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2 years ago

Look at diagram above. In what direction does the medium move relative to the direction of the wave? Explain.

Physics

1 answer:

Ainat [17]2 years ago

7 0

Answer:
It goes upward and downward
Explanation:
The way the man is moving the string indicates it goes movement of up and sown, I believe the straight line(B) is a distraction.

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A baseball is thrown horizontally at a rate of 40m/s toward a home plate 18.4 m away. How far below the launch height is the bal

sammy [17]

Answer:

Explanation:

Using the formula for calculating range expressed as;

R = U√2H/g where

R is the distance moves in horizontal direction = 18.4m

H is the height

U is the velocity of the baseball = 40m/s

g is the acceleration due to gravity = 9.8m/s²

Substitute the given parameters into the formula and calculate H as shown;

18.4 = 40√2H/9.8

18.4/40 = √2H/9.8

0.46 = √2H/9.8

square both sides;

(0.46)² = (√2H/9.8)²

0.2116 = 2H/9.8

2H = 9.8*0.2116

2H = 2.07368

H = 2.07368/2

H = 1.03684m

Hence the ball is 1.03684m below the launch height when it reached home plate.

8 0

2 years ago

In which city did the wright brothers take off on their very first flight

Nezavi [6.7K]

<h2>Answer: Kitty Hawk, North Carolina </h2>

The Wright brothers, Wilbur and Orville, were pioneers of aviation, since they flew in a device heavier than air, which was inconceivable at that time.

Their first successful flight was on December 17th, 1903 in Kitty Hawk, North Carolina, which lasted only 12 seconds in which their plane (the Flyer I, with 341 kg, 6.4 m long and a wingspan of 12.3 m) traveled 37 m without touching the ground. This was achieved through the help of an external catapult that "threw" them into the air.

It should be noted that the Wright brothers only studied until high school, however, their passion for solving the problem of the human inability to fly, their perseverance and experience acquired over the years in their bicycle company, led them to reach that goal. An achievement that marked the beginning of the aviation era.

7 0

3 years ago

After exercise your heart rate is

harina [27]

Your heart rate is higher than normal.

7 0

3 years ago

Read 2 more answers

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be:

$\displaystyle q(t) = q_0 \, e^{-t / \tau}$ .

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$ .

$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$ .

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$ .

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$ .

7 0

3 years ago

Two sticky spheres are suspended from light ropes of length LL that are attached to the ceiling at a common point. Sphere AA has

a_sh-v [17]

Answer:

h’ = 1/9 h

Explanation:

This exercise must be solved in parts:

* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy

starting point. Higher

Em₀ = U = m g h

final point. Lower, just before the crash

Em_f = K = ½ m $v_{b}^2$

energy is conserved

Em₀ = Em_f

m g h = ½ m v²

v_b = $\sqrt{2gh}$

* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

p₀ = 2m 0 + m v_b

final instant. Right after the crash

p_f = (2m + m) v

the moment is preserved

p₀ = p_f

m v_b = 3m v

v = v_b / 3

v = ⅓ $\sqrt{2gh}$

* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together

Starting point. Lower

Em₀ = K = ½ 3m v²

Final point. Higher

Em_f = U = (3m) g h'

Em₀ = Em_f

½ 3m v² = 3m g h’

we substitute

h’= $\frac{v^2}{2g}$

h’ = $\frac{1}{3^2} \ \frac{ 2gh}{2g}$

h’ = 1/9 h

6 0

3 years ago