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3 years ago

What is the motion of the particles in this kind of wave?

Physics

2 answers:

Musya8 [376]3 years ago

7 0

The answer to this question is B I think

adelina 88 [10]3 years ago

3 0

I think B. Let me know if I’m wrong. Have a great day❤️

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Standing and holding a barbell that has a mass of 100kg at a height of 2m involves being done on the barbell to maintain

Basile [38]

Answer:

1960Joules

Explanation:

Since we are not told what too find, we can as well find the Gravitational Potential Energy.

GPE = mass * acceleration due to gravity * height

GPE = 100*9.8 * 2

GPE = 980*2

GPE = 1960Joules

Hence the gravitational potential Energy is 1960Joules

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3 years ago

Practicing good human relations skills and having a positive attitude will

skad [1K]

Have a positive effect on you as a human being. Your attitude will improve, your charisma will spread, and you will a better, more well rounded person.

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4 years ago

Read 2 more answers

When operated on a household 110.0 V line, typical hair dryers draw about 1650 W of power. The current can be modeled as a long,

Andre45 [30]

Explanation:

Given that,

Voltage of household line, V = 110 V

Power of the hairdryer, P = 1650 W

During use, the current is about 1.95 cm from the user's hand.

(a) Power is given by :

$P=V\times I\\\\I=\dfrac{P}{V}\\\\I=\dfrac{1650\ W}{110\ V}\\\\I=15\ A$

(b) Again the power is given by :

$P=\dfrac{V^2}{R}$

R is resistance of the dryer

$R=\dfrac{V^2}{P}\\\\R=\dfrac{(110)^2}{1650}\\\\R=7.34\ \Omega$

(c) The magnetic field produced by the dryer at the user's hand is given by :

$B=\dfrac{\mu_o I}{2\pi r}\\\\B=\dfrac{4\pi \times 10^{-7}\times 15}{2\pi \times 1.95\times 10^{-2}}\\\\B=1.53\times 10^{-4}\ T$

Hence, this is the required solution.

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3 years ago

Unscramble the bolded letters to guess the 6 letter word code. Did you get the poses or exercises correct? MOUNTAIN POSE TRIANGL

nikitadnepr [17]

Explanation:

I don't see a question...

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3 years ago

A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre

Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

$h=\sqrt{(36m)^2+(227m)^2}=229.83m$ (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

$y=y_o+v_ot+\frac{1}{2}gt^2$ (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

$3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s$

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

$v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}$

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

$\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°$

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

4 0

3 years ago