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2 years ago

15

A Top Fuel Dragster initially at rest undergoes an average acceleration of 39 m/s2 for 3.97 seconds. How far will it go in this

time?

1 answer:

tigry1 [53]2 years ago

5 0

Answer:

s = 307.34 m

Explanation:

In order to find the distance covered by the dragster during the given time, we will use second equation of motion. The second equation of motion is written as follows:

s = Vi t + (0.5)at²

where,

s = distance covered by the dragster = ?

Vi = Initial Velocity = 0 m/s

t = time interval = 3.97 s

a = acceleration = 39 m/s²

Therefore,

s = (0 m/s)(3.97 s) + (0.5)(39 m/s²)(3.97 s)²

<u>s = 307.34 m</u>

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FrozenT [24]

Answer:

f=15.5 Hz

Explanation:

Let's determine the internal resistance:

$R=\frac{(p*L)}{A}$

ρ = 1.68*10^-8 Ω m

$L=0.060m*4*60 = 14.4m$

$A=\pi*r^2\\A= \pi*(5.9x10^-4m/2)^2=2.734*10^-7m^2$

$R=(1.68*10^-8)*(14.4m)/(2.734*10^-7m^2)= 0.884$ Ω

Since the bulb is rated at 12.0 V and 25.0 W,

Current

$I=\frac{25W}{12.0v}=2.08 A$

Therefore, voltage drop inside generator =

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Actual EMF required is

$E_{mf}=12.0v+2.35v=14.35v$

Note that this is an RMS value.

The peak voltage is

$v_{peak}=14.15v*\sqrt{2} =20.29v$

For a generator, by Faraday's Law,

$E_{(max)}=N*B*A*w$

$20.29v=(60)*(0.650T)*(0.06m)^2$ *ω

ω $=144.5\frac{rad}{s}$

f=ω/(2π)=

f=144.5 rad/s/(2π)

f=23.001 Hz

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2 years ago

The rate of change of atmospheric pressure P with respect to altitude h is proportional to P, provided that the temperature is c

puteri [66]

Answer:

64.59kpa

Explanation:

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2 years ago

Differences between Pressure and upthrust

Angelina_Jolie [31]

Answer:

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2 years ago

Andy is waiting at the signal. As soon as the light turns green, he accelerates his car at a uniform rate of 8.00 meters/second2

Tatiana [17]

You can reason it out like this:

-- The car starts from rest, and goes 8 m/s faster every second.

-- After 30 seconds, it's going (30 x 8) = 240 m/s.

-- Its average speed during that 30 sec is (1/2) (0 + 240) = 120 m/s

-- Distance covered in 30 sec at an average speed of 120 m/s

   = 3,600 meters .
___________________________________

The formula that has all of this in it is the formula for
distance covered when accelerating from rest:

   Distance = (1/2) · (acceleration) · (time)²

   = (1/2) · (8 m/s²)    · (30 sec)²

   = (4 m/s²) · (900 sec²)

   = 3600 meters.

_________________________________

When you translate these numbers into units for which
we have an intuitive feeling, you find that this problem is
quite bogus, but entertaining nonetheless.

When the light turns green, Andy mashes the pedal to the metal
and covers almost 2.25 miles in 30 seconds.

How does he do that ?

By accelerating at 8 m/s². That's about 0.82 G !

He does zero to 60 mph in 3.4 seconds, and at the end
of the 30 seconds, he's moving at 534 mph !

He doesn't need to worry about getting a speeding ticket.
Police cars and helicopters can't go that fast, and his local
police department doesn't have a jet fighter plane to chase
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3 0

2 years ago

Two wires are stretched between two fixed supports and have the same length. One wire A there is a second-harmonic standing wave

lina2011 [118]

(a) Greater

The frequency of the nth-harmonic on a string is an integer multiple of the fundamental frequency, $f_1$ :

$f_n = n f_1$

So we have:

- On wire A, the second-harmonic has frequency of $f_2 = 660 Hz$ , so the fundamental frequency is:

$f_1 = \frac{f_2}{2}=\frac{660 Hz}{2}=330 Hz$

- On wire B, the third-harmonic has frequency of $f_3 = 660 Hz$ , so the fundamental frequency is

$f_1 = \frac{f_3}{3}=\frac{660 Hz}{3}=220 Hz$

So, the fundamental frequency of wire A is greater than the fundamental frequency of wire B.

(b) $f_1 = \frac{v}{2L}$

For standing waves on a string, the fundamental frequency is given by the formula:

$f_1 = \frac{v}{2L}$

where

v is the speed at which the waves travel back and forth on the wire

L is the length of the string

(c) Greater speed on wire A

We can solve the formula of the fundamental frequency for v, the speed of the wave:

$v=2Lf_1$

We know that the two wires have same length L. For wire A, $f_1 = 330 Hz$ , while for wave B, $f_B = 220 Hz$ , so we can write the ratio between the speeds of the waves in the two wires:

$\frac{v_A}{v_B}=\frac{2L(330 Hz)}{2L(220 Hz)}=\frac{3}{2}$

So, the waves travel faster on wire A.

7 0

2 years ago