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TiliK225 [7]
3 years ago
11

a train engineer started the train from a standstill and sped up to 5 meters per second, she then rounded a corner at a constant

speed. She then sped the train up to 10 meters per second on the next straightaway. Next, the train chugged around another bend at 10 m/sec, and then its speed slowed to 5 meters per second after coming out of the bend. Finally, the train reached the station, and the engineer applied the brakes and brought the engine to a stop. How many times did the train accelerate? Once a carousel is in motion, the constant movement of the carousel horse around the center of the circle can best be described as
Physics
1 answer:
olasank [31]3 years ago
3 0

for acceleration we can define that rate of change in velocity is know as acceleration

So whenever velocity of train is changing with time we can say train is accelerating

Now here if initially train is standstill then after some time its speed is 5 m/s

so here the train is accelerated first time

Then on straight path its speed changed from 5 m/s to 10 m/s so here train gets accelerated second time

After this train chugged around a curve with same speed 10 m/s

SO here since train is moving in curve so here its direction of velocity is continuously changing and this type of acceleration is known as centripetal acceleration

SO this is accelerated Third time

Then its speed decreases and it comes to speed of 5 m/s from 10 m/s

So here it is acceleration of train for Fourth time

Then finally train comes to stop so again its speed changed from 5 m/s to 0

so this is acceleration of train Fifth time

So total train will accelerate 5 times in whole path

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The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistanc
qaws [65]

Answer:

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

Voltmeter read 6 volts

ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

R = \frac{6}{2.5} - 2.4 ohm

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

IR = V

\frac{3R}{R+3} \times 2.5 = 6

R= 12 ohm

2) circuit Q

I\times (R+0.1) =V

R+0.1 =\frac{6}{2.5}

R = 2.3 ohm

5 0
3 years ago
How does friction make it possible for you to walk across the floor?
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5 0
3 years ago
Moons that show few craters probably
ch4aika [34]

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<h3>Can Earth have Two moons?</h3>

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1 year ago
How to convert grams to liters?? V.V? so if your using the weigh machine...and the machine reads in grams and you want liters...
masya89 [10]
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6 0
3 years ago
The area of a rectangular park is 4 mi^2. The park has a width that is equal to "w", and a length that is 3 mi longer than the w
goldfiish [28.3K]

Answer:

l= 4 mi   : width of the park

w= 1 mi  : length of the park

Explanation:

Formula to find the area of ​​the rectangle:

A= w*l       Formula(1)

Where,

A is the area of the  rectangle in mi²

w is the  width of the rectangle in mi

l is the  width of the rectangle in mi

Known data

A =  4 mi²

l = (w+3)mi    Equation (1)

Problem development

We replace the data in the formula (1)

A= w*l  

4 = w* (w+3)

4= w²+3w

w²+3w-4= 0

We factor the equation:

We look for two numbers whose sum is 3 and whose multiplication is -4

(w-1)(w+4) = 0 Equation (2)

The values ​​of w for which the equation (2) is zero are:

w = 1 and w = -4

We take the positive value w = 1 because w is a dimension and cannot be negative.

w  = 1 mi  :width of the park

We replace w  = 1 mi  in the equation (1) to calculate the length of the park:

l=  (w+3) mi

l= ( 1+3) mi

l= 4 mi

8 0
3 years ago
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