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3 years ago

PLEASE HELP!!! 30 POINTS

1 answer:

6 0

The impulse and the action and reaction law emit explain the sensation of the foot when kicking two different objects the answer in each case

a) kick the ball the sensation is not painful

b) kick the wall the feeling is painful

Newton's second law establishes a linear relationship between force and mass

F = m a

Where the bold letters indicate vectors, F is the force, m the mass and the acceleration

Newton's third law or action and reaction law stable that the force acts on a pairs, the kicking of an object with a force F the object responds with a force of equal magnitude but opposite direction on us.

From Newton's second law we can derive a relationship between the impulse and the variation of the momentum

I = ∫ F dt = Δp

Where I is the impulse, t the time and Δp the variation of the momentum

The impulse is related to the sensation that the body receives due to the variation of the amount of movement, in this case we have two situations

a) Kick a soccer ball

The mass of the ball is small so the force applied by the foot creates a large acceleration without the foot having a significant decrease in foot speed. Consequently the impulse applied by the ball on the foot is small, the sensation is not painful

b) Kick a brick wall

In this case, the mass of the wall is high and when kicked it does not acquire any movement, consequently the reaction of the wall creates a high impulse on the foot that causes its speed to quickly decrease to zero, the physical sensation of this reduction in quantity movement is painful.

In conclusion, the impulse and the law of action and reaction allow us to find the answer in each case.

a) kick the ball feeling painless

b) kick the wall painful feeling

learn more about the action and reaction law and impulse here:

brainly.com/question/6677486

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Which is more 3 tons or 6001 pounds

Mnenie [13.5K]

3 tonnes

The answer is 3 tonnes, because 3 tonnes equal 6613.87 pounds, which is higher than 6001.

I hope this helps! I would love it if you could help me answer my question too!

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3 years ago

A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.

mash [69]

a) The acceleration of gravity is $4.96 m/s^2$

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

The acceleration of gravity for an object near a planet is given by

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2$

The critical velocity for a satellite orbiting around a planet is given by

$v=\sqrt{\frac{GM}{R+h}}$

where we have again:

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s$

Converting into km/h,

$v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h$

The period of the orbit is given by the circumference of the orbit divided by the velocity:

$T=\frac{2\pi (R+h)}{v}$

where

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

v = 6668 m/s

Substituting,

$T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s$

One day consists of:

$t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s$

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

$n=\frac{t}{T}=\frac{86400}{8452}=10.2$

The escape velocity for an object in the gravitational field of a planet is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where here we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

Substituting, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s$

We can apply again the formula to find the escape velocity for the rocket:

$v=\sqrt{\frac{2GM}{R+h}}$

Where this time we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=0$ , because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s$

Learn more about gravitational force:

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4 years ago

A train travels 120 km in 2 hours and 30 minutes. what is it average speed?

Solnce55 [7]

The answer is 48Kmh because it is 120km divided by 2.5 or 2 and a half hours

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3 years ago

At a certain time a particle had a speed of 43 m/s in the positive x direction, and 5.8 s later its speed was 95 m/s in the oppo

Vanyuwa [196]

Answer:

$-23.8\frac{m}{s^{2}}$

Explanation:

Average acceleration is defined as the change on velocity on an interval of time:

$a_{avg}=\frac{V_f-V_i}{t}$

with Vf the final velocity, Vi the initial velocity and t the time interval. Note that the initial velocity is positive because is in the positive x direction and final velocity is negative because is in the opposite direction, so:

$a_{avg}=\frac{-95\frac{m}{s}-43\frac{m}{s}}{5.8s}$

$a_{avg}=-23.8\frac{m}{s^{2}}$

The negative sign indicates that the acceleration is in the negative x direction.

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4 years ago

3. An object of mass 90 kg travels down a slide.

Zolol [24]

Gpe= 90* 9.81 * 15 = 13243.5 Joules

KE= 13243.5 =1/2 m v^2
v^2= 295.3
v= 17.2 m/s

6 0

4 years ago