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g100num [7]
3 years ago
9

An automobile with an initial speed of 4.92 m/s accelerates uniformly at the rate of 3.2 m/s2 . Find the final speed of the car

after 4.5 s. Answer in units of m/s. 008 (part 2 of 2) 10.0 points Find the displacement of the car after 4.5 s.
Physics
1 answer:
Rudik [331]3 years ago
4 0

Answer:19.32 m/s

Explanation:

Given

initial speed of car(u)=4.92 m/s

acceleration(a)=3.2 m/s^2

Speed of car after 4.5 s

using equation of motion

v=u+at

v=4.92+3.2\times 4.5=4.92+14.4

v=19.32 m/s

Displacement of the car after 4.5 s

v^2-u^2=2as

19.32^2-4.92^2=2\times 3.2\times s

349.05=2\times 3.2\times s

s=54.54 m

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A heater is being designed that uses a coil of 14-gauge nichrome wire to generate 300 W using a voltage of V = 110 V . How long
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Answer:

The length of the wire is 83.2 m.

Explanation:

Given that,

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Voltage, V = 110 V

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The electric power of a wire is given by :

P=\dfrac{V^2}{R}\\\\R=\dfrac{V^2}{P}\\\\R=\dfrac{(110)^2}{300}\\\\R=40.34\ \Omega

Area of cross section of the wire is :

A=\pi r^2\\\\A=\pi (0.000815)^2\\\\A=2.08\times 10^{-6}\ m^2

Resistance of a material is given by :

R=\rho \dfrac{L}{A}\\\\L=\dfrac{RA}{\rho}\\\\L=\dfrac{40\times 2.08\times 10^{-6}}{10^{-6}}\\\\L=83.2\ m

So, the length of the wire is 83.2 m. Hence, this is the required solution.

5 0
3 years ago
In each of two coils the rate of change of the magnetic flux in a single loop is the same. The emf induced in coil 1, which has
irinina [24]

Answer:

 Coil 2 have  235  loops

Explanation:

Given  

The number of loops in coil 1 is n ₁= 159

The emf induced in coil 1 is  ε ₁ = 2.78 V

The emf induced in coil 2 is  ε ₂ = 4.11 V

Let

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Given, the emf in a single loop in two coils are same. That is,

ϕ ₁/n ₁= ϕ ₂ n ₂⟹ 2.78/159 = 4.11/ n ₂

n₂=\frac{159 * 4.11}{2.78}

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A driver in a 2290-kg car car traveling at 42.7 m/s slams on the brakes and skids to a stop. If the coefficient of friction betw
8_murik_8 [283]

Answer:

126.56 m

Explanation:

Applying,

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

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

4 0
3 years ago
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