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Phoenix [80]
3 years ago
12

A skateboarder rolls off a ledge that is 1.12 m high, and lands 1.48 m from the base of the edge. How much time was he in the ai

r?
Physics
1 answer:
nignag [31]3 years ago
3 0

The rider's horizontal motion, and how much ground he covers before he hits it, have nothing to do with how long he takes to hit the ground.  The problem is simply:  "How long does it take an object to fall 1.12 m from rest ?"

This seems like a good time to use this formula:

Distance fallen from rest = (1/2) (acceleration) (time)²

The problem doesn't tell us what planet the skateboarder is exercising  on. I'm going to assume it's on Earth, where the acceleration of gravity is 9.8 m/s².  And now, here's the solution to the problem I just invented:

1.12 m = (1/2) (9.8 m/s²) (time)²

Time²  = (1.12 m) / (9.8 m/s²)

Time²  = 0.1143 sec²

Time = √(0.1143 sec² )

<em>Time = 0.34 second</em>

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iren2701 [21]

Answer:

K.E=29.403125J

Explanation:

From the question we are told that

Mass M=100

Height 50-20=30m

Generally the equation for velocity before impact is is is mathematically given by

v=\sqrt{2gh}

v=\sqrt{2*9.8*30}

v=24.25

Generally the equation for Kinetic Energy is is mathematically given by

K.E=\frac{1}{2}mv^2

K.E=\frac{1}{2}*100*(24.25)^2\\

K.E=29403.125J

K.E=29.403125J

8 0
3 years ago
A wildlife researcher is tracking a flock of geese. The geese fly 4.0 km due west, then turn toward the north by 40º and fly ano
Kobotan [32]

Wildlife researcher starts from a and then reaches b, he turns towards north 40 degree to move towards c.

Total displacement is ac

Total horizontal displacement = 4+4 cos40 =7.06 km

Total vertical displacement = 4 sin40 =2.57 km

Total displacement = \sqrt{7.06^2+2.57^2} = 7.51 km

7 0
3 years ago
Which energy transformation converts energy from the sun’s core to light energy needed by plants?
Assoli18 [71]
Nuclear to electromagnetic
3 0
3 years ago
Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar sy
BaLLatris [955]

Answer:

E) are almost circular, with low eccentricities.

Explanation:

Kepler's laws establish that:

All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).

A planet describes equal areas in equal times (Kepler's second law).

The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).

T^{2} = a^{3}

Where T is the period of revolution and a is the semi-major axis.

Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.

However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)

In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).

Therefore, option A and B can not be true.

In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.  

4 0
4 years ago
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