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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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PLEASE HELP ASAP!!!!!!!
devlian [24]
Remember that the left is NEGATIVE
and RIGHT is POSITIVE

16. +3N - 6N = -3N

therefore 3N to the left (because left = -)



17. +5N - 5N = 0N , zero resultant force



18. +6N - 4N = +2N

therefore 2N to the right (because right = +)
6 0
3 years ago
A canon is tilled back 30.0 degrees and shoots a cannon ball at 155 m/s. What is the
Umnica [9.8K]

Answer:

= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m

Therefore, highest point that the cannon ball reaches is 168.7m

Explanation:

the cannon is fired at an angle 30 o to the horizonatal with a speed of 155 m/s

highest point that the cannon ball reaches?

H_{max}=\frac{V^2\sin ^2 \theta}{2g}

g = 9.8m/s2

= \frac{(115^2)(\sin 30)^2}{2\times 9,8} \\\\= 168.7m

Therefore, highest point that the cannon ball reaches is 168.7m

6 0
3 years ago
Match the correct sentence together.
Brut [27]
I’m pretty sure you times them so 1 with A, 2 with e, 3 with C, and 4 with B
6 0
2 years ago
Read 2 more answers
What fundamental frequency would you expect from blowing across the top of an empty soda bottle that is 24 cm deep
kari74 [83]

Answer:

708.3 Hz

Explanation:

For an open-air column, like the empty can, the fundamental frequency is given by

f_1 = \frac{v}{2L}

where

v = 340 m/s is the speed of sound

L is the length of the column

In this problem, the length of the bottle is

L = 24 cm = 0.24 m

Therefore, the fundamental frequency is

f_1 = \frac{340 m/s}{2(0.24 m)}=708.3 Hz

8 0
3 years ago
Someone please help me
Trava [24]

Answer:

The answer to your question is: D) Ф₂ = 49.71°

Explanation:

Data

n₁ = 1.33

Ф₁ = 35°C

n₂ = 1

Ф₂ = sin⁻¹ (n₁ sinФ₁/n₂)

Process

Substitution

Ф₂ = sin⁻¹ (n₁ sinФ₁/n₂)

Ф₂ = sin⁻¹ (1.33 sin 35/1)

Ф₂ = sin⁻¹ (1.33 x 0.574/ 1)

Ф₂ = sin⁻¹ ( 0.7628 / 1)

Ф₂ = sin⁻¹ (0.7628)

Ф₂ = 49.71°

8 0
4 years ago
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