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3 years ago

The idea that species change over time is called

Physics

2 answers:

Free_Kalibri [48]3 years ago

7 0

Evolution

hope this helps

qaws [65]3 years ago

4 0

The answer is evolution. When a specifies evolves over time they change and adapt to their environment.

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Which lists types of materials from most conductive to least conductive?

Lilit [14]

Superconductor, conductor, semiconductor, insulator

7 0

3 years ago

Read 2 more answers

What is the mass of a basketball that has Connecticut energy of 102 J and it's travelling at 5 m/s

Stella [2.4K]

Answer:

8.16Kg

Explanation:

K.E= 1/2 mv²

102=1/2 m 5²

m=8.16 Kg

8 0

3 years ago

A wheel has a radius of 5.9 m. How far (path length) does a point on the circumference travel if the wheel is rotated through an

posledela

Answer:

(a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

Explanation:

Given that,

Radius = 5.9 m

(a). Angle $\theta=30°$

We need to calculate the angle in radian

$\theta=30\times\dfrac{\pi}{180}$

$\theta=0.523\ rad$

We need to calculate the path length

Using formula of path length

$Path\ length =angle\times radius$

$Path\ length=0.523\times5.9$

$Path\ length =3.09\ m$

(b). Angle = 30 rad

We need to calculate the path length

$Path\ length=30\times5.9$

$Path\ length=177\ m$

(c). Angle = 30 rev

We need to calculate the angle in rad

$\theta=30\times2\pi$

$\theta=188.4\ rad$

We need to calculate the path length

$Path\ length=188.4\times5.9$

$Path\ length =1111.56\ m$

Hence, (a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

8 0

3 years ago

A rectangular block of steel measures 4cm*2cm*1.5cm*. its mass is 93.6g

nata0808 [166]

Answer:

Explanation:

5 0

3 years ago

7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surf

Serga [27]

Answer:

z = 0.8 (approx)

Explanation:

given,

Amplitude of 1 GHz incident wave in air = 20 V/m

Water has,

μr = 1

at 1 GHz, r = 80 and σ = 1 S/m.

depth of water when amplitude is down to 1 μV/m

Intrinsic impedance of air = 120 π Ω

Intrinsic impedance of water = $\dfrac{120\pi}{\epsilon_r}$

Using equation to solve the problem

$E(z) = E_0 e^{-\alpha\ z}$

E(z) is the amplitude under water at z depth

E_o is the amplitude of wave on the surface of water

z is the depth under water

$\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}$

$\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}$

$\alpha =21.07\ Np/m$

now ,

$1 \times 10^{-6} = 20 e^{-21.07\times z}$

$e^{21.07\times z}= 20\times 10^{6}$

taking ln both side

21.07 x z = 16.81

z = 0.797

z = 0.8 (approx)

5 0

4 years ago