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m_a_m_a [10]
3 years ago
14

In Example Suppose the rod stays on the wall in the same position, but the hinge is moved to be at the center of mass position o

f the rod. The rod is then held out horizontally and released. How do the energiesof the system change after the rod is released? Check all that apply.
a. The change in the system’s gravitational potential energy would be zero
b. The change in the system’s gravitational potential energy is positive
c. The change in the system’s gravitational potential energy is negative
d. The change in the system’s kinetic energy would zeroThe change in the system’s kinetic energy is positive
e. The change in the system’s Kinetic energy is negative
Physics
1 answer:
navik [9.2K]3 years ago
4 0

Answer:

A

Explanation:

The change in the system’s gravitational potential energy would be zero

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Una esfera de radio 0.4m tiene una masa de 300kg, se desea sumergir en agua para saber si flota o no. En este ejercicio use la d
iragen [17]

Answer:

La esfera no flotará pero se hundirá cuando se coloque en el agua porque su densidad es mayor que la del agua.

Explanation:

De la pregunta anterior, se obtuvieron los siguientes datos:

Radio (r) de la esfera = 0,4 m

Masa de esfera = 300 Kg

Densidad del agua = 1000 Kg / m³

A continuación, determinaremos el volumen de la esfera. Esto se puede obtener de la siguiente manera:

Radio (r) de la esfera = 0,4 m

Pi (π) = 3,14

Volumen (V) de la esfera =?

V = 4/3 πr³

V = 4/3 × 3,14 × 0,4³

V = 12,56 / 3 × 0,064

V = 0,27 m³

A continuación, determinaremos la densidad de la esfera. Esto se puede obtener como se ilustra a continuación:

Masa de esfera = 300 Kg

Volumen de esfera = 0,27 m³

Densidad de esfera =?

Densidad = masa / volumen

Densidad de la esfera = 300 / 0,27

Densidad de la esfera = 1111,11 Kg / m³

Comparando la densidad del agua y la de la esfera.

Sustancia >>>>>>> Densidad

Agua >>>>>>>>>>> 1000 Kg / m³

Esfera >>>>>>>>>> 1111,11 Kg / m³

De la ilustración anterior, podemos ver que la densidad de la esfera es mayor que la del agua.

Por lo tanto, la esfera no flotará sino que se hundirá cuando se coloque en el agua porque su densidad es mayor que la del agua.

4 0
3 years ago
Please help I don’t get this give me answers please
Lelu [443]

Answer:

c

Explanation:

3 0
3 years ago
Beginning 145 miles directly south of the city of Hartville, a car travels due west. If the car is travelling at a speed of 42 m
ziro4ka [17]

Answer:

The rate of change of the distance is 14.89.

Explanation:

Given that,

Distance = 145 miles

Speed of car = 42 miles/hr

Distance covered by car = 55 miles

We need to calculate the the rate of change of the distance

According to figure,

Let OA is x, and AB is y.

Now, using Pythagorean theorem

x^2=y^2+145^2

On differentiating

2x\dfrac{dx}{dt}=2y\dfrac{dy}{dt}

\dfrac{dx}{dt}=\dfrac{y}{x}\dfrac{dy}{dt}

\dfrac{dx}{dt}=\dfrac{55\times42}{\sqrt{55^2+145^2}}

\dfrac{dx}{dt}=14.89\ miles/hr

Hence, The rate of change of the distance is 14.89.

8 0
4 years ago
A jet lands at 80.0 m/s, the pilot applying the brakes 2.0 s after landing.
asambeis [7]

Answer:

Schiff by chili I hoop hookup

6 0
4 years ago
A hummingbird 3.4m above the ground flies 1.2 m along a straight line path. Upon spotting a flower below, the hummingbird drops
Anit [1.1K]

A = horizontal displacement of the humming bird = 1.2 m

B = vertical displacement of the humming bird = 1.4 m

C = net displacement of the humming bird from initial to final position = ?

In the triangle drawn , Using Pythagorean theorem

C = √(A² + B²)

inserting the values

C = √(1.2² + 1.4²)

C = √(1.44 + 1.96)

C = √(3.4)

C = 1.4 m

Hence the net displacement of hummingbird comes out to be 1.4 m

4 0
3 years ago
Read 2 more answers
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