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viva [34]
3 years ago
13

If you're driving 55 MPH and you suddenly need to stop, how many feet will you travel before the car comes to a stop

Physics
1 answer:
lys-0071 [83]3 years ago
5 0
ANSWER: 170 Feet
REASON: with good breaks and a dry road your car should stop and skip 170 feet, with perception and reaction time of stopping you should stop within 170 feet
You might be interested in
Work of 2 joules is done in stretching a spring from its natural length to 11 cm beyond its natural length. what is the force (i
Sidana [21]
The law applied here is Hooke's Law which describes the force exerted by the spring with a given distance. The equation for this is F = kΔx, where F is the force in Newtons, k is the spring constant in N/m while Δx is the displacement in meters.

If you want to find work done by a spring, this can be solved by using differential equations. However, derived equations are already ready for use. The equation is

W = k[{x₂-x₁)² - (x₁-xn)²],

where 
xn is the natural length
x₁ is the stretched length 
x₂ is also the stretched length when stretched even further than x₁

In this case xn =x₁. So, that means that (x₁-xn) = 0 and (x₂-x₁) = 11 cm or 0.11 m.

Then, substituting the values,

2 J = k (0.11² -0²)
k = 165.29 N/m

Finally, we use the value of k to the Hooke's Law to determine the Force.

F = kΔx = (165.29 N/m)(0.11 m)
F = 18.18 Newtons
5 0
3 years ago
A concrete piling of 50 kg is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. How much will the wire stretch?
pentagon [3]

Explanation:

It is given that,

Mass of concrete pilling, m = 50 kg

Diameter of wire, d = 1 mm

Radius of wire, r = 0.0005 m

Length of wire, L = 11.2

Young modulus of steel, Y=20\times 10^{10}\ N/m^2

The young modulus of a wire is given by :

Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}

Y=\dfrac{F.L}{A\Delta L}

\Delta L=\dfrac{F.L}{A.Y}

\Delta L=\dfrac{50\ kg\times 9.8\ m/s^2\times 11.2\ m}{\pi (0.0005\ m)^2\times 20\times 10^{10}\ N/m^2}

\Delta L=0.034\ m

So, the wire will stretch 0.034 meters. Hence, this is the required solution.

8 0
3 years ago
A student is watching waves come in from the ocean. He noticed that the first wave he saw (Wave A) had twice the amplitude of th
Alika [10]

Answer:

Wave A

<em>I</em><em> </em><em>hope this</em><em> </em><em>helps</em><em> </em>

8 0
2 years ago
Andy is entering early adolescence. although he used to be a big help to his mother by doing household chores, he now refuses, s
o-na [289]

Answer:

a. gender intensification

Explanation:

This theory states that in the adolescence both girls and boys are pressured to follow gender roles established culturally.

8 0
3 years ago
Consider two thin, coaxial, coplanar, uniformly charged rings with radii a and b푏 (a
Wittaler [7]

Answer:

electric potential, V = -q(a²- b²)/8π∈₀r³

Explanation:

Question (in proper order)

Consider two thin coaxial, coplanar, uniformly charged rings with radii a and b (b < a) and charges q and -q, respectively. Determine the potential at large distances from the rings

<em>consider the attached diagram below</em>

the electric potential at point p, distance r from the center of the outer charged ring with radius a is as given below

Va = q/4π∈₀ [1/(a² + b²)¹/²]

Va = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} }

Also

the electric potential at point p, distance r from the center of the inner charged ring with radius b is

Vb = \frac{-q}{4\pi e0} * \frac{1}{(b^{2} + r^{2} )^{1/2} }

Sum of the potential at point p is

V = Va + Vb

that is

V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } + \frac{-q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }

V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }

V = \frac{q}{4\pi e0} * [\frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{1}{(b^{2} + r^{2} )^{1/2} }]

the expression below can be written as the equivalent

\frac{1}{(a^{2} + r^{2} )^{1/2} }  = \frac{1}{(r^{2} + a^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} }

likewise,

\frac{1}{(b^{2} + r^{2} )^{1/2} }  = \frac{1}{(r^{2} + b^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }

hence,

V = \frac{q}{4\pi e0} * [\frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]

1/r is common to both equation

hence, we have it out and joined to the 4π∈₀ denominator that is outside

V = \frac{q}{4\pi e0 r} * [\frac{1}{{(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]

by reciprocal rule

1/a² = a⁻²

V = \frac{q}{4\pi e0 r} * [{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} - {(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2}]

by binomial expansion of fractional powers

where (1+a)^{n} =1+na+\frac{n(n-1)a^{2} }{2!}+ \frac{n(n-1)(n-2)a^{3}}{3!}+...

if we expand the expression we have the equivalent as shown

{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} = (1-\frac{a^{2} }{2r^{2} } )

also,

{(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2} = (1-\frac{b^{2} }{2r^{2} } )

the above equation becomes

V = \frac{q}{4\pi e0 r} * [((1-\frac{a^{2} }{2r^{2} } ) - (1-\frac{b^{2} }{2r^{2} } )]

V = \frac{q}{4\pi e0 r} * [1-\frac{a^{2} }{2r^{2} } - 1+\frac{b^{2} }{2r^{2} }]

V = \frac{q}{4\pi e0 r} * [-\frac{a^{2} }{2r^{2} } +\frac{b^{2} }{2r^{2} }]\\\\V = \frac{q}{4\pi e0 r} * [\frac{b^{2} }{2r^{2} } -\frac{a^{2} }{2r^{2} }]

V = \frac{q}{4\pi e0 r} * \frac{1}{2r^{2} } *(b^{2} -a^{2} )

V = \frac{q}{8\pi e0 r^{3} } * (b^{2} -a^{2} )

Answer

V = \frac{q (b^{2} -a^{2} )}{8\pi e0 r^{3} }

OR

V = \frac{-q (a^{2} -b^{2} )}{8\pi e0 r^{3} }

8 0
3 years ago
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