What are the four elements that Aristotle included in his model of matter

Si el cuerpo no se mueve halle T

Aleks [24]

Well sorry but this is the wrong language.

3 0

2 years ago

Read 2 more answers

A hopper jumps straight up to a height of 1.3 m. With what velocity did he leave the floor

xxMikexx [17]

The velocity with which the jumper leaves the floor is 5.1 m/s.

<h3>What is the initial velocity of the jumper?</h3>

The initial velocity of the jumper or the velocity with which the jumper leaves the floor is calculated by applying the principle of conservation of energy as shown below.

Kinetic energy of the jumper at the floor = Potential energy of the jumper at the maximum height

¹/₂mv² = mgh

v² = 2gh

v = √2gh

where;

v is the initial velocity of the jumper on the floor
h is the maximum height reached by the jumper
g is acceleration due to gravity

v = √(2 x 9.8 x 1.3)

v = 5.1 m/s

Learn more about initial velocity here: brainly.com/question/19365526
#SPJ1

3 0

1 year ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0

3 years ago

Can Hotspots can be found at the edges of tectonic plates?

PolarNik [594]

Answer:

No.

Explanation:

Hot spots don't appear at the barriers of tectonic plates, instead, they originate at hot centres known as mantle plumes. Mantle plumes exist below the tectonic plates and may develop a string of volcanoes on the Earths surface.

6 0

2 years ago

A silver sphere with radius 1.3611 cm at 23.0°C must slip through a brass ring that has an internal radius of 1.3590 cm at the s

Readme [11.4K]

Answer:

The temperature must the ring be heated so that the sphere can just slip through is 106.165 °C.

Explanation:

For brass:

Radius = 1.3590 cm

Initial temperature = 23.0 °C

The sphere of radius 1.3611 cm must have to slip through the brass. Thus, on heating the brass must have to attain radius of 1.3611 cm

So,

Δ r = 1.3611 cm - 1.3590 cm = 0.0021 cm

<u>The linear thermal expansion coefficient of a metal is the ratio of the change in the length per 1 degree temperature to its length.</u>

<u>Thermal expansion for brass = 19×10⁻⁶ °C⁻¹</u>

Thus,

$\alpha=\frac {\Delta r}{r\times \Delta T}$

Also,

$\Delta T=T_{final}-T_{Initial}$

So,

$19\times 10^{-6}=\frac {0.0021}{1.3290\times (T_{final}-23.0)}$

Solving for final temperature as:

$(T_{final}-23.0)=\frac {0.0021}{1.3290\times 9\times 10^{-6}}$

<u>Final temperature = 106.165 °C</u>

5 0

3 years ago