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loris [4]
3 years ago
9

The gravitional energy of an object is always measured realative to the ?

Physics
2 answers:
mel-nik [20]3 years ago
6 0
Gravitational potential energy is

                       (mass) x (gravity) x (height) .

When you see this formula, you should be asking yourself:
"Self !  'Height' above WHAT ?

The answer to that question is:  It doesn't matter.  It can be the height
above anything, as long as you make very clear what the reference
(zero) level is, so that it doesn't change during the course of working
with the problem.

There is no particular level relative to which the gravitational potential
energy must always be measured.

In a large number of cases ... like when the problem involves something
going up and down hills, or roller coasters making loops, or soccer balls
or golf balls being launched ... the reference level for gravitational energy
is the ground, because it's THERE and it's convenient.

But if the action in the problem takes place in an office on the 80th floor
of the Aon building in Chicago, or down in a coal-mine shaft in Kentucky
where da sun don't shan, then the floor of the room you're in would be a
much wiser and more convenient level to adopt as the the zero-reference
level.

There's no law.
zaharov [31]3 years ago
4 0
I would say that the answer would be MASS.
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3 0
3 years ago
Read 2 more answers
A bucket of water of mass 14.0 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.260
ruslelena [56]

Answer:

The tension in the rope is 41.38 N.

Explanation:

Given that,

Mass of bucket of water = 14.0 kg

Diameter of cylinder = 0.260 m

Mass of cylinder = 12.1 kg

Distance = 10.7 m

Suppose we need to find that,

What is the tension in the rope while the bucket is falling

We need to calculate the acceleration

Using relation of torque

\tau=F\times r

I\times\alpha=F\times r

Where, I = moment of inertia

\alpha = angular acceleration

\dfrac{Mr^2}{2}\times\dfrac{a}{r}=F\times r

F=\dfrac{M}{2}a...(I)

Here, F = tension

The force is

F=m(g-a)...(II)

Where, F = tension

a = acceleration

From equation (I) and (II)

\dfrac{M}{2}a=m(g-a)

a=\dfrac{g}{1+\dfrac{M}{2m}}

Put the value into the formula

a=\dfrac{9.8}{1+\dfrac{12.1}{2\times14.0}}

a=6.84\ m/s^2

We need to calculate the tension in the rope

Using equation (I)

F=\dfrac{M}{2}a

Put the value into the formula

F=\dfrac{12.1}{2}\times6.84

F=41.38\ N

Hence, The tension in the rope is 41.38 N.

6 0
3 years ago
Question 4. Calculate the kinetic energy of a Year 11 pupil (mass 55Kg)
bogdanovich [222]

Answer:

990 J

Explanation:

Kinetic energy is:

KE = ½ mv²

Given m = 55 kg and v = 6 m/s:

KE = ½ (55 kg) (6 m/s)²

KE = 990 J

4 0
3 years ago
2. A student drives 7.8-km trip to school and averages a speed of
Alekssandra [29.7K]

Answer:

<em>The total time is: t=451.22 sec</em>

<em>The average speed is: V=34.57 m/s</em>

Explanation:

<u>Average speed</u>

The average speed is calculated by dividing the total distance traveled by an object (x) by the total time it took it to travel that distance (t).

\displaystyle V=\frac{x}{t}

Since the student makes the trip in two parts, we have to calculate the total distance and the total time.

We know the distance to school is 7.8 Km = 7,800 m. The student makes his way home over the same distance, thus the total distance is

x=2*7,800 m=15,600 m

The first trip to school was done at an average speed of v1=32.6 m/s. Knowing the distance and speed, we can calculate the time:

\displaystyle t1=\frac{x1}{v1}=\frac{7,800}{32.6}=239.26\ sec

The second trip back home was done at an average speed of v2=36.8 m/s. Let's calculate the second time:

\displaystyle t2=\frac{x2}{v2}=\frac{7,800}{36.8}=211.96\ sec

The total time is:

t=239.26\ sec+211.96\ sec=451.22\ sec

\boxed{t=451.22\ sec}

The average speed is:

\displaystyle V=\frac{15,600}{451.22}=34.57\ m/s

\boxed{\displaystyle V=34.57\ m/s}

6 0
3 years ago
I attempted to answer and got 0m, please explain how to get to the answer.
mafiozo [28]

The maximum height to which the ball attain before falling back down is 1147.96 m

<h3>Data obtained from the question</h3>

The following data were obtained from the question:

  • Initial velocity (u) = 150 m/s
  • Final velocity (v) = 0 m/s (at maximum height)
  • Acceleration due to gravity (g) = 9.8 m/s²
  • Maximum height (h) =?

<h3>How to determine the maximum height </h3>

The maximum height reached by the ball can be obtained as illustrated below:

v² = u² – 2gh (since the ball is going against gravity)

0² = 150² – (2 × 9.8 × h)

0 = 22500 – 19.6h

Collect like terms

0 – 22500 = –19.6h

–22500 = –19.6h

Divide both side by –19.6

h = –22500 / –19.6

h = 1147.96 m

Thus, the maximum height reached by the ball is 1147.96 m

Learn more about motion under gravity:

brainly.com/question/22719691

#SPJ1

5 0
1 year ago
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