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pishuonlain [190]
3 years ago
13

A car is initially moving at 20 m/s east and a little while later it is moving at 10 m/s north. Which of the following best desc

ribes the orientation of the average acceleration during this time interval?
a.northeast

b.northwest

c.west

d.north of west
Physics
1 answer:
Nina [5.8K]3 years ago
5 0

Answer:d

Explanation:

Given

First car is moving  towards east with velocity 20 m/s

\vec{v_1}=20\hat{i}

then it turns towards north then velocity is  

\vec{v_2}=10\hat{j}

suppose car takes t sec to change its path so average acceleration is given by

a=\frac{v_2-v_1}{t}

a=\frac{1}{t}(10\hat{j}-20\hat{i})

So average  acceleration is towards North of west.

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RideAnS [48]

Answer:

-414.96 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-3.85^2}{2\times 0.75}\\\Rightarrow a=-9.88\ m/s^2

F=ma\\\Rightarrow F=42\times -9.88\\\Rightarrow F=-414.96\ N

The force the ground exerts on the parachutist is -414.96 N

If the distance is shorter than 0.75 m then the acceleration will increase causing the force to increase

5 0
3 years ago
Plz help
katrin2010 [14]

The particles of the medium (slinky in this case) move up and down (choice #2) in a transverse wave scenario.

This is the defining characteristic of transverse waves, like particles on the surface of water while a wave travels on it, or like particles in a slack rope when someone sends a wave through by giving it a jolt.

The other kind of waves is longitudinal, where the particles of the medium move "left-and-right" along the direction of the wave propagation. In the case of the slinky, this would be achieved by giving a tensioned slinky an "inward" jolt. You would see that such a jolt would give rise to a longitudinal wave traveling along the length of the tensioned slinky. Another example of longitudinal waves are sound waves.

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3 years ago
Two parakeets sit on a swing with their combined center of mass 10.0 cm below the pivot. At what frequency do they swing?
Oksanka [162]

Answer:

1.58 Hz

Explanation:

The frequency of the simple pendulum is given by

f = 1/T

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In this problem, I = 10.0 cm = 0.1 m  

f = 1/2π√9.8/0.1

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3 years ago
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Setler [38]

Answer:

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Explanation:

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This law relates the index of refraction of the water (n2), the index of refraction of the air (n1), the incidence angle relative to the vertical (theta1) and the refraction angle relative to the vertical (theta2) by using the next equation:

(n1)*(sin(theta1))=(n2)*(sin(theta2))

Then we will find the refraction angle relative to the vertical this way:

(n1/n2)*(sin(theta1))=sin(theta2)

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Then, theta2=32.12°

Now that we have this information we can imagine a triangle with a 30cm height and a 32.12° angle. This way we can find how much X is, this X will be the distance between the vertical line and the spot the beam hits the bottom, so we can use some trigonometry to find it, this way:

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tan(32.12)=(24cm/Y)

Y=(24cm/tan(32.12))

Then, Y=38cm

In this case, you must add 8cm of water to the tank to move the beam on the bottom 5cm

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