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Romashka [77]
3 years ago
14

Determine the minimum angle at which a frictionless road should be banked so that a car traveling at 20.0 m/s can safely negotia

te the curve if the radius of the curve is 200.0 m.

Physics
1 answer:
EleoNora [17]3 years ago
3 0

Answer:

Minimum angle at which frictionless road should be banked is 11.53°

Explanation:

Consider the figure below to understand the banking curve. θ is  minimum angle at which frictionless road should be banked.

Forces acting along x-axis:

F_{N}sin\theta=\frac{mv^{2}}{R}---(1)\\\\

Forces acting along y-axis:

F_{N}cos\theta-mg=0---(2)\\\\

Dividing (1) and (2)

\frac{F_{N}sin\theta}{F_{N}cos\theta}=\frac{\frac{mv^{2}}{R}}{mg}\\\\tan\,\theta=\frac{v^{2}}{rg}\\\\v=20\,m/s,\,r=200\,m\\\\tan\,\theta=\frac{(20)^{2}}{(200)(9.8)}\\\\tan\,\theta=0.204\\\\\theta=tan^{-1}(0.204)\\\theta=11.53^{o}\\

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Find the magnitude of the vector v given its initial and terminal points. Round your answer to four decimal places.
Sliva [168]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is |v| = 6.93

Explanation:

From the question we are told that

    The initial point is (x_1 , y_1 , z_1 ) = (-1 , 7 , 4 )

    The  terminal point is  (x_2 , y_2 , z_2) = (-5 , 11, 8 )

Generally the magnitude of the vector is mathematically represented as

     |v| = \sqrt{(x_2 -x_1 )^2 + (y_2 - y_1 )^2 + (z_2 -z_1)^2}

=>   |v| = \sqrt{(-5 -(-1) )^2 + (11 - 7 )^2 + (8 -4)^2}

=>   |v| = 6.93

3 0
3 years ago
2. Two charged particles as shown in figure below. QP = +10 μC and Qq = +20 μC are separated by a distance r = 10 cm. What is th
netineya [11]

Answer:

F = 180 N

Explanation:

Given that,

Charge, q₁ = 10 μC

Charge, q₂ = 20 μC

The distance between the charges, r = 10 cm = 0.1 m

We need to find the magnitude of the electrostatic force. The formula for the electrostatic force is given by :

F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times 10\times 10^{-6}\times 20\times 10^{-6}}{(0.1)^2}\\F=180\ N

So, the magnitude of the electrostatic force is 180 N.

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3 years ago
Two long, parallel wires separated by 2.00 cm carry currents in opposite directions. The current in one wire is 1.75 A, and the
Natasha_Volkova [10]

The force per unit length between the two wires is 6.0\cdot 10^{-5} N/m

Explanation:

The magnitude of the force per unit length exerted between two current-carrying wires is given by

\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}

where

\mu_0 = 4\pi \cdot 10^{-7} Tm/A is the vacuum permeability

I_1, I_2 are the currents in the two wires

r is the separation between the two wires

For the wires in this problem, we have

I_1 = 1.75 A

I_2 = 3.45 A

r = 2.00 cm = 0.02 m

Substituting into the equation, we find

\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(1.75)(3.45)}{2\pi (0.02)}=6.0\cdot 10^{-5} N/m

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4 0
4 years ago
You observe a hockey puck of mass 0.12 kg, traveling across the ice at speed 18.3 m/sec. The interaction of the puck and the ice
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The stopping distance is 143.1 m

Explanation:

First of all, we have to find the acceleration of the hockey puck. This can be done by using Newton's second law of motion:

\sum F =ma

where

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m = 0.12 kg is the mass of the puck

a is the acceleration

Solving for a,

a=\frac{\sum F}{m}=\frac{-0.14}{0.12}=-1.17 m/s^2

The motion of the puck is a uniformly accelerated motion, therefore we can use the following suvat equation:

v^2-u^2=2as

where:

v = 0 is the final velocity (the puck comes to a stop)

u = 18.3 m/s is the initial velocity

a=-1.17 m/s^2 is the acceleration

s is the stopping distance

And solving for s, we find

s=\frac{v^2-u^2}{2a}=\frac{0-(18.3)^2}{2(-1.17)}=143.1 m

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