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2 years ago

What is the amount of displacement of a runner who runs exactly 2 laps around a 400 meter track?

2 answers:

5 0

Displacement is the distance and direction from the start point to the end point. Our runner finished exactly where he started. His displacement is zero.

BartSMP [9]2 years ago

5 0

Answer:

Total displacement is 0.

Explanation:

The displacement of runner is equal to the shortest path covered by it. It is equal to final position minus initial position.

While distance covered by runner is equal to the total path covered by it.

Here, we have to find the displacement of a runner who runs exactly 2 laps around a 400 meter track.

In this case, the amount of displacement in 2 laps is zero because the runner is exactly at initial point after 2 laps.

Hence, the correct option is (b).

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A car travels 500m in 50s, then 1,500m in 75s. Calculate its averages speed for the whole journey

SIZIF [17.4K]

Answer:

15m/s

Explanation:

500 ÷ 50 = 10m/s

1500 ÷ 75 = 20m/s

10 + 20 = 30

30 ÷ 2 = 15m/s

8 0

2 years ago

Read 2 more answers

What is the total energy q released in a single fusion reaction event for the equation given in the problem introduction? use c2

kkurt [141]

Answer:

$4.3\cdot 10^{-12} J$

Explanation:

The fusion reaction in this problem is

$4^1_1H \rightarrow ^4_2He +2e^+$

The total energy released in the fusion reaction is given by

$\Delta E = c^2 \Delta m$

where

$c=3.0\cdot 10^8 m/s$ is the speed of light

$\Delta m$ is the mass defect, which is the mass difference between the mass of the reactants and the mass of the products

For this fusion reaction we have:

$m(^1_1H)=1.007825u$ is the mass of one nucleus of hydrogen

$m(^4_2 He)=4.002603u$ is the mass of one nucleus of helium

So the mass defect is:

$\Delta m =4m(^1_1 H)-m(^4_2 He)=4(1.007825u)-4.002603u=0.028697u$

The conversion factor between atomic mass units and kilograms is

$1u=1.66054\cdot 10^{-27}kg$

So the mass defect is

$\Delta m =(0.028697)(1.66054\cdot 10^{-27})=4.765\cdot 10^{-29}kg$

And so, the energy released is:

$\Delta E=(3.0\cdot 10^8)^2(4.765\cdot 10^{-29})=4.3\cdot 10^{-12} J$

6 0

2 years ago

The wind blows a jay bird south with a force of 300 Newtons. The

adoni [48]

Answer:

F = 316.22 N

Explanation:

Given that,

The wind blows a jay bird south with a force of 300 Newtons.

The jay bird flies north, against the wind, with a force of 100 newtons.

Both the forces are acting perpendicular to each other. The net force is given by the resultant of forces as follows :

$F=\sqrt{300^2+100^2} \\\\F=316.22\ N$

Hence, the net force on the jay bird is 316.22 N.

6 0

2 years ago

A 0.56 kg model rocket produces 53 newtons of upward thrust as it shoots upward off the launch pad. With what acceleration meter

Yanka [14]

resultant force = thrust – weight

acceleration = resultant force (newtons, N) divided by mass (kilograms, kg).

Acceleration = resultant force divided by mass

53N/0.56

=94.64 approximately 95

= 95m/s^2

This means that, every second, the speed of the rocket increases by 95m/s2

the S.I unit of Acceleration is meter per second square.

6 0

1 year ago

Let this oscillator have the same energy as a mass on a spring, with the same k and m, released from rest at a displacement of 5

allochka39001 [22]

There are a lot of same examples that you may have worked before, where the mass on a spring uses a classics when it comes to mechanics. So in this system, always put in your mind that there is an enormous quantum standard that one can use in the equation. It should be 2.10x10 raise to a negative sixth. J.

3 0

1 year ago