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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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Answer:

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<u>Geologists combine both laboratory and field data to illustrate the results of their research. Some observations that can the geologist make by working outdoors instead of in a lab are as follows:</u>

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Lactic acid is produced during low-moderate exercise <br> True <br> False
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damaskus [11]

Answer:

a=3125000 m/s^2\\a=3.125*10^6 m/s^2

Acceleration, in m/s, of such a rock fragment = 3.125*10^6m/s^2

Explanation:

According to Newton's Third Equation of motion

V_f^2-V_i^2=2as

Where:

V_f is the final velocity

V_i is the initial velocity

a is the acceleration

s is the distance

In our case:

V_f=V_{escape},  V_i=0,s=4 m

So Equation will become:

V_{escape}^2-V_i^2=2as\\V_{escape}^2-0=2as\\V_{escape}^2=2as\\a=\frac{V_{escape}^2}{2s}\\a=\frac{(5*10^3m)^2}{2*4}\\a=3125000 m/s^2\\a=3.125*10^6 m/s^2

Acceleration, in m/s, of such a rock fragment = 3.125*10^6m/s^2

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3 years ago
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A student releases a marble from the top of a ramp. The marble increases
Fed [463]

Answer:

Vf = 69.56 cm/s

Explanation:

In order to find the final speed of the ramp, we will use the equations of motion. First we use second equation of motion to find out the acceleration of marble:

s = Vi t + (1/2)at²

where,

s = distance traveled = 160 cm

Vi = Initial Speed = 0 cm/s (since, marble starts from rest)

t = time interval = 4.6 s

a = acceleration = ?

Therefore,

160 cm = (0 cm/s)(4.6 s) + (1/2)(a)(4.6 s)²

a = (320 cm)/(4.6 s)²

a = 15.12 cm/s²

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