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Burka [1]
3 years ago
6

If a transverse wave of energy travels through the water in this pond ,how wil the floating leaf be moved

Physics
2 answers:
STatiana [176]3 years ago
8 0
Move up and down but stay in the same position
Anna71 [15]3 years ago
4 0
The floating leaf will move up and down but it will remain in horizontal translation equilibrium
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An airplane pilot wishes to fly due west. A wind of 80.0 km/h (about 50 mi/h) is blowing toward the south. (a) If the airspeed o
Vesnalui [34]

Answer:

a) 75.5 degree relative to the North in north-west direction

b) 309.84 km/h

Explanation:

a)If the pilot wants to fly due west while there's wind of 80km/h due south. The north-component of the airplane velocity relative to the air must be equal to the wind speed to the south, 80km/h in order to counter balance it

So the pilot should head to the West-North direction at an angle of

cos(\alpha) = 80/320 = 0.25

\alpha = cos^{-1}(0.25) = 1.32 rad = 180\frac{1.32}{\pi} = 75.5^0 relative to the North-bound.

b) As the North component of the airplane velocity cancel out the wind south-bound speed. The speed of the plane over the ground would be the West component of the airplane velocity, which is

320sin(\alpha) = 320sin(75.5^0) = 309.84 km/h

7 0
3 years ago
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Even when the head is held erect, as in the figure below, its center of mass is not directly over the principal point of support
alexandr1967 [171]

We are asked to determine the force required by the neck muscle in order to keep the head in equilibrium. To do that we will add the torques produced by the muscle force and the weight of the head. We will use torque in the clockwise direction to be negative, therefore, we have:

\Sigma T=r_{M\perp}(F_M)-r_{W\perp}(W)

Since we want to determine the forces when the system is at equilibrium this means that the total sum of torque is zero:

r_{M\perp}(F_M)-r_{W\perp}(W)=0

Now, we solve for the force of the muscle. First, we add the torque of the weight to both sides:

r_{M\perp}(F_M)=r_{W\perp}(W)

Now, we divide by the distance of the muscle:

(F_M)=\frac{r_{W\perp}(W)}{r_{M\perp}}

Now, we substitute the values:

F_M=\frac{(2.4cm)(50N)}{5.1cm}

Now, we solve the operations:

F_M=23.53N

Therefore, the force exerted by the muscles is 23.53 Newtons.

Part B. To determine the force on the pivot we will add the forces we add the vertical forces:

\Sigma F_v=F_j-F_M-W

Since there is no vertical movement the sum of vertical forces is zero:

F_j-F_M-W=0

Now, we add the force of the muscle and the weight to both sides to solve for the force on the pivot:

F_j=F_M+W

Now, we plug in the values:

F_j=23.53N+50N

Solving the operations:

F_j=73.53N

Therefore, the force is 73.53 Newtons.

8 0
1 year ago
A truck is driving north at 35 miles per hour and passes a car going south at 40 miles per hour. What is the speed of the car fr
inna [77]

Answer:

70

Explanation:

right...

7 0
2 years ago
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A physical property of halite (table salt) is its _____.
nexus9112 [7]
Taste: salty
color: varies. ex: white, clear, purple, yellow, etc.
status: mineral
4 0
3 years ago
What is projectile motion​
MakcuM [25]

<h2><u>Projectile</u><u> </u><u>motion</u><u>:</u></h2>

<em>If</em><em> </em><em>an</em><em> </em><em>object is given an initial velocity</em><em> </em><em>in any direction and then allowed</em><em> </em><em>to travel freely under gravity</em><em>, </em><em>it</em><em> </em><em>is</em><em> </em><em>called a projectile motion</em><em>. </em>

It is basically 3 types.

  1. horizontally projectile motion
  2. oblique projectile motion
  3. included plane projectile motion

7 0
3 years ago
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