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marishachu [46]
2 years ago
8

A person jumps from a plane in is falling the person releases a parish you in continues to fall the person lays 35.2 kg and ther

e is a driving force of 45N what is the net for acting on the person
Physics
1 answer:
irakobra [83]2 years ago
7 0

Answer:

cookie

Explanation:

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You leave on a 450 miles trip in order to attend a meeting that will start 10.8 hours after you begin your trip. Along the way y
konstantin123 [22]

Answer:

c. 2.6 h

Explanation:

The longest time spent over dinner is the time that you have available minus the minimum possible time spent in the trip.

The time of the trip is found using:

t = \frac{d}{v}

Where distance is d and velocity is v. The time will be minimum at maximum velocity. Replacing with the data we have:

Ttrip = \frac{450}{55} = 8.1818 h

Tdinner = 10.8h - 8.1818 h = 2.6181h

that aproximates 2.6 h.

4 0
3 years ago
A net force acting on an object will always cause a change in the object's _________.
Nataly [62]
 cause a change in the object's velocity
3 0
3 years ago
why do we not notice a difference in the fall and winter in the amount of oxygen in the air even when leaves fall off
sveticcg [70]
Earths tilt making the sun go haywire lol XD
6 0
3 years ago
A runner starts from rest and achieves a maximum speed of 8.85 m/s.
ycow [4]

Answer:

The appropriate answer is "0.9152 seconds".

Explanation:

The given values are:

Maximum speed,

v = 8.85 m/s

Acceleration,

u = 9.67 m/s²

Now,

⇒ v = u+at

By putting the values, we get

⇒ 8.85=0+9.67t

⇒      t=\frac{8.85}{9.67}

⇒         =0.9152 \ second    

5 0
3 years ago
A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth.
Maksim231197 [3]

Answer:

Explanation:

We shall apply law of  conservation of mechanical energy for projectile being thrown .

Total energy on the surface = total energy at height h required

a ) At height h , velocity = .351 x ( 2 GM/R x h )

\frac{-GM}{R} + \frac{m\times(.351\times\sqrt{2GM})^2 }{2R } = \frac{-GMm}{R+h} + 0

\frac{-GMm}{R} +\frac{1}{2}\times  \frac{-2GMm}{R} \times0.123=\frac{-GMm}{R+h}

\frac{0.877GMm}{R} =\frac{-GMm}{R+h}

h = .14 R

b )

\frac{-GM}{R} + \frac{m\times(.351\times2GM) }{2R } = \frac{-GMm}{R+h} + 0

\frac{-0.649GMm}{R} = \frac{-GMm}{R+h}

h = .54 R

c ) least initial mechanical energy required at launch if the projectile is to escape Earth

= GMm / R + 1/2 m (2GM/R)

= 0

5 0
3 years ago
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