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swat32
3 years ago
12

Runner A is initially 5.7 km west of a flagpole and is running with a constant velocity of 8.8 km/h due east. Runner B is initia

lly 4.2 km east of the flagpole and is running with a constant velocity of 7.6 km/h due west. How far are the runners from the flagpole when their paths cross? Answer in units of km.
Physics
1 answer:
vampirchik [111]3 years ago
4 0

Answer:

0.39 km west

Explanation:

The position of Runner A is:

x = -5.7 + 8.8 t

The position of Runner B is:

x = 4.2 − 7.6 t

When the positions are equal:

-5.7 + 8.8 t = 4.2 − 7.6 t

16.4 t = 9.9

t = 0.604

Plug into either equation to find the position at this time:

x = -5.7 + 8.8 (0.604)

x = -0.39

The runners are 0.39 km west of the flagpole when they meet.

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Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain d=2.46\times10^{3}\ m

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Distance of the enemy ship from the mountain d'=6.10\times10^{2}\ m

Initial velocity v=2.55\times10^{2}\ m/s

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

v_{x}=v\cos\theta

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v_{x}=2.55\times10^{2}\cos74.9

v_{x}=66.42\ m/s

We need to calculate the vertical component of initial velocity

Using formula of vertical component

v_{y}=v\sin\theta

Put the value into the formula

v_{y}=2.55\times10^{2}\sin74.9

v_{y}=246.19\ m/s

We need to calculate the time

Using formula of time

t=\dfrac{d}{v_{x}}

t=\dfrac{2.46\times10^{3}}{66.42}

t=37.03\ sec

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

H= v_{y}t-\dfrac{1}{2}gt^2

Put the value in the equation

H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2

H=2397.4\ m

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Question 1 (13 marks) A charge Q is located on the top-left corner of a square. Charges of 2Q, 3Q and 4Q are located on the othe
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Answer:

Explanation:

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Similarly electric field due to charge 3Q near 2Q

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E₂ = 3KQ x 10⁴  j / 25

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= 108.63 ( i + 2 j ) N .

Magnitude of this force

= 108.63 x √5

= 243 N approx .

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3 years ago
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