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2 years ago

Jane, looking for Tarzan, is running at top speed (6.0 m/s ) and grabs a vine hanging vertically from a tall tree in the jungle.

Physics

1 answer:

elena-s [515]2 years ago

8 0

(a) The maximum height reached by Jane is 1.8 m.

(b) The length of the vine will affect the time of her motion, which will impact on speed and maximum height attained.

<h3>Maximum height Jane can swing</h3>

apply the principle of conservation of energy;

P.E = K.E

mgh = ¹/₂mv²

h = v²/2g

where;

v is speed of jane
g is acceleration due to gravity

h = (6²)/(2 x 9.8)

h = 1.84 m

<h3>Time of motion of Jane</h3>

Assuming Jane to be in simple harmonic motion, the time of motion is calculated as;

T = 2π√(L/g)

where;

L is the length of the vine

Thus, the length of the vine will affect the time of her motion, which will impact on speed and maximum height attained.

Learn more about maximum height here: brainly.com/question/12446886

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Explanation:

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If two balls have the same volume,

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Here, we are required to find the relationship between balls of different mass(a measure of weight) and different volumes.

1. Ball A will have the greater density
2. Ball C and Ball D have the same density.
3. Ball Q will have the greater density.
4. Ball X and Y will have the same density

The density of an object is given as its mass per unit volume of the object.

Mathematically;.

Density = Mass/Volume.

For Case 1:

Va = Vb and Ma = 2Mb
D(b) = (Mb)/(Vb) and D(a) = 2(Mb)/Vb
Therefore, the density of ball A,
D(a) = 2D(b).
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For Case 2:

Vc = 3Vd,

Vd = (1/3)Vc

Md = (1/3)Mc

D(c) = (Mc)/(Vc) and D(d) = (1/3)Md/(1/3)Vd

D(c) = D(d).

Therefore, ball C and D have the same density

For Case 3:

Vp = 2Vq and Mp = Mq
D(p) = (Mq)/2(Vq) and D(q) = (Mq)/Vq
Therefore, the density of ball P is half the density of ball Q
Therefore, ball Q has the greater density.

For case 4:

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Therefore, Ball X and Ball Y have the same density.

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3 years ago

A projectile fired from a gun has initial horizontal and vertical components of velocity equal to 60 m/s and 80 m/s, respectivel

DerKrebs [107]

This question involves the concepts of projectile motion and launch speed.

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(b) The launch angle of the projectile is "53.13°".

<h3>(a) LAUNCH SPEED</h3>

A projectile motion is a motion that takes place on both x and y axes, simultaneously. In this motion the initial launch speed is given by the following formula:

$v_o=\sqrt{v_{ox}^2+v_{oy}^2}$