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2 years ago

A box of mass 3.6 kg is lifted 5.4 m above the

Physics

1 answer:

horsena [70]2 years ago

7 0

So, the energy change that occurs is 190.512 J.

<h3>Introduction</h3>

Hello ! I am Deva from Brainly Indonesia will help you regarding energy and its transformation. In this case, it's the use of energy from the lifter to be equivalent to the change in the object's potential energy. Why potential energy? Because the box undergoes a change in height and the potential energy specializes at a certain height. Work (W) due to change in potential energy ( $\sf{\Delta PE}$ ) can be realized in the equation :

$\sf{W = \Delta PE}$

$\sf{W = m \cdot g \cdot h_2 - m \cdot g \cdot h_2}$

$\boxed{\sf{\bold{W = m \cdot g \cdot (h_2 - h_1)}}}$

With the following condition :

W = work of subject (J)
$\sf{\Delta PE}$ = change of potential energy (J)
m = mass (kg)
g = acceleration of the gravity (m/s²)
$\sf{h_2}$ = final height (m)
$\sf{h_1}$ = initial height (m)

<h3>Problem Solving</h3>

We know that :

m = mass = 3.6 kg
g = acceleration of the gravity = 9.8 m/s²
$\sf{h_2}$ = final height = 5.4 m
$\sf{h_1}$ = initial height = 0 m

What was asked :

W = work of subject = ... J

Step by step :

$\sf{W = \Delta PE}$

$\sf{W = m \cdot g \cdot (h_2 - h_1)}$

$\sf{W = 3.6 \cdot 9.8 \cdot (5.4 - 0)}$

$\sf{W = 3.6 \cdot 9.8 \cdot 5,4}$

$\boxed{\sf{W = 190.512 \: J}}$

So, the energy change that occurs is 190.512 J.

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3 years ago

Along the remote Racetrack Playa in Death valley, California, stones sometimes gouge out prominent trails in the desert floor, a

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Answer:

F = 196 N

Explanation:

For this exercise we will use Newton's second law, we define a reference system with the x axis in the direction of movement of the stones and the y axis vertically

Y axis

N- W = 0

N = mg

X axis

F -fr = ma

In this case, they ask us for the force to keep moving, so the stones go at constant speed, which implies that the acceleration is zero.

F- fr = 0

F = fr

the friction force has the equation

fr = μ N

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we substitute

F = μ mg

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2 years ago

In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st

bekas [8.4K]

Answer:

a) t = 746 s

b) t = 666 s

Explanation:

Total time will be the sum of the partial times between stations plus the time stopped at the stations.
Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

$v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s (1)$

Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

$t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)$

Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

$x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m (3)$

In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

$t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)$

We can find the distance traveled while the train was decelerating as follows:

$x_{3} = (v_{1} * t_{3}) + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m (5)$

Finally, we need to know the time traveled at constant speed.
So, we need to find first the distance traveled at the constant speed of 26.4m/s.
This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)

The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

$t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s (7)$

Total time between two stations is simply the sum of the three times we have just found:
t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
tm = t*5 = 131.6 * 5 = 658.2 s (9)
Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

Using all the same premises that for a) we know that the only difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
We can find the distance traveled at constant speed, rewriting (6) as follows:

x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)

The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

$t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s (12)$

Total time between two stations is simply the sum of the three times we have just found:
t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
tm = t*3 = 207.4 * 3 = 622.2 s (14)
Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)

7 0

2 years ago

The first mechanized industry was

Oxana [17]

C. Textiles

It was the first thing mechanized in the Industrial Revolution

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2 years ago

Convert the following in to SI units without changing its values<br> a. 4000 g <br>b. 2.4 km

lana66690 [7]

hi <3

i believe i explained this answer properly in my last answer but it would be 4kg and 2400m as these are the SI units for these values.

hope this helps :)