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Naddik [55]
1 year ago
9

A highway curve with radius R = 274 m is to be banked so that a car traveling v = 25.0 m/s will not skid sideways even in the ab

sence of friction. At what angle should the curve be banked? 2. what is the correct equation for the sum of the forces in the x direction? What is the correct equation for the sum of the forces in the y direction? A. ΣFx:−Nsinθ=mv2/R; ΣFy:Ncosθ−mg=0 B. ΣFx:−Ncosθ=−mv2/R; ΣFy:Nsinθ−mg=0 C. ΣFx:Nsinθ=mv2/R; ΣFy:Ncosθ+mg=0 D. ΣFx:−Nsinθ=mv2/R; ΣFy:Ncosθ=0 E. ΣFx:−Nsinθ=−mv2/R; ΣFy:Ncosθ−mg=0
Physics
1 answer:
belka [17]1 year ago
6 0

Answer:

A)   θ = 13.1º  , B)  E

Explanation:

A) For this exercise, let's use Newton's second law, let's set a reference frame where the axis ax is in the radial direction and is horizontal, the axis y is vertical.

In this reference system the only force that we must decompose is the Normal one, let's use trigonometry

        sin θ = Nₓ / N

        cos θ = N_{y} / N

        Nₓ = N sin θ

       Ny = N cos θ

x-axis (radial)

        Nₓ = m a

where the acceleration is centripetal

         a = v² / R

we substitute

        -N sin θ = -m v² / R                   (1)

the negative sign indicates that the force and acceleration towards the center of the circle

y-axis (Vertical)

          Ny - W = 0

           N cos θ = mg

           N = mg / cos θ

we substitute in 1

          mg / cos θ  sin θ = m v² / R

          g tan θ = v² / R

          θ = tan⁻¹ (v² / gR)

we calculate

        θ = tan⁻¹ (25² / 9.8 274)

        θ = 13.1º

B) when comparing the equations the correct one is E

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Answer:

Explanation:

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The diameters of bolts produced by a certain machine are normally distributed with a mean of 0.30 inches and a standard deviatio
enot [183]

Answer:

2.25 %

Explanation:

65-95-99.7 is a rule to remember the precentages that lies around the mean.

at the range of mean (\mu) plus or minus one standard deviation (\sigma), P([\mu-\sigma \leq X \leq \mu+\sigma])\approx 68.3%

at the range of mean plus or minus two standard deviation, P([\mu -2\sigma \leq X \leq \mu+2\sigma])\approx 95.5%

at the range of mean plus or minus three standard deviation, P([\mu - 3\sigma\leq X \leq \mu+3\sigma])\approx 99.7%

So, note that  they are asking about the probability that it is greater than 0.32, that is the mean (0.3) plus two times the standard deviation (0.1) (P(X \leq \mu+2\sigma))  

So we know that the 95.5% is between \mu - 2\sigma = 0.3 -2*0.1 = 0.28 and \mu + 2\sigma = 0.3 +2*0.1 = 0.32, hence approximately the 4.5% (100%-95.5%) is greater than 0.32 or less than  0.28. But half (4.5%/2=2.25%) is greater than 0.32 and the other half is less than 0.28.

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8 0
1 year ago
A runner runs 300 m at an average speed of 3.0 m/s. She then runs another 300m at an average
Kaylis [27]

Answer:

B. 4 m/s

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v=d/t

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3 0
2 years ago
A particle of charge 2.3 ✕ 10−8 C experiences an upward force of magnitude 4.6 ✕ 10−6 N when it is placed in a particular point
Marysya12 [62]
<h2>Answer:</h2>

(a) +2 x 10² N/C (upwards)

(b) -2.2μN or -2.2 x 10⁻⁶N (downwards)

<h2>Explanation:</h2>

The force (F) acting on a particle of charge (Q) at a particular point is related to its electric field (E) by the following;

F = Q x E   ----------------------(i)

This means that the force acting on the charged particle is the product of its charge and the electric field at that point.

<em>(a) Given</em>;

Q = charge of the particle = 2.3 x 10⁻⁸ C

F = force acting on the particle = 4.6 x 10⁻⁶N

<em>Substitute these values into equation (i) as follows;</em>

=> 4.6 x 10⁻⁶  = 2.3 x 10⁻⁸ x E

=> E = 4.6 x 10⁻⁶ ÷ 2.3 x 10⁻⁸

=> E =  2 x 10² N/C

Since the value is positive, the electric field is directed upwards.

Therefore, the electric field at that point is +2 x 10² N/C

<em>(b) If a charge of q = -1.1 x 10⁻⁸ is placed there, the force (F) acting is calculated as follows;</em>

<em>Substitute Q = q into equation (i) as follows;</em>

F = q x E

<em>Substitute the value of q and E = 2 x 10² N/C into the equation above as follows;</em>

F = -1.1 x 10⁻⁸ x 2 x 10²

F = -2.2 x 10⁻⁶ N

F = -2.2μN

Since the value of the force, F, is negative, it means it is directed downwards.

Therefore the force on the charge is -2.2μN

3 0
2 years ago
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