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3 years ago

A turtoise moves a distance of 100 metres in 15 minutes. What is the average speed of the tortoise in km/h

Physics

1 answer:

Amanda [17]3 years ago

7 0

Answer:

0.4 km/h

Explanation:

Speed is defined as distance moved per unit time and can be expressed in either m/s or Km/h or miles/h

Mathematically,

$s=\frac {d}{t}$

where d represent distance covered and t is time taken

Taking d as 100 m and converting it into km, we divide by 1000 hence d= 0.1 km

Time is 15 minutes and to convert to hours we divide by 60 hence 0.25 hours

Speed, $s=\frac {0.1}{0.25}=0.4 Km/h$

Therefore, the speed is equivalent to 0.4 km/h

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Yes carbon dioxide is needed for photosynthesis while cellular respiration needs oxygen and dispurses carbon dioxide

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3 years ago

What is an effector strain of bacteria

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3 years ago

1. Simplify the following expression<br> 8-6/4-12+3^2

DanielleElmas [232]

Answer:

-5/2

Explanation:

8-6/4-12+3^2

8-6/4-12+9

2/4-12+9

LCM =4

(2-48+36)/4

(-46+36)/4

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3 years ago

If a box of supplies is dropped from the cargo hold of an airplane travelling at an altitude of 4,410 meters how long will it ta

Paraphin [41]

Answer:

It will take 30 seconds to reach the ground, and it will be travelling at 294 m/s when it does so. This means that its average velocity was 147 m/s.

Explanation:

$d=v_ot+\dfrac{1}{2}at^2$

Since the initial velocity of a dropped object is 0, we can make this the equation:

$d=\dfrac{1}{2}at^2 \\\\4410=\dfac{1}{2}(9.8)t^2 \\\\t^2=900$

$t=30\text{ seconds}$

The final velocity can be calculated with the formula:

$v_f=v_o+at$

Once again, since there is no initial velocity:

$v_f=at \\\\v_f=(9.8)(30)=294m/s$

Since the initial velocity is 0, the average vertical velocity is 294/2=147 m/s.

Hope this helps!

5 0

3 years ago

A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit

Masja [62]

Answer:

Wavelength of the incident wave in air = 1 m

Wavelength of the incident wave in medium 2 = 0.33 m

Intrinsic impedance of media 1 = 377 ohms

Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

a) Wavelength of the incident wave in air

The frequency of the electromagnetic wave in air, f = 300 MHz = 3 * 10⁸ Hz

Speed of light in air, c = 3 * 10⁸ Hz

Wavelength of the incident wave in air:

$\lambda_{air} = \frac{c}{f} \\\lambda_{air} = \frac{3 * 10^{8} }{3 * 10^{8}} \\\lambda_{air} = 1 m$

Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

$n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3$

$\lambda_{2} = \frac{c}{nf}\\\lambda_{2} = \frac{3 * 10^{6} }{3 * 3 * 10^{6}}\\\lambda_{2} = 1/3\\\lambda_{2} = 0.33 m$

b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

$n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

$n_1 = \sqrt{\frac{4\pi * 10^{-7} }{8.84 * 10^{-12} } }$

$n_1 = 377 \Omega$

The intrinsic impedance of media 2 is given as:

$n_2 = \sqrt{\frac{\mu_r \mu_0}{\epsilon_r \epsilon_{0} } }$

Permeability of free space, $\mu_{0} = 4 \pi * 10^{-7} H/m$

Permittivity for air, $\epsilon_{0} = 8.84 * 10^{-12} F/m$

ϵr = 9

$n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }$

$n_2 = 125.68 \Omega$

c) The reflection coefficient,r and the transmission coefficient,t at the boundary.

Reflection coefficient, $r = \frac{n - n_{0} }{n + n_{0} }$

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

$r = \frac{3 - n_{0} }{3 + n_{0} }$

Transmission coefficient at the boundary, t = r -1

d) The amplitude of the incident electric field is $E_{0} = 10 V/m$

Maximum amplitudes in the total field is given by:

$E = tE_{0}$ and $E = r E_{0}$

E = 10r, E = 10t

3 0

3 years ago