressure = $P_{gauge}$

z₂ - z₁ = z = 50 cm = 0.5 m

∴ $P_{gauge}$ = 997 kg/m³ × 9.8 m/s² × 0.5 m = 4,885.3 Pa

The gauge pressure, $P_{gauge}$ = 4,885.3 Pa

(b) The forces acting on the elbow are;

$F_{Rx}$ + $P_{gauge}$ ·A = -β· $\dot m$ ·v

$F_{Ry}$ = β· $\dot m$ ·v

∴ $F_{Rx}$ = -β· $\dot m$ ·v - $P_{gauge}$ ·A

$F_{Rx}$ = -1.03 × 40 kg/s × 5.1083 m/s - 4,885.3 Pa × 0.00785398163 m² ≈ -208.83 N

$F_{Ry}$ = 1.03 × 40 kg/s × 5.1083 m/s = 210.46196 N

The resultant force, $F_R$ , is given as follows;

$F_R$ = √( $F_{Rx}$ ² + $F_{Rz}$ ²)

∴ $F_R$ = √((-208.83)² + (210.46196)²) ≈ 296.486433934

Therefore;

The anchoring force needed to hold the elbow in place, $F_R$ ≈ 296.5 N

The direction of the anchoring force, θ

$\theta = tan^{-1}\left( \dfrac{F_{Ry}}{F_{Rx}} \right)$

∴ θ = arctan(210.46196/(-208.83)) = -45..2230044°

θ = 180° + -45.2230044° = 134.7769956°

∴ The direction of the anchoring force is approximately, θ ≈ 134.8°

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2 years ago

For an object sliding at velocity v with a force of friction Fk and a normal force FNs which equation can be usedto calculate th

musickatia [10]

Answer:

C.

Explanation:

Apex says so

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2 years ago

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Gala2k [10]

Answer:

d. 0.3 N

Explanation:

Force: This is defined as the product of mass of a body and its acceleration.

The S.I unit of Force is Newton (N).

Mathematically Force can be represented as,

F = ma .................. Equation 1

Where F = force, m = mass, a = acceleration.

also

a = (v-u)/t............... Equation 2

Where v = final velocity, u = initial velocity, t = time.

Given: v = 0.60 m/s, u = 0 m/s ( From rest), t= 0.16 s.

Substitute into equation 2

a = (0.60-0)/0.16

a = 3.75 m/s²

Also given: m = 80 g = 0.08 kg.

Substitute into equation 1,

F = 0.08(3.75)

F = 0.3 N.

Hence he average force = 0.3 N

The right option is d. 0.3 N

6 0

2 years ago