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3 years ago

Okay I better get a good answer for this guy I'm puttin' all my money on it.

Physics

2 answers:

Black_prince [1.1K]3 years ago

3 0

Answer:

His third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A.

Explanation:

klio [65]3 years ago

3 0

Answer: Explanation:

Newton's third law of motion is simply whenever two objects interact,

they exert equal and opposite forces on each other.

This is often worded as 'every action has an equal and opposite reaction'. you have to remember two forces: 1 is act on two different objects , 2 are of the same type (i.e. both contact forces)

The statement simply means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs - equal and opposite action-reaction force pairs.

the attached is an example of "in every action, there is an equal and opposite and reaction.

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ehidna [41]

Use the Inverse square law, Intensity (I)<span> of a light </span>is inversely proportional to the square of the distance(d).

I=1/(d*d)

Let Intensity for lamp 1 is L1 distance be D1 so on, L2 D2 for Intensity for lamp 2 and its distance.

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L1/15=(200*200)/(400*400)
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2 years ago

A 2000 kg truck traveling north at 34 km/h turns east and accelerates to 58 km/h. (a) What is the change in the truck's kinetic

barxatty [35]

Explanation:

It is given that,

Mass of the truck, m = 2000 kg

Initial velocity of the truck, u = 34 km/h = 9.44 m/s

Final velocity of the truck, v = 58 km/h = 16.11 m/s

(a) Change in truck's kinetic energy, $\Delta E=\dfrac{1}{2}m(v^2-u^2)$

$\Delta E=\dfrac{1}{2}\times 2000\ kg\times (16.11^2-9.44^2)$

$\Delta E=170418.5\ J$

$\Delta E=1.7\times 10^5\ J$

(b) Change in momentum of the truck, $\Delta p=m(v-u)$

$\Delta p=2000\ kg\times (16.11-9.44)$

$\Delta p=13340\ kg-m/s$

Hence, this is the required solution.

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3 years ago

PLEASE HELPP I NEED TO PASS THIS !!

goldfiish [28.3K]

Option D is correct.

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2 years ago

The position of the front bumper of a test car under microprocessor control is given by x(t) = 2.31 m + (4.90 m/s2)t2 - (0.100 m

vovangra [49]

The equation of the car is given by the equation,

x(t) = 2.31 + 4.90t² - 0.10t⁶

If we are going to differentiate the equation in terms of x, we get the value for velocity.

dx/dt = 9.8t - 0.6t⁵

Calculate for the value of t when dx/dt = 0.

dx/dt = 0 = (9.8 - 0.6t⁴)(t)

The values of t from the equation is approximately equal to 0 and 2.

If we substitute these values to the equation for displacement,

(0) , x = 2.31 + 4.90(0²) - 0.1(0⁶) = 2.31

(2) , x = 2.31 + 4.90(2²) - 0.1(2⁶) = 15.51

Thus, the positions at the instants where velocity is zero are 2.31 and 15.51 meters.

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3 years ago

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2 years ago