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Naddik [55]
2 years 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]2 years 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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2 years ago
Two 5.0 mm × 5.0 mm electrodes are held 0.10 mm apart and are attached to 7.5 V battery. Without disconnecting the battery, a 0.
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Answer:

A) V = 7.5 V

B) E = 75,000 V/m

C) Q = 16.6 pC

D) V = 7.5 V

E) E = 24,000 V/m

F) Q = 52 pC

Explanation:

Given:

- The Area of plate A = ( 5 x 5 ) mm^2

- The distance between plates d = 0.10 mm

- The thickness of Mylar added t = 0.10 mm

- Voltage supplied by battery V = 7.5 V

Solution:

A) What is the capacitor's potential difference before the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

B) What is the capacitor's electric field before the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

                                E = V / k*d

k = 1 (air)                  E = 7.5 / 0.10*10^-3

                                E = 75,000 V/m

C) What is the capacitor's charge Q before the Mylar is inserted?

                                C = k*A*ε / d

k = 1 (air)                   C = ( 0.005^2 * 8.85*10^-12 ) / 0.0001

                                C = 2.213 pF

                                Q = C*V

                                Q = 7.5*(2.213)

                                Q = 16.6 pC

D) What is the capacitor's potential difference after the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

E) What is the capacitor's electric field after the Mylar is inserted?    

- The Electric Field E between the capacitor plates is given by:

                                E = V / k*d

k = 3.13                     E = 7.5 / (3.13)0.10*10^-3

                                E = 24,000 V/m              

F) What is the capacitor's charge after the Mylar is inserted?      

                                C = k*A*ε / d

k = 3.13                    C = 3.13*( 0.005^2 * 8.85*10^-12 ) / 0.0001

                                C = 6.927 pF

                                Q = C*V

                                Q = 7.5*(6.927)

                                Q = 52 pC                                      

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Answer:

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The total time required for the ball to return to the child’s hand is 3.2654 s.

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The final velocity (v) = 0 m/s

Initial velocity (u) = 4 m/s

a = -0.25g = -0.25*9.8 = -2.45 m/s². (Negative because it is in opposite of the velocity and also it deaccelerates while going up).

Let time be t  to reach the top.

Using

v = u + a×t

0 = 4 - 2.45*t

t = 1.6327 s

Since, this is the same time the ball will come back. So,

<u>Total time to go and come back = 2* 1.6327 = 3.2654 s </u>

To find the distance, using:

v² = u² + 2×a×s

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s = 3.2653 m

<u>Thus, the distance the ball moves up the incline before reversing its direction is 3.2653 m.</u>

5 0
2 years ago
A person is working on a steel structure while standing on the ground. An accident occurred where 5 A pass through the structure
matrenka [14]

Answer:

35mA

Explanation:

Hello!

To solve this problem we must use the following steps

1. Find the electrical resistance of the metal rod using the following equation

R=\alpha  \frac{l}{a}

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α=

metal rod resistivity=2x10^-4 Ωm

l=leght=2m

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solving

R=(2x10^-4)\frac{2}{0.00283} =0.14

2. Now we model the system as a circuit with parallel resistors, where we will call 1 the metal rod and 2 the man(see attached image)

3.we know that the sum of the currents in 1 and 2 must be equal to 5A, by the law of conservation of energy

I1+I2=5

4.as the voltage on both nodes is the same we can use ohm's law in resitance 1 and 2 (V=IR)

V1=V2

(0.14I1)=2000(i2)

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I1=14285.7i2

5.Now we use the equation found in step 3

14285.7i2+i2=5

i2=\frac{5}{14285.7+1} =3.5x10^-4A=35mA

6 0
2 years ago
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