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3 years ago

What is Newton's 2nd law of motion? * Your answer

Physics

2 answers:

Novay_Z [31]3 years ago

4 0

Acceleration of an object is depended upon the net force acting open the object and the mass of the object

Andrei [34K]3 years ago

3 0

Answer:

Newton's second law describes the affect of net force and mass upon the acceleration of an object.

F = force

m = mass of an object

a = acceleration

Explanation:

f=m.a

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We think of our muscular system as the system that helps us move around. It also functions on an internal level by helping food

trapecia [35]

Stomach;small intenstines

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3 years ago

Suppose you pour water into a container until it reaches a depth of 12 cm. Next, you carefully pour in an 8.5 cm thickness of ol

Phantasy [73]

Answer:

the pressure at the bottom is approximately 103264 Pa

Explanation:

From Pascal's law, the pressure at the bottom of the container is the pressure from the atmosphere and the columns of water and olive , therefore

Pressure at the bottom = Pressure at the surface of the liquid ( atmospheric pressure) + pressure of the column of water + pressure of the column of olive oil

since

- Pressure at the surface of the liquid = atmospheric pressure = 101325 Pa

- pressure of the column of water = density of water * gravity * level of water

= 1000 kg/m³ * 9.8 m/s² * 0.12 m = 1176 Pa

- pressure of the column of olive oil= density of olive oil* gravity * level of olive oil = 916 kg/m³ * 9.8 m/s² * 0.085 m = 763 Pa

therefore

Pressure at the bottom = 101325 Pa + 1176 Pa + 763 Pa = 103264 Pa

density of olive oil was taken from internet sources ( we can check that is lower than the one of water, and thus it floats )

3 0

3 years ago

A space probe is coasting through space at a steady speed of 100p feet per second. the booster rocket fires for 1/2 seconds so t

aleksandr82 [10.1K]

Answer:

Acceleration: $9800 ft/s^2$

Explanation:

The acceleration of an object is equal to the rate of change of velocity:

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity to change from u to v

For the space probe in this problem, we have:

u = 100 ft/s (initial velocity)

v = 5000 ft/s (final velocity)

t = 0.5 s (time taken)

Therefore, the acceleration is

$a=\frac{5000-100}{0.5}=9800 ft/s^2$

6 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

Represent the following sentence as an algebraic expression, where "a number" is the

agasfer [191]

Answer:

2 (x+3)

Explanation:

8 0

3 years ago