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3 years ago

Will a sled gain more speed sliding down a 10 m hill or a 40 m hill and why?

Physics

1 answer:

Mrac [35]3 years ago

6 0

Answer:

the 40 m hill

Explanation:

because the longer it goes downhill the more speed it will pick up.

You might be interested in

A 2.72 A current flows through a wire for

agasfer [191]

Current=2.72A=I
Time=t=9.88s
Charge=Q

$\\ \rm\longmapsto I=\dfrac{Q}{t}$

$\\ \rm\longmapsto Q=It$

$\\ \rm\longmapsto Q=2.72\times 9.88$

$\\ \rm\longmapsto Q=26.87C$

8 0

2 years ago

Who has greater displacement, an astronaut who has just completed an orbit of the earth or you when you have just traveled from

Arte-miy333 [17]

Answer:

Explanation:

This is going to sound like an absurd answer, but sometimes physics can be a little strange.

This answer is weird because of the definition of displacement. It means the distance from the starting point to the ending point, disregarding what happened in between. The point is that the astronaut is at the starting point of his orbit. By definition the starting and ending points are the same. His displacement is 0.

So the answer is you have the greater displacement when you walked one way to school. The starting point and the ending point are different. You have gone further.

However just to make things a little nasty, when you walk home again, your displacement will be the same as the astronaut's -- 0 meters because you will be right back where you started from.

7 0

3 years ago

Find your acceleration from 8.3 m/s to 12.5 m/s in 1.24 seconds.

balu736 [363]

The average acceleration for an object undergoing this change in velocity is

(12.5 m/s - 8.3 m/s) / (1.24 s) = (4.2 m/s) / (1.24 s) ≈ 3.4 m/s²

4 0

2 years ago

What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.

valkas [14]

<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (1)

Where;

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M=6.39(10)^{23}kg$ is the mass of Mars

$a=3.4(10)^{6}m$ is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a circular orbit and a low orbit near the surface as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}$ (2)

$T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}$ (3)

$T=\sqrt{11581157.44 s^{2}}$ (4)

Finally:

$T=3403.1099s=56.718min$ This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0

3 years ago

Fighter jet starting from airbase A flies 300 km east , then 350 km at 30° west of north and then 150km north to arrive finally

Tems11 [23]

A jet fighter flies from the airbase A 300 km East to the point M. Then 350 km at 30° West of North.
It means : at 60° North of West. So the distance from the final point to the line AM is :
350 · cos 60° = 350 · 0.866 = 303.1 km
Let`s assume that there is a line N on AM.
AN = 125 km and NM = 175 km.
And finally jet fighter flies 150 km North to arrive at airbase B.
NB = 303.1 + 150 = 453.1 km
Then we can use the Pythagorean theorem.
d ( AB ) = √(453.1² + 125²) = √(205,299.61 + 15,625) = 470 km
Also foe a direction: cos α = 125 / 470 = 0.266
α = cos^(-1) 0.266 = 74.6°
90° - 74.6° = 15.4°
Answer: The distance between the airbase A and B is 470 km.
Direction is : 15.4° East from the North.

4 0

3 years ago