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2 years ago

7

If a man travels from 0m to 5m then back to -3m, all in four seconds, what is the mans displacement?

1 answer:

yawa3891 [41]2 years ago

7 0

Answer:

-3 m

Explanation:

Displacement is the final position minus the initial position.

Δx = x − x₀

Δx = -3 m − 0 m

Δx = -3 m

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Weight is influenced by which of the following

nignag [31]

Out of the given options, weight is influenced by mass and gravity

Answer: Option A

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A wave is traveling through a medium ad it travels it displaces particles of matter in the same direction as the wave is traveli

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How long does a car (1000 kg) have a speed of 30 m/s from a rest if the engine power is 10kw

Lemur [1.5K]

Answer: 90.1 s

Explanation:

Use equation for power:

P=F*V

Use eqation for force:

F=ma

F---force

V---velocity

Vr=om/s

V=30m/s

m=1000kg

P=10000W

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P=FV

F=P/V

F=10000W/30m/s

F=333.33N

Use equation for force to find accelartaion.

F=ma

a=F/m

a=333.33N/1000kg

a=0.333 m/s²

Use equation for accelaration to find out time:

a=(V-Vs)/t

t=(V-Vs)/a

t=(30m/s)/(0.333m/s²)

t=90.09 s≈90.1 s

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3 years ago

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You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration

jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time $t$ the ball is in air, we can use the following equation:

$y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}$

Where:

$y=0m$ is the final height of the ball

$y_{o}=40 m$ is the initial height of the ball

$V_{o}=10 m/s$ is the initial velocity of the ball

$t$ is the time the ball is in air

$g=9.8 m/s^{2}$ is the acceleration due to gravity

$\theta=30\°$

Then:

$0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}$

$0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}$

Multiplying both sides of the equation by -1 and rearranging:

$4.9 m/s^{2}t^{2}-5m/s t-40 m=0$

At this point we have a quadratic equation of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Where:

$a=4.9$

$b=-5$

$c=-40$

Substituting the known values:

$t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}$

Solving the equation and choosing the positive result we have:

$t=3.41 s$ This is the time the ball is in air

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3 years ago