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drek231 [11]
3 years ago
5

A train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a forward velocity of The train the

n moves at a constant velocity of for 420 s. The train then slows down uniformly at until it is brought to a halt. The acceleration during the first 5.6 km of travel is closest to which of the following?
a. 0.19 m/s^2
b. 0.14 m/s^2
c. 0.16 m/s^2
d. 0.20 m/s^2
e. 0.17 m/s^2
Physics
1 answer:
Korvikt [17]3 years ago
7 0

Answer:

The acceleration during the first 5.6 km of travel is closest to 0.16 m/s²    

Option c) 0.16 m/s² is the correct answer.

Explanation:

Given the data in the question;

since the train starts from rest,

Initial velocity; u = 0 m/s

final velocity; v = 42 m/s

distance covered S = 5.6 km = ( 5.6 × 1000 )m = 5600 m

acceleration a = ?

From the third equation of motion;

v² = u² + 2as

we substitute in our values

( 42 )² = ( 0 )² + [ 2 × a × 5600 ]

1764 = 0 + [ 11200 × a ]

1764 = 11200 × a

a = 1764 / 11200

a = 0.1575 ≈ 0.16 m/s²          { two decimal place }

Therefore, The acceleration during the first 5.6 km of travel is closest to 0.16 m/s²    

Option c) 0.16 m/s² is the correct answer.

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a) At a position of 2.0m, the Initial energy is all made up of the potential energy=m*g*hi<span>
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The percentage of energy remaining is E=m*g*hi/m*g*hf x 100 

and since mass and gravity are constant so it leaves us with just E=hi/hf 
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b) Here use the equation vf^2=vi^2+2gd 

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vi =sqrt(-2gd) <span>
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3 years ago
Explain why elements in the same family have similar physical and chemical properties
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They have similar physical and chemical properties because of thier valence electrons
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A camera takes a properly exposed photo with a 4.0 mm diameter aperture and a shutter speed of 1/1000 s. If the photographer foc
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Answer:

option E

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given,

diameter = 4 mm

shutter speed = 1/1000 s

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E V = log_2(\dfrac{N^2}{t})

N is the diameter of the aperture and t is the time of exposure

now,

log_2(\dfrac{N^2}{t_1})=log_2(\dfrac{N^2}{t_2})

\dfrac{N_1^2}{t_1}=\dfrac{N_2^2}{t_2}

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\dfrac{4^2}{1000}=\dfrac{N_2^2}{250}

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4 0
3 years ago
The starter motor of a car engine draws a current of 140 A from the battery. The copper wire to the motor is 4.20 mm in diameter
GenaCL600 [577]

Answer:

(a)106.4C

b)0.5676mm

Explanation:

(a)To get the charge that have passed through the starter then The current will be multiplied by the duration

I= current

t= time taken

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Q= I*t = 140*0.760 = 106.C

(b) b. How far does an electron travel along the wire while the starter motor is on?(mm)

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We can now calculate the area of the conductor which is

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Where e= 1.6*10^-19

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Therefore , the distance traveled is:

x = v*t = 0.7468 * 0.760 = 0.5676mm

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