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arsen [322]
3 years ago
13

a rocket, initially at rest, steadily gains speed at a rate of 13.0m/s^2 for 6.40 during take-off. How far did the rocket travel

during take off?​
Physics
1 answer:
Aliun [14]3 years ago
3 0

Explanation:

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lutik1710 [3]

Answer:20cm

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3 years ago
A wave passes along the surface of the water in a ripple tank. Describe the motion of a molecule on the surface as the water pas
laila [671]

Answer:

Explanation:

Water waves are generally a transverse wave which do not cause permanent displacement of molecules of the medium. Transverse waves are waves in which the direction of propagation of the wave is perpendicular to the direction of vibration of the particles of the medium.

As the wave propagates from one point to another on the surface of water transferring energy, a molecule of water on its surface vibrates upwards and downwards. Its motion is perpendicular to the direction of propagation of the wave. After the vibration, it comes back to its initial position.

8 0
3 years ago
In Linnean taxonomy, the most unique level is _____.<br><br> species<br> genus<br> kingdom<br> breed
Nookie1986 [14]
The answer is species, I hope this helps!
6 0
3 years ago
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What is electricity
fgiga [73]
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7 0
3 years ago
Arocket launches at an angle of 33.6 degrees from the horizontal at a
babymother [125]

Answer:

Y component = 32.37

Explanation:

Given:

Angle of projection of the rocket is, \theta=33.6

Initial velocity of the rocket is, u=58.5

A vector at an angle \theta with the horizontal can be resolved into mutually perpendicular components; one along the horizontal direction and the other along the vertical direction.

If a vector 'A' makes angle \theta with the horizontal, then the horizontal and vertical components are given as:

A_x=A\cos \theta(\textrm{Horizontal or X component})\\A_y=A\sin \theta(\textrm{Vertical or Y component})

Here, as the velocity is a vector quantity and makes an angle of 33.6 with the horizontal, its Y component is given as:

u_y=u\sin \theta

Plug in the given values and solve for u_y. This gives,

u_y=(58.5)(\sin 33.6)\\u_y=58.5\times 0.55339\\u_y=32.373\approx32.37(\textrm{Rounded to two decimal places})

Therefore, the Y component of initial velocity is 32.37.

4 0
3 years ago
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