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Maurinko [17]
2 years ago
13

A moving van travels 10km North, then 4 km east, drops off some furniture and then drives 8 km south. (a) Sketch the path of the

moving van. (2) How far is the van from its starting point? (Hint: it is the magnitude of the displacement vector).

Physics
1 answer:
Juli2301 [7.4K]2 years ago
3 0

Answer:

4.47 km

Explanation:

If we draw the path of the van then we get a shape with two exposed points A and D. If we draw a line from point D perpendicular to BA we get point E. This gives us a right angled triangle ADE.

From Pythagoras theorem

AD² = AE² + ED²

AD=\sqrt{AE^2+ED^2}\\\Rightarrow AD=\sqrt{2^2+4^2}\\\Rightarrow AD=\sqrt{20}\\\Rightarrow AD=4.47\ km

Hence, the van is 4.47 km from its initial point

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While driving, your car has an initial position of 3.2 m, an initial velocity of -8.4 m/s, and
KIM [24]

Answer:

The position of the car at t = 1.5 s is at -8.1625 meters

Explanation:

The initial position of the car is 3.2 meters

The initial velocity is -8.4 m/s

The constant acceleration is 1.1 m/s²

We need to find the final position of the car at the time t = 1.5 seconds

The displacement <em>s</em> = final position - initial position

s=ut+\frac{1}{2}at^{2}, where <em>u</em> is the initial velocity, <em>a</em> is the

constant acceleration and <em>t</em> is the time

So we can find the final velocity by using the rule:

final position - initial position = ut+\frac{1}{2}at^{2}

initial position = 3.2 meters , u = -8.4 m/s , a = 1.1 ²m/s , t = 1.5 s

Substitute these values in the rule

final position - 3.2 = (-8.4)(1.5)+\frac{1}{2}(1.1)(1.5)^{2}

final position - 3.2 = -12.6 + 1.2375

final position - 3.2 = -11.3625

add 3.2 for both sides

final position = -8.1625

<em>That means the car is at 8.1625 meters in opposite direction</em>

<em>The position of the car at t = 1.5 s is at -8.1625 meters </em>

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3 years ago
Why you do scientists collect multiple trials?
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electrical energy sometimes.

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Explanation:

The expansion of the universe is the increase in distance between any two given ... Metric expansion is a key feature of Big Bang cosmology, is modeled mathematically with the ...

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2 years ago
Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8
umka2103 [35]

Answer:

a) F=2.048\times 10^{-7}\ N

b) a=0.1138\ m.s^{-2}

Explanation:

Given:

  • mass of raindrops, m=1.8\times 10^{-6}\ kg
  • charge on the raindrops, q=+21\times 10^{-12}\ C
  • horizontal distance between the raindrops, r=0.0044\ m

A)

<u>From the Coulomb's Law the force between the charges is given as:</u>

F=\frac{1}{4\pi.\epsilon_0} .\frac{q_1.q_2}{r^2}

we have:

\epsilon_0=8.854\times 10^{-12}\ C^2.N^{-1}.m^{-2}

<em>Now force:</em>

F=\frac{1}{4\pi\times 8.854\times 10^{-12}} .\frac{21\times 10^{-12}\times 21\times 10^{-12}}{0.0044^2}

F=2.048\times 10^{-7}\ N

B)

<u>Now the acceleration on the raindrops due to the electrostatic force:</u>

a=\frac{F}{m}

a=\frac{2.048\times 10^{-7}}{1.8\times 10^{-6}}

a=0.1138\ m.s^{-2}

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2 years ago
Read 2 more answers
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