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3 years ago

Where does potential energy come from

1 answer:

6 0

If earth were to every pull back on that object, this is where its kinetic energy will come from. Regarding some other planet pulling it, note that there is currently potential energy stored in the object(when it is onearth), in the other planet's gravitational field.

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An object with little mass won't require a lot of force to move.

Anna35 [415]

Answer:

True

Explanation:

Heavier objects (objects with more mass) are more difficult to move and stop. Heavier objects (greater mass) resist change more than lighter objects. Example: Pushing a bicycle or a Cadillac, or stopping them once moving. The more massive the object (more inertia) the harder it is to start or stop.

7 0

3 years ago

Listed following are the names and mirror diameters for six of the world’s greatest reflecting telescopes used to gather visible

ziro4ka [17]

Answer:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope

Explanation:

How much light a telescope can collect depends on its diameter, since in a bigger area more photons will be collected.

Remember that in a circle the area is defined as:

$A = \pi r^{2}$ (1)

Where A is the area and r is its radius.

However, the radius can be determined by means of its diameter.

$d = 2r$

$r = \frac{d}{2}$ (1)

Where d is its diameter.

An example of this is when a person is collecting raindrops with a bucket and with a cup. Since the bucket has a bigger area than the cup, it will collect more raindrops by unit of time. In this scenario the raindrops represent the photons.

To determine the light collecting area of each telescope, equation 2 will be replaced in equation 1.

$A = \pi (\frac{d}{2})^{2}$ (3)

Case for Large binocular telescope:

$A_{mirror1} = \pi (\frac{8.4m}{2})^{2}$

$A_{mirror1} = 55.41m$

For the second mirror will be the same value

$A = A_{mirror1}+A_{mirror2}$

$A = 55.41m+55.41m$

$A= 110.82m$

Case for Keck 1 telescope:

$A = \pi (\frac{10m}{2})^{2}$

$A = 78.53m$

Case for Hobby-Ebberly telescope:

$A = \pi (\frac{9.2m}{2})^{2}$

$A = 66.47m$

Case for Subaru telescope:

$A = \pi (\frac{8.3m}{2})^{2}$

$A = 54.10m$

Case for Gemini North telescope:

$A = \pi (\frac{8m}{2})^{2}$

$A = 50.26m$

Case for Magellan 2 telescope:

$A = \pi (\frac{6.5m}{2})^{2}$

$A = 33.18m$

Hence, they may be rank in the following way:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope.

Key term:

Photons: particles that constitute light. 

3 0

3 years ago

A thin rod of length 0.75 m and mass 0.42 kg is suspended

MrRissso [65]

Answer:

a) K = 0.63 J, b) h = 0.153 m

Explanation:

a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is

w² = $\frac{m g d}{I}$

where d is the distance from the pivot point to the center of mass and I is the moment of inertia.

The rod is a homogeneous body so its center of mass is at the geometric center of the rod.

d = L / 2

the moment of inertia of the rod is the moment of a rod supported at one end

I = ⅓ m L²

we substitute

w = $\sqrt{\frac{mgL}{2} \ \frac{1}{\frac{1}{3} mL^2} }$

w = $\sqrt{\frac{3}{2} \ \frac{g}{L} }$

w = $\sqrt{ \frac{3}{2} \ \frac{9.8}{0.75} }$

w = 4.427 rad / s

an oscillatory system is described by the expression

θ = θ₀ cos (wt + Φ)

the angular velocity is

w = dθ /dt

w = - θ₀ w sin (wt + Ф)

In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1

In the exercise it is indicated that at the lowest point the angular velocity is

w = 4.0 rad / s

the kinetic energy is

K = ½ I w²

K = ½ (⅓ m L²) w²

K = 1/6 m L² w²

K = 1/6 0.42 0.75² 4.0²

K = 0.63 J

b) for this part let's use conservation of energy

starting point. Lowest point

Em₀ = K = ½ I w²

final point. Highest point

Em_f = U = m g h

energy is conserved

Em₀ = Em_f

½ I w² = m g h

½ (⅓ m L²) w² = m g h

h = 1/6 L² w² / g

h = 1/6 0.75² 4.0² / 9.8

h = 0.153 m

5 0

3 years ago

Daniel takes his two dogs, Pauli the Pointer and Newton the Newfoundland, out to a field and lets them loose to exercise. Both d

DedPeter [7]

Answer:

Explanation:

3.4 m/s due North, -1.1 m/s due East

7 0

4 years ago

Help quick physics area question

zlopas [31]

Answer: The area of brick in contact with the floor is 1539 $cm^{3}$ .

Explanation:

Given: Length = 19 cm

Width = 9 cm

Height = 9 cm

As the brick is rectangular in shape. Hence, its area will be calculated as follows.

$Area = length \times width \times height$

Substitute the values into above formula as follows.

$Area = length \times width \times height\\= 19 cm \times 9 cm \times 9 cm\\= 1539 cm^{3}$

Thus, we can conclude that area of brick in contact with the floor is 1539 $cm^{3}$ .

8 0

3 years ago