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Andreas93 [3]
3 years ago
11

think about something that has happened to you physically—a fall, a jump, an accident, or something you may have done hundreds o

f times in your favorite sport. Analyze the action and describe it in terms of Newton’s laws. Identify the initial conditions and the forces involved.
Physics
1 answer:
timurjin [86]3 years ago
5 0

My best hobby is driving. Driving has something to do with Newton's first law of motion, which states that an object will continue to be in its state of rest or in uniform motion in a straight line unless it is acted upon by an external force. This law means that an object will continue to be in motion in the same direction unless it is acted upon by a force. Newton's first law of motion is also called the law of inertia.

I usually experience the law of inertia when I am driving my car.

Every morning, for me to move the car from its state of rest to a state of uniform motion, I have to switch on the ignition, which represent an unbalanced force that move the car out of its states of rest. When I am driving, the car continue in motion and in the same direction, unless I apply the brake. The application of the brake is an example of applying an unbalanced force to stop a body in motion.

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Science requires theories that can be tested by research. true false
dezoksy [38]
True Requires the development of theories that can be tested by systematic research.
3 0
3 years ago
In the ground state of hydrogen, according to the Bohr model, an electron orbits 5.3 x 10-11 m from the nucleus. It undergoes a
Readme [11.4K]

Answer:

Explanation:

Given

radius of electron(r)=5.3\times 10^{-11} m

centripetal acceleration (a_c)=9\times 10^22 m/s^2

we know

a_c=\frac{v^2}{r}

v=\sqrt{r\times a_c}

v=\sqrt{5.3\times 10^{-11}\times 9\times 10^{22}}

v=\sqrt{47.7\times 10^11}

v=21.84\times 10^5 m/s

(b)For n=10

r=100\times 5.3\times 10^{-11} m\approx 5.3\times 10^{-9} m

a_c=10^4\times 9\times 10^{22} m/s^2

a_c=9\times 10^{26} m/s^2

v=\sqrt{r\times a_c}

v=\sqrt{9\times 10^{26}\times 5.3\times 10^{-9}}

v=21.84\times 10^8 m/s

8 0
3 years ago
In what way could a random mutation provide an organism with an advantage? With a example please
saveliy_v [14]

Answer:

They are called beneficial mutations. They lead to new versions of proteins that help organisms adapt to changes in their environment. Beneficial mutations are essential for evolution to occur. They increase an organism's changes of surviving or reproducing, so they are likely to become more common over time.

Explanation:

7 0
2 years ago
on aircraft carriers, catapults are used to accelerate jet air craft to flight speeds in a short distance. One such catapult tak
sineoko [7]

Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) - (speed at the beginning)

The jet's change in speed = (70 m/s) - (zero) = 70 m/s

So acceleration = (70 m/s) / (2.5 s)

Acceleration = (70 / 2.5) m/s²

<em>Acceleration = 28 m/s²</em>  

That's about 2.9 G's .  Jet pilots can endure a lot more than that, but maybe the catapult or the hook on the airplane can't.  Let's look a little closer:

F = m A (Newton #2)

The force on the airplane = (18,000 kg) x (28 m/s²)

Force on the airplane = 504,000 Newtons

That's about 113,000 pounds !  Maybe the part of the airplane that the catapult pushes on can't handle any more force than that.  Or maybe that's the most force the catapult can deliver.

Also, the REACTION force on the catapult is the same 113,000 pounds.  Maybe the hooks or the chains or the struts on the catapult can't handle any more force than that.

That's almost 57 tons for gosh sakes !  Maybe the DECK of the carrier can't handle more force than that, and that's why they can't launch the airplane with acceleration of more than 2.9 G's .

8 0
3 years ago
How far does a car travel in 90 seconds if it’s traveling 55 m/s? Show equation
podryga [215]

You just said the car is traveling at the speed of 55 m/s.  If I understand this  correctly, that means the car will cover:

55 meters in the first second,

55 meters in the 2nd second,

55 meters in the 3rd second,

55 meters in the 4th second,

55 meters in the 5th second,

.

.

.

55 meters in the 87th second,

55 meters in the 88th second,

55 meters in the 89th second, and

55 meters in the 90th second.

That's 55 meters 90 times.  If you just move these words around a little bit, it says "90 times 55 meters" . . . a pretty simple arithmetic problem.

The equation is . . . <em>Distance = (55 m/s) times (time, in seconds)</em> .

I get <em>4,953 meters</em>.  You should check me on this.

8 0
3 years ago
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