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elena-14-01-66 [18.8K]
3 years ago
11

Thalia is drafting a plan to move a large, perfect sphere concrete sculpture that is in front of her office building. Describe t

he considerations Thalia would need to make if she was drafting a plan to move the sculpture.
Physics
2 answers:
kobusy [5.1K]3 years ago
5 0

thalia would need to consider that the sculpture's inertia is keeping the object in place. Newton's first law applies to this situation. The inertia would have to be overcome to enable the sculpture to move, but the inertia would also keep the sculpture moving after its movement was started. The force that is required to move the sculpture would depend on the mass and acceleration at which Thalia wanted the sculpture to move. Finally, any force applied to the sculpture would be resisted by the sculpture in an equal and opposite manner.

Marina86 [1]3 years ago
4 0
For the answer to this question, 
Thalia must consider the weight of the object and the mass of the sculpture. Weight and mass are different things. She should also consider the time on how long it will take to move it and where she'll move it.
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Note: The rope is 20 m long. Answer like this: (1.<br> 2._____ etc)
qaws [65]

Answer:

1. <u>Potential energy</u>, 2. <u>Potential and kinetic energy</u>, 3. <u>Potential and kinetic energy</u>, 4. <u>Potential and kinetic energy</u>, 5. <u>Potential energy</u>

Explanation:

We note that the total mechanical energy (M.E.) of the body is given as follows;

M.E. = K.E. + P.E. = Constant

Where;

K.E. = The kinetic energy of the body = (1/2)·m·v²

P.E. = The potential energy of the body = m·g·h

m = The mass of the person

v = The velocity with which the person is in motion

g = The acceleration due to gravity ≈ 9.81 m/s²

h = The height of the person above the ground

The length of the rope = 20 m

The initial height at location 1, h₁ = 40.0 m

At location 1, the velocity, v₁ = 0.00 m/s

The mechanical energy, M.E. = K.E.₁ + P.E.₁

∴  K.E.₁ = 0 and P.E.₁ = m ×9.81×40

M.E. = (1/2) ×m ×0² + m ×9.81×40

∴ M.E. = 0 + P.E.₁ the type of energy present at location 1 is only potential energy

At location 2, the velocity, v₂ = 10.0 m/s

The mechanical energy, M.E. = K.E.₂ + P.E.₂ = (1/2) ×m ×10² + m ×9.81×40

∴  K.E.₂ = 50·m and P.E.₂ = m ×9.81×35 = 343.35·m

M.E. = 50·m + 343.35·m the type of energy at location 2 is both kinetic energy, K.E. and potential energy, P.E.

At location 3, the velocity, v₃ = 20.0 m/s

The mechanical energy, M.E. = K.E.₃ + P.E.₃ = (1/2) ×m ×20² + m ×9.81×20

∴  K.E.₃ = 200·m and P.E.₃ = m ×9.81×20 = 196.2·m

M.E. = 200·m + 196.2·m the type of energy at location 3 is both kinetic energy, K.E. and potential energy, P.E.

At location 4, the velocity, v₄² = 350.0 m²/s², h₄ = 15.0 m

The mechanical energy, M.E. = K.E.₄ + P.E.₄ = (1/2) × m ×350 + m ×9.81×15

∴  K.E.₄ = 175·m and P.E.₄ = m×9.81×15 = 147.15·m

M.E. = 175·m + 147.15·m the type of energy at location 4 is both kinetic energy, K.E. and potential energy, P.E.

At location 5, the velocity, v₅ = 0 m/s, h₅ = 10.0 m

The mechanical energy, M.E. = K.E.₅ + P.E.₅ = (1/2) × m × 0 + m ×9.81×10

∴  K.E.₅ = 0·m and P.E.₅ = m×98.1 = 98.1·m

M.E. = 0·m + 98.1·m the type of energy at location 5 is only potential energy, P.E.

Therefore, we have;

\left|\begin{array}{ccc}Location&&Type(s) \ of \ Energy \ Presents\\1&&Potential \ Energy\\2&&Potential  \ and \ Kinetic \ Energy\\3&&Potential  \ and \ Kinetic \ Energy\\4&&Potential  \ and \ Kinetic \ Energy\\5&&Potential  \  Energy\end{array} \right |

5 0
2 years ago
15. Sandra decided to talk a walk at her neighborhood park. She walks 20 meters
FinnZ [79.3K]

Answer:

Long question good luck:)..............

Explanation:

8 0
3 years ago
How long will it take for a car at rest to accelerate at 7m/s^2 to a speed of 45 m/s
Gennadij [26K]
Speed/acceleration = time
45/7
6.4
You have to round the decimal to 6.4
8 0
2 years ago
Using our understanding of the law of gravity. What happens to the gravity as we triple the distance between two objects?
MA_775_DIABLO [31]

Answer:

1/9

Explanation:

<em>Newton’s Law of Universal Gravitation </em>

Objects with mass feel an attractive force that is proportional to their masses and inversely proportional to the square of the distance.

F = GMm/r²

where  

F - the gravitatioal force in Newtons,  

M   and m  -two masses in kilograms  

r  - the separation in meters.  

G  - the gravitational constant (6.674*10 ⁻¹¹ N (m/kg) ² )

Because of the magnitude of  G , gravitational force is very small unless large masses are involved.

So according to above equation , when the masses are not changing , force is inversely propotional to the square of distance

F1 ∝ 1/r² ---------------(1)

F2 ∝ 1(3r)²

F2 ∝ 1/9r²--------------(2)

(2)/(1)

\frac{F_2}{F_1} =\frac{1}{9}\\ F_2 =\frac{F_1}{9}

From their you get as the distance tripled, Force reduce by a factor of 9(3³)

for example , assume the distance get doubled ,Force reduce by a factor of 4 (2²)

3 0
3 years ago
How much mass should be attached to a vertical ideal spring having a spring constant (force constant) of 39.5 n/m so that it wil
mrs_skeptik [129]
The frequency of a simple harmonic oscillator such as a spring-mass system is given by
f= \frac{1}{2 \pi}   \sqrt{ \frac{k}{m} }
where 
k is the spring constant
m is the mass attached to the spring.

Re-arranging the formula, we get:
m= \frac{k}{4 \pi^2 f^2}
and since we know the constant of the spring:
k=39.5 N/m
and the frequency of oscillation:
f=1.00 Hz
we can find the value of the mass attached to it:
m= \frac{39.5 Hz}{4 \pi^2 (1.00 Hz)^2} = 1.00 kg
7 0
3 years ago
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