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2 years ago

Tortoise and the Hare

Physics

2 answers:

DedPeter [7]2 years ago

4 0

D.......................

Tanya [424]2 years ago

3 0

The answer would be D. The tortoise moved at a constant velocity throughout the race; the hare stopped to rest periodically.

Hope this helped. Good luck!

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Two forces (4N and 3N) pull your the left, while a 12N force pulls to the right. What is the net force

Lapatulllka [165]

Answer:

The answer is 11N to the right

Explanation:

Because 4N-3N= 1N

Therefore, 12N-1N=11N

The netforce is 11N to the right, because the greatest force is 12N to the right so it is more likely that the object is being pulled to the right.

8 0

2 years ago

The x component of vector is -27.3 m and the y component is +43.6 m. (a) What is the magnitude of ? (b) What is the angle betwee

lara [203]

Answer:51.44 units

Explanation:

Given

x component of vector is $-27.3\hat{i}$

y component of vector is $43.6\hat{j}$

so position vector is

$r=-27.3\hat{i}+43.6\hat{j}$

Magnitude of vector is

$|r|=\sqrt{27.3^2+43.6^2}$

$|r|=\sqrt{2646.25}$

|r|=51.44 units

Direction

$tan\theta =\frac{43.6}{-27.3}=-1.597$

vector is in 2nd quadrant thus

$180-\theta =57.94$

$\theta =122.06^{\circ}$

4 0

2 years ago

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity v is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$

Solving for M, we get

$M = \dfrac{4\pi^2 r^3}{GT^2}$

Putting this expression back into Eqn(1), we get

$R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}$

$\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}$

$\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}$

$\:\:\:\:= 2.6×10^6\:\text{m}$

5 0

2 years ago

Mario rdoll a coin up a slope at 2 m/s. It travels 2.7 m, comes to a stop and rolls back down. What is the coin's acceleration?

leva [86]

The coin's acceleration is 0.37 m/s²

Acceleration is the rate of change of the velocity of an item with appreciation to time. Accelerations are vector portions. The orientation of an item's acceleration is given by the orientation of the net pressure appearing on that object.

Calculation:-

V² = U -2aS

a = U/2S

= 2/2×2.7

= 0.37 m/s²

Acceleration is the charge at which velocity modifications with time, in terms of each speed and route. A factor or an object moving in a straight line is accelerated if it quickens or slows down. movement on a circle is extended despite the fact that the rate is consistent because the course is continually changing.

Learn more about acceleration here:- brainly.com/question/29110429

#SPJ9

8 0

11 months ago

The particles in an object move in (many/ two directions)

pentagon [3]

Answer:

They are negitive

Explanation:

Im pretty sure

8 0

3 years ago