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2 years ago

7

this is how i made the map , i’m not sure where to put the symbols but can someone help me now answer the last few questions i p

osted for it question 3,4,5 and 6 please

1 answer:

N76 [4]2 years ago

8 0

Answer:

hi my name is bill im here to help no problem 1+1=2

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Mr. hitch taught us about sedimentary, metamorphic, and igneous rocks. he described how they were formed, what they contain, and

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Mr. Hitch taught us about sedimentary, metamorphic, and igneous rocks. He described how they were formed, what they contain, and showed us samples of each. He is a good geologist.

The missing word and answer is: geologist.

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2 years ago

If the ball that is thrown downward has an acceleration of magnitude aaa at the instant of its release (i.e., when there is no l

kakasveta [241]

Explanation:

At the instant of release there is no force but an acceleration of a, this means the ball is falling freely under the force of gravity. Then the acceleration would be due to force of gravity and acceleration a = g =9.81 m/s^2.

g= acceleration due to gravity

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2 years ago

What is the shape of the orbit of satellites?

Rainbow [258]

All objects in orbit must follow the path of an ellipse (one of Keplers laws)

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3 years ago

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

A particle starts from the origin at t = 0 with an initial velocity of 4.8 m/s along the positive x axis.If the acceleration is

tia_tia [17]

Answer:

Part a)

Velocity = 6.9 m/s

Part b)

Position = (3.6 m, 5.175 m)

Explanation:

Initial position of the particle is ORIGIN

also it initial speed is along +X direction given as

$v_x = 4.8 m/s$

now the acceleration is given as

$\vec a = -3.2 \hat i + 4.6 \hat j$

when particle reaches to its maximum x coordinate then its velocity in x direction will become zero

so we will have

$v_f = v_i + at$

$0 = 4.8 - 3.2 t$

$t = 1.5 s$

Part a)

the velocity of the particle at this moment in Y direction is given as

$v_f_y = v_i + at$

$v_f_y = 0 + 4.6(1.5)$

$v_f_y = 6.9 m/s$

Part b)

X coordinate of the particle at this time

$x = v_x t + \frac{1}{2}a_x t^2$

$x = 4.8(1.5) - \frac{1}{2}(3.2)(1.5^2)$

$x = 3.6 m$

Y coordinate of the particle at this time

$y = v_y t + \frac{1}{2}a_y t^2$

$y = 0(1.5) + \frac{1}{2}(4.6)(1.5^2)$

$y = +5.175 m$

so position is given as (3.6 m, 5.175 m)

5 0

3 years ago