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jeka94
2 years ago
5

A light-rail commuter train accelerates at a rate of 1.35 m/s. D A 33% Part (a) How long does it take to reach its top speed of

85 km/h, in seconds, starting from rest? Hints deduction per hint. Hints remaining 1 Feedback: deduction per feedback A 33 Part (b) The same train ordinarily brakes at a rate of 1.8 m/s. How long does it take to come to a stop, in seconds, from its top speed? 433 Part (c) In emergencies the train can slow more rapidly. coming to rest from 85 km/h in 7.9 5. What is the magnitude of the acceleration during emergency braking, in meters per square second?
Physics
1 answer:
Dennis_Churaev [7]2 years ago
6 0

Answer:

a) 17.49 seconds

b) 13.12 seconds

c) 2.99 m/s²

Explanation:

a) Acceleration = a = 1.35 m/s²

Final velocity = v = 85 km/h = 85\frac{1000}{3600}=23.61\ m/s

Initial velocity = u = 0

Equation of motion

v=u+at\\\Rightarrow 23.61=0+1.35t\\\Rightarrow t=\frac{23.61}{1.35}=17.49\ s

Time taken to accelerate to top speed is 17.49 seconds.

b) Acceleration = a = -1.8 m/s²

Initial velocity = u = 23.61\ m/s

Final velocity = v = 0

v=u+at\\\Rightarrow 0=23.61-1.8t\\\Rightarrow t=\frac{23.61}{1.8}=13.12\ s

Time taken to stop the train from top speed is 13.12 seconds

c) Initial velocity = u = 23.61 m/s

Time taken = t = 7.9 s

Final velocity = v = 0

v=u+at\\\Rightarrow 0=23.61+a7.9\\\Rightarrow a=\frac{-23.61}{7.9}=-2.99\ m/s^2

Emergency acceleration is 2.99 m/s² (magnitude)

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1. State the law of conservation of energy and what it means for you as a human considering how energy works.
Arturiano [62]

Answer: 1. The law of consevation of energy sates that energy can neither be created nor destroyed. It can only be transformed or transfered from one form to another. The law of conservation of energy is found everywhere for example, Water falls from the sky, converting potential energy to kinetic energy.

2. Different forms of energy are related because energy cannot be created or destroyed. they can all be transformed into from one form to another.

Explanation:

5 0
2 years ago
A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica
-BARSIC- [3]

Answer:

33.2 m

Explanation:

For the first object:

y₀ = 81.5 m

v₀ = 0 m/s

a = -9.8 m/s²

t₀ = 0 s

y = y₀ + v₀ t + ½ at²

y = 81.5 − 4.9t²

For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

y = 33.2

7 0
3 years ago
Question 1(Multiple Choice Worth 4 points) The star named Canopus has a declination of approximately –52°. Which of these statem
Y_Kistochka [10]

Answer:

It is 52° below the celestial equator.

Explanation:

The declination is the angle in degrees measured north (+) or south (-) of the an imaginary line called the celestial equator.

The celestial equator is a projection of the earth's equator on the celestial sphere. imaginary

The star named Canopus has a declination of approximately –52°.

Since the angle is negative, this shows that it is south or below the celestial equator and at 52° south of the celestial equator.

Thus, the star named Caponus is 52° below the celestial equator.

8 0
2 years ago
Use appropriate units and significant figures.  USE THE LAW OF COSINES AND LAW OF SINES.
N76 [4]

Answer:

The resultant velocity is 86.1 mi/h.    

Explanation:

The law of cosines is given by:

c^{2} = a^{2} + b^{2} - 2abcos(\theta)

Where:

c: is the resultant velocity =?

a: is the velocity of the plane = 75.0 mi/h

b: is the velocity of the wind = 15.0 mi/h  

θ: is the angle between "a" and "b"                          

The angle between "a" and "b" can be found as follows:

\theta = 180.0 - 46.0 = 134.0 ^{\circ}

Now, by using the law of cosines we have:

c^{2} = (75.0)^{2} + (15.0)^{2} - 2*75.0*15.0*cos(134.0) = 7413.0

c = 86.1 mi/h    

Therefore, the resultant velocity is 86.1 mi/h.    

The law of sines is:

\frac{a}{sin(\gamma)} = \frac{b}{sin(\alpha)} = \frac{c}{sin(\theta)}

Where:

γ: is the angle between "b" and "c"

α: is the angle between "a" and "c"

So, if we want to find "c" by using the law of sines, we need to know another angle besides θ (γ or α), and the statement does not give us.

I hope it helps you!        

3 0
2 years ago
Pam, wearing a rocket pack, stands on frictionless ice. She has a mass of 49 kg. The rocket supplies a constant force for 22.0 m
ss7ja [257]

Answer:

Magnitude of the force is 4350N

Explanation:

As the woman accelerates at a distance of 22 m to go from rest to 62.5 m / s, we can use the kinematics to find the acceleration

v² = v₀² + 2 a x

v₀ = 0

a = v² / 2x

 a = 62.5²/(2 × 22)

 a = 88.78m/s²

the time you need to get this speed

     v = v₀ + a t

     t = v / a

     t = 62.5 / 88.78

     t = 0.704s

Let's caculate the magnitude of the force

F = ma

= 49 × 88.78

= 4350.22

≅ 4350N

Magnitude of the force is 4350N

     t = 1,025 s

      a = 55.43 m / s²

4 0
2 years ago
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