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3 years ago

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You throw a ball into a box and close the lid. You hear the ball bouncing around inside as the ball’s energy changes from gravit

ational potential energy to kinetic energy, to elastic potential energy, and so on. If you wait a minute or two, what will have happened to the ball’s energy?

1 answer:

beks73 [17]3 years ago

5 0

Answer:

Thermal energy.

Explanation:

We know that energy is conserved and it only transform one form to another form ,Like potential energy transform in to kinetic energy when a ball fall down from a height and obtain some velocity.

In this case energy changes from gravitational energy to kinetic energy then elastic energy and after some time this energy will convert in to thermal energy.

So the answer is thermal energy.

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The table below shows two types of electromagnetic waves and three random applications of electromagnetic waves.

Musya8 [376]

Answer:

a. Microwaves—3 and infrared waves—1

Explanation:

Microwaves and infrared waves are both part of the electromagnetic spectrum, but they have different frequency and wavelength.

In particular:

- Microwaves are long-wavelength electromagnetic waves, with wavelength between 1 mm and 1 m. Their wavelength is longer than visible light

- Infrared waves are also long-wavelength electromagnetic waves, but their wavelength is shorter than microwaves: between 700 nm and 1 mm. Their wavelength is also longer than visible light.

The two types of waves are also used for different purposes. In particular:

- Infrared waves are emitted by any hot object, and their intensity depends on the temperature of the object. Therefore, they are used in astronomy to show the heat released by astronomical objects (option 1)

- Microwaves are used to study the Cosmic Microwave Background (CMB). This is electromagnetic radiation that permeates the whole universe, and its wavelength depends inversely on the local temperature. Therefore, areas with longer wavelength have lower temperature, and viceversa. Therefore, microwaves are used to measure temperature differences in space (option 3).

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3 years ago

How a suit of armor might be a good analogy for a function of the skeleton system

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How might a suit of armor be a good analogy for a function of the skeletal system?

It's a frame for your body and protects organs and armor protects your body from injury

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3 years ago

Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad

love history [14]

Answer:

a) The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ , b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

$\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}$

Where:

$a_{r}$ - Magnitude of the radial acceleration, measured in meters per square second.

$a_{t}$ - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

$\| \vec a \| = a_{t}$

$\| \vec a \| = r \cdot \alpha$

Where:

$\alpha$ - Angular acceleration, measured in radians per square second.

$r$ - Radius of rotation (Radius of a tire), measured in meters.

Given that $\alpha = 15\,\frac{rad}{s^{2}}$ and $r = 0.31\,m$ . The linear acceleration experimented by the car is:

$\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)$

$\| \vec a \| = 4.65\,\frac{m}{s^{2}}$

The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ .

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

Where:

$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

If $\theta_{o} = 0\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\alpha = 15\,\frac{rad}{s^{2}}$ , the final angular position is:

$\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}$

$\theta = 46.875\,rad$

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

$\theta = 7.46\,rev$

The tires did 7.46 revolutions in 2.50 seconds from rest.

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4 years ago

You own a high speed digital camera that can take a picture every 0.5 seconds. You decide to take a picture every 0.5 seconds of

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-- There is no need to develop the pictures. They are available immediately in a digital camera.

-- There is no change in the teacher from one picture to the next.

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3 years ago