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4 years ago

» Accelerating Lear Jet

Physics

1 answer:

adell [148]4 years ago

5 0

The equation relevant to use here is:

v^2 = v0^2 + 2 a d

where,

v is final velocity = ?

v0 is intial velocity = 193.8 m/s

a is acceleration = 13.5 m/s^2

d is distance = 2370 m

v = sqrt [193.8^2 + 2 (13.5) 2370]

v = 318.67 m/s

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Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the fastest-accelerating animals,

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Answer:

Explanation:

given,

speed of cheetah = 27 m/s

time taken = 4 s

mass of cheetah = 110 kg

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3 years ago

The magnetic field inside a 5.0-cm-diameter solenoid is 2.0 T and decreasing at 4.60 T/s. Part A What is the electric field stre

cluponka [151]

Complete Question

The magnetic field inside a 5.0-cm-diameter solenoid is 2.0 T and decreasing at 4.60 T/s.

Part A What is the electric field strength inside the solenoid at a point on the axis?

Part B

What is the electric field strength inside the solenoid at a point 1.50 cm from the axis?

Answer:

Part A

$E = 0 \ V/m$

Part B

$E_{15} = 0.0345 \ V/m$

Explanation:

From the question we are told that

The diameter of the solenoid is $d = 5.0 \ cm = 0.05 \ m$

The magnetic field is $B = 2.0 \ T$

The rate of the change of the magnetic field is $\frac{dB}{dt} = 4.60 \ T/s$

The radius of the solenoid is mathematically represented as

$R = \frac{ d}{2}$

substituting values

$R = \frac{ 5.0 *10^{-2}}{2} = 0.025 \ m$

Generally the of the solenoid is mathematically represented as

$E = \frac{ r}{2} * |\frac{dB}{dt} |$

Now at the point on axis is r = 0 given that the axis is the origin so

$E = \frac{ 0}{2} * |\frac{dB}{dt} |$

$E = 0 \ V/m$

Now the electric field strength inside the solenoid at a point 1.50cm from the axis is mathematically represented as

$E_{15} = \frac{ 15*10^{-2 }}{2} * |4.60 |$

$E_{15} = 0.0345 \ V/m$

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4 years ago