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3 years ago

Different forms of the same element are called what

Physics

2 answers:

GarryVolchara [31]3 years ago

8 0

Answer:

Explanation:

I think its Allotropes because Allotropes are different physical forms of the same element.

solmaris [256]3 years ago

4 0

Answer:

Allotropes

Explanation:

Because it is the same element the chemical conposition is the same i.e. The atoms that make the substances are identical. Allotropes can be crystalline or amorphous and sometimes they can be varied in properties.

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Suppose that a freely falling object were somehow equipped with an odometer. Would the readings of distance fallen each second i

maks197457 [2]

Answer:

1 greater distances fallen in successive seconds

Explanation:

When a body falls freely it is subjected to the action of the force of gravity, which gives an acceleration of 9.8 m / s2, consequently, we are in an accelerated movement

If we use the kinematic formula we can find the position of the body

Y = Vo t + ½ to t2

Where the initial velocity is zero or constant and the acceleration is the acceleration of gravity

Y = - ½ g t2 = - ½ 9.8 t2 = -4.9 t2

Let's look for the position for successive times

t (s) Y (m)

1 -4.9

2 -19.6

3 -43.2

The sign indicates that the positive sense is up

It can be clearly seen that the distance is greatly increased every second that passes

3 0

3 years ago

How much time would it take for the sound of thunder to travel 2000 meters if sound travels of 330 meters per sec

Lubov Fominskaja [6]

2000÷330=6.06 repatant so the answer would be about 6.06 seconds

4 0

3 years ago

Read 2 more answers

The animation shows a ball which has been kicked upward at an angle. Run the animation to watch the motion of the ball. Click in

sergejj [24]

Answer:

The acceleration of the ball is $a_y = - 0.3672 \ m/s^2$

Explanation:

From the question we are told that

The maximum height the ball reachs is $H_{max} = 42.24 \ m$

The horizontal component of the initial velocity of the ball is $v_{ix} = 5.57 \ m/s$

The vertical component of the initial velocity of the ball is $v_{iy} = = 16.18 m/s$

The vertically motion of the ball can be mathematically represented as

$v_{fy}^2 = v_{iy} ^2 + 2 a_{y} H_{max}$

Here the final velocity at the maximum height is zero so $v_{fy} = 0 \ m/s$

Making the acceleration $a_y$ the subject we have

$a_y = \frac{v_{iy} ^2}{2H_{max}}$

substituting values

$a_y = - \frac{5.57^2}{2* 42.24}$

$a_y = - 0.3672 \ m/s^2$

The negative sign shows that the direction of the acceleration is in the negative y-axis

6 0

3 years ago

A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa

andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

$\omega=\frac{v}{r}$

where

$\omega$ is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s$

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

$\omega=\frac{v}{r}$

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s$

(c) 5550 m

The maximum playing time of the CD is

$t =74.0 min \cdot 60 s/min = 4,440 s$

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

$d=vt=(1.25 m/s)(4,440 s)=5,550 m$

(d) $-6.4\cdot 10^{-3} rad/s^2$

The angular acceleration of the CD is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 21.6 rad/s$ is the final angular speed (when the CD is scanned at the outermost part)

$\omega_i = 50.0 rad/s$ is the initial angular speed (when the CD is scanned at the innermost part)

$t=4440 s$ is the time elapsed

Substituting into the equation, we find

$\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2$

5 0

3 years ago

Read 2 more answers

Consider 2 converging lenses of focal lengths 5 mm (objective) and 50 mm.(eyepiece) An object 0.1 mm in size is placed a distanc

AleksandrR [38]

Answer:

1) q₁ = 12.987 cm , b) L = 17.987 cm , c) m = 179.87

Explanation:

We can solve the geometric optics exercises with the equation of the constructor

1 / f = 1 / p + 1 / q

where f is the focal length, p and q are the distance to the object and the image respectively.

Let's apply this equation to our case

1) f = 5mm = 0.5 cm

p₁ = 5.2 mm = 0.52 cm

h = 0.1 mm = 0.01 cm

1 / q₁ = 1 / f- 1 / p

1 / q₁ = 1 / 0.5 - 1 / 0.52 = 2 - 1.923

1 / q₁ = 0.077

q₁ = 12.987 cm

2) in this part they tell us that the eyepiece creates an image at infinity, therefore the object that comes from being at the focal length of the eyepiece

p₂ = 5 cm

The absolute thing that goes through the two lenses is

L = q₁ + p₂

L = 12.987 +5

L = 17.987 cm

3) This lens configuration forms the so-called microscope, whose expression for the magnifications

m = -L / f_target 25 cm / f_ocular

m = - 17.987 / 0.5 25 / 5.0

m = 179.87

8 0

3 years ago

Different forms of the same element are called what​

Different forms of the same element are called what