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3 years ago

which of these waves requires a medium to travel through light wave electromagnetic waves sound wave microwave

Physics

2 answers:

natita [175]3 years ago

9 0

Answer:

hello! the answer is c sound wave

Explanation:

i took the test ;)

navik [9.2K]3 years ago

7 0

Sound waves require a medium to travel through. :D

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on a standard position vs time graph, the distance between two points on the horizontal axis represents the time interval, and t

Stels [109]

Answer: true

Explanation: I got the question and guessed true and it was right

3 0

2 years ago

Determine the magnitude of the force between two 42 m-long parallel wires separated by 0.03 m, both carrying 6.3 A in the same d

xz_007 [3.2K]

Answer:

The magnitude of the force between the two parallel wires is 0.0111 N.

Explanation:

Given;

length of the two parallel wires, L = 42 m

distance between the two wires, r = 0.03 m

current in both wires, I₁, I₂ = 6.3 A

Therefore, the magnitude of the repulsive force between the two parallel wires is given by;

$F = \frac{\mu_0 I_1I_2l}{2\pi r}\\\\where;\\\mu_0 \ is \ permeability \ of \ free \ space = 4\pi *10^{-7} \ T.m/A \\\\F = \frac{(4\pi *10^{-7})(6.3)^2(42)}{2\pi (0.03)}\\\\F = 0.0111 \ N$

Therefore, the magnitude of the force between the two parallel wires is 0.0111 N.

6 0

3 years ago

A farmer lifts his hay bales into the top loft of his barn by walking his horse forward with a constant velocity of 1 ft/s. Dete

Lesechka [4]

Answer:

The velocity of the hay bale is - 0.5 ft/s and the acceleration is $6.25\times 10^{- 3} ft/s^{2}$

Solution:

As per the question:

Constant velocity of the horse in the horizontal, $v_{x} = 1 ft/s$

Distance of the horse on the horizontal axis, x = 10 ft

Vertical distance, y = 20 ft

Now,

Apply Pythagoras theorem to find the length:

$20^{2} + 10^{2} = l^{2}$

$l^{2}= 500$

Now,

$x^{2} + y^{2} = 500$ (1)

Differentiating equation (1) w.r.t 't':

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$

$x\frac{dx}{dt} = - y\frac{dy}{dt}$

where

$\frac{dx}{dt}$ = Rate of change of displacement along the horizontal

$\frac{dy}{dt}$ = Rate of change of displacement along the vertical

$v_{x}$ = velocity along the x-axis.

$v_{y}$ = velocity along the y-axis

$xv_{x} = -yv_{y}$

$v_{y} = - 10\times \frac{1}{20} = - 0.5 ft/s$

$|v_{y}| = 0.5\ ft/s$

Acceleration of the hay bale is given by the kinematic equation:

$v_{y}^{2} = u_{y} + 2ay$

$(-0.5)^{2} =0 + 2ay$

$0.25 = 2ay$

$\frac{0.25}{2y} = a$

$a = \frac{0.25}{2\times 20} = 6.25\times 10^{- 3} ft/s^{2}$

7 0

3 years ago

What kind of water temperature lowers the dissolved oxygen

ehidna [41]

<h3>Answer</h3>

At a high temperature above 20° oxygen solubility starts to decrease.

<h3>Explanation</h3>

Oxygen, O2 is a very essential component of water as we can see in its chemical formula h2O.

The solubility of oxygen decreases as temperature increases. This means that warmer water will have less dissolved oxygen than does cooler water.

<h3>Other factors that affects oxygen solubility in water</h3>

Salt levels

higher the salt levels in water, lower will be oxygen in it.

Pressure

Water at lower altitudes can hold more dissolved oxygen than water at higher altitudes because dissolved oxygen will increase as pressure increases.

5 0

3 years ago

Read 2 more answers

A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the

BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

$F_g=\frac{Gm_1m_2}{r^2}$ where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

$F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2}$ I'm going to do the math on the top and then on the bottom and divide at the end.

$F_g=\frac{2.4012*10^{40}}{3.721*10^{23}}$ and now when I divide I will express my answer to the correct number of sig dig's:

$Fg=$ 6.45 × 10¹⁶ N

8 0

3 years ago