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vodka [1.7K]
2 years ago
5

WAVES AND SOUND

Physics
1 answer:
lawyer [7]2 years ago
6 0

Answer:

i think it is B

Explanation:

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A car travels 82 meters do North and 14 seconds the car turns around and travels 44 m due south in four seconds what is the magn
Yakvenalex [24]

Answer:

2.11 m/s

Explanation:

Take north to be positive and south to be negative.

Average velocity = displacement / time

v = (82 m + -44 m) / (14 s + 4 s)

v = 2.11 m/s

The velocity is positive, so it is 2.11 m/s north.  The magnitude of the velocity is 2.11 m/s.

8 0
3 years ago
An object that is moving must have a change in A) its speed. B) its position. C) its acceleration. D) its applied force.
denis-greek [22]

An object that's moving doesn't necessarily change its speed or acceleration. Also, the force applied to it doesn't need to change ... in fact, a moving object doesn't need ANY force applied to it in order to keep moving.

But any moving object WILL have a change in its position ... THAT's how you know it's moving, and that's WHY you say "It's moving !". (choice-B)

8 0
3 years ago
An ant crawls along a sidewalk with a velocity of 0.1 m/s in a direction that is 45 degrees relative to the edge of the sidewalk
ahrayia [7]

Answer:

C) 1.0 m

Explanation:

The component of the velocity parallel to the sidewalk is:

vₓ = v cos θ

vₓ = 0.1 m/s cos 45°

vₓ = 0.0707 m/s

The distance traveled after 14 seconds is:

d = vₓ t

d = (0.0707 m/s) (14 s)

d = 0.99 m

Closest answer is C) 1.0 m.

6 0
3 years ago
Read 2 more answers
A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part
mestny [16]

Differentiate the components of position to get the corresponding components of velocity :

v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t

v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}

At <em>t</em> = 5.0 s, the particle has velocity

v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}

v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}

The speed at this time is the magnitude of the velocity :

\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}

The direction of motion at this time is the angle \theta that the velocity vector makes with the positive <em>x</em>-axis, such that

\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}

4 0
2 years ago
What is the collective noun of chocolate​
Vedmedyk [2.9K]

Answer:

creamy with milk and chocolate

4 0
3 years ago
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