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scoray [572]
4 years ago
11

At highway speeds, a particular car is capable of an acceleration of about 1.6 m/s?. At this rate,

Physics
1 answer:
kicyunya [14]4 years ago
5 0

Given parameters:

Acceleration of the car = 1.6m/s

Initial speed  = 80km/hr

Final speed  = 110km/hr

Solution:

Time taken to achieve this speed = ?

Solution:

Acceleration is the rate of change of velocity with the time taken.

  Mathematically;

     a  = \frac{V - U}{T}

where a is the acceleration

           V is the final velocity

           U is the initial velocity

           T is the time taken

Now make the unknown time the subject of the expression;

      aT  = V - U

         T = \frac{V - U}{a}  

Convert the given acceleration to km/hr;

       1.6m/s  = 1.6 x \frac{m}{s}  x \frac{1km}{1000m} x \frac{3600s}{1hr} = 5.76km/hr

Input the parameters and solve;

       T = \frac{110 - 80}{5.76}  = 5.2hrs

The time taken is 5.2hrs

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A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calcu
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Question:

A wire 2.80 m in length carries a current of 5.20 A in a region where a uniform magnetic field has a magnitude of 0.430 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

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Answer:

(a)5.42 N (b)6.26 N (c)5.42 N

Explanation:

From the question

Length of wire (L) = 2.80 m

Current in wire (I) = 5.20 A

Magnetic field (B) = 0.430 T

Angle are different in each part.

The magnetic force is given by

F=I \times B \times L \times sin(\theta)

So from data

F = 5.20 A \times 0.430 T \times 2.80 sin(\theta)\\\\F=6.2608 sin(\theta) N

Now sub parts

(a)

\theta=60^{o}\\\\Force = 6.2608 sin(60^{o}) N\\\\Force = 5.42 N

(b)

\theta=90^{o}\\\\Force = 6.2608 sin(90^{o}) N\\\\Force = 6.26 N

(c)

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3 0
3 years ago
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