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umka21 [38]
3 years ago
15

Pick an activity you enjoy, such as running or riding a scooter, and describe how newton's laws apply to that activity.

Physics
2 answers:
deff fn [24]3 years ago
5 0
I like playing basketball. So I'm the object in motion. Until an unbalanced force comes and hits me I fall and stay at rest. 
Solnce55 [7]3 years ago
4 0
If you are running, your feet always go down because of gravity, same with scootering, the scooter stays on the ground because of gravity, same with volleyball and any other sport. How you score in volleyball is when the ball hits the ground on the other team's side, when the ball finally falls to the ground, it is because of gravity
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At time t=0, a particle is located at the point (3,6,9). It travels in a straight line to the point (5,2,7), has speed 8 at (3,6
Elis [28]

The particle has constant acceleration according to

\vec a(t)=2\,\vec\imath-4\,\vec\jmath-2\,\vec k

Its velocity at time t is

\displaystyle\vec v(t)=\vec v(0)+\int_0^t\vec a(u)\,\mathrm du

\vec v(t)=\vec v(0)+(2\,\vec\imath-4\,\vec\jmath-2\,\vec k)t

\vec v(t)=(v_{0x}+2t)\,\vec\imath+(v_{0y}-4t)\,\vec\jmath+(v_{0z}-2t)\,\vec k

Then the particle has position at time t according to

\displaystyle\vec r(t)=\vec r(0)+\int_0^t\vec v(u)\,\mathrm du

\vec r(t)=(3+v_{0x}t+t^2)\,\vec\imath+(6+v_{0y}t-2t^2)\,\vec\jmath+(9+v_{0z}t-t^2)\,\vec k

At at the point (3, 6, 9), i.e. when t=0, it has speed 8, so that

\|\vec v(0)\|=8\iff{v_{0x}}^2+{v_{0y}}^2+{v_{0z}}^2=64

We know that at some time t=T, the particle is at the point (5, 2, 7), which tells us

\begin{cases}3+v_{0x}T+T^2=5\\6+v_{0y}T-2T^2=2\\9+v_{0z}T-T^2=7\end{cases}\implies\begin{cases}v_{0x}=\dfrac{2-T^2}T\\\\v_{0y}=\dfrac{2T^2-4}T\\\\v_{0z}=\dfrac{T^2-2}T\end{cases}

and in particular we see that

v_{0y}=-2v_{0x}

and

v_{0z}=-v_{0x}

Then

{v_{0x}}^2+(-2v_{0x})^2+(-v_{0x})^2=6{v_{0x}}^2=64\implies v_{0x}=\pm\dfrac{4\sqrt6}3

\implies v_{0y}=\mp\dfrac{8\sqrt6}3

\implies v_{0z}=\mp\dfrac{4\sqrt6}3

That is, there are two possible initial velocities for which the particle can travel between (3, 6, 9) and (5, 2, 7) with the given acceleration vector and given that it starts with a speed of 8. Then there are two possible solutions for its position vector; one of them is

\vec r(t)=\left(3+\dfrac{4\sqrt6}3t+t^2\right)\,\vec\imath+\left(6-\dfrac{8\sqrt6}3t-2t^2\right)\,\vec\jmath+\left(9-\dfrac{4\sqrt6}3t-t^2\right)\,\vec k

4 0
3 years ago
Question 2 (ID=81813)
zaharov [31]

Answer:

The answer is B

Explanation:

5/2=2.5

2.5x2=5

Hope this helps ik its kinda confusing lol

7 0
2 years ago
Read 2 more answers
Is the substance an atom or a molecule?
ANTONII [103]

Answer:

A pair of oxygen atoms is a molecule of oxygen. A molecule is the smallest particle of a substance that exists independently. Molecules of most elements are made up of only one of atom of that element. Oxygen, along with nitrogen, hydrogen, and chlorine are made up of two atoms.

Explanation:

6 0
2 years ago
If a cannonball is fired horizontally it will not go in a straight line why?
goldfiish [28.3K]

Explanation:

A projectile (Cannon ball) is launched at an angle to the horizontal and rises up to a peak while moving horizontally. When it reaches the peak, the projectile starts to fall.

5 0
2 years ago
Hydraulic engineers often use, as a unit of volume of water, the "acre-foot", defined as the volume of water that will cover 1 a
Alex17521 [72]

Answer:

Volume = 1,015 acre-feet (Approx)

Explanation:

Given:

Rain = 1.7 in

Time = 30 min

Area = 29 km²

Find:

Volume in acre-feet

Computation:

1 km = 1,000 m

1 m = 3.28 feet

1 km² = 247.105 acre

d = 1.7 in = 1.7 / 12 = 0.14167 ft

Area = 29 × 247.105 = 7,166.045 acre

Volume = 7,166.045 acre × 0.14167 ft

Volume = 1,015 acre-feet (Approx)

7 0
3 years ago
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