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3 years ago

Explain your observation in terms of force and Newtons 3rd law

Physics

1 answer:

Simora [160]3 years ago

4 0

Answer:

If an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A

Explanation:

the law of force show that focuses will always come in 2 or pairs and it also says that one body cannot exert a force on another without experiencing a force itself.

( hope that helps :) #pk1208 is here to help )

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A projectile is fired at v 0 = 381.0 m/s at an angle of θ = 73.5 ∘ , with respect to the horizontal. Assume that air friction wi

Ipatiy [6.2K]

Answer:

Explanation:

velocity of projection, vo = 381 m/s

angle of projection, θ = 73.5°

The formula for the range is

$R=\frac{u^{2}Sin2\theta }{g}$

$R=\frac{381^{2}Sin147 }{9.8}$

R = 8067.4 m

Range in shorten by 34.1 %

So, the new range is

R' = 8067.4 - 34.1 x 8067.4/100

R' = 5316.4 m

5 0

3 years ago

If you place a glass cylinder in Wessin Oil, the view of the cylinder nearly disappeared when the eyedropper was full of Wessin

Viefleur [7K]

Answer:

A. the indices of refraction matched

Explanation:

The index of refraction, or refractive

index, is a measure of how fast light

rays travel through a given medium.

Alternatively, it could be said that

the refractive index is the measure of

the bending of a light ray when

passing from one medium to

another. Mathematically, it can be

represented as a ratio between two

different velocities – the velocity of

light in vacuum and the velocity of

light in a given medium.

For example, try putting a pencil in a jar full of water. If you look at the pencil from above, it would look as though the pencil has bent in the water. That happens due to the refraction of light. It occurs because as light rays enter water, they slow down, as the speed of light in water is lower than the speed of light in air. The magnitude of how much a medium refracts a light ray is determined by the index

3 0

3 years ago

The y-position of a damped oscillator as a function of time is shown in the figure.

NISA [10]

(1) The period of the oscillator is 1 second.

(2) The damping coefficient is 0.93.

<h3>What is period of oscillation?</h3>

The period of oscillation is the time taken to make one complete cycle.

From the graph, the time taken to make one complete oscillation is 1 second.

<h3>Damping coefficient</h3>

equation of the wave is given as;

y(t) = Ae^(-btx) cos(ωt)

3.5 = Ae^(-0) cos(0)

3.5 = A x 1

A = 3.5 cm

-3 = 3.5e^(-bx) cos(ω)

-3/3.5 = e^(-bx) cos(ω)

-0.857 = e^(-bx) cos(ω)

-0.857 / cos(ω) = e^(-bx)

ln[-0.857 / cos(ω)] = -bx

ln[-0.857 / cos(ω)] / b = - x ---- (1)

-2 = 3.5e^(-2bx) cos(2ω)

-0.57 = e^(-2bx) cos(2ω)

ln[-0.57 / cos(2ω)] = -2bx

ln[-0.57 / cos(2ω)] /2b = - x ------(2)

solve (1) and (2)

ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b

-0.57 / cos(ω) = 2(-0.857 / cos(ω))

2(-0.857/cosω) = -0.57/cos2ω

-(2 x 0.857) / (-0.57) = cosω/cos 2ω

3 = cosω/cos 2ω

3(cos 2ω) = cosω

3(2cos²ω - 1) = cos ω

6cos²ω - 6 = cosω

6cos²ω - cosω - 6 = 0

let cosω = y

6y² - y - 6 = 0

solve the quadratic equation;

y = 1.1 or -0.92

cosω = -0.92

ω = arc cos(-0.92)

ω = 2.74 rad/s

From equation (1)

ln[-0.857 / cos(ω)] / x = -b ---- (1)

let x = 1

ln(-0.857/cos(2.74) = -b

-0.93 = -b

b = 0.93

Thus, the damping coefficient is 0.93.

Learn more about damping coefficient here: brainly.com/question/14058210

#SPJ1

4 0

2 years ago

Calculate the work done by a 50.0 N force on an object as it moves 9.00 m, if the force is oriented at an angle of 135° from the

aivan3 [116]

Answer:

Work done, W = -318.19 Joules

Explanation:

It is given that,

Force acting on the object, F = 50 N

Distance covered by the force, d = 9 m

Angle between the force and the distance traveled, $\theta=135^{\circ}$

The work done by an object is equal to the product of force and distance traveled. It is equal to the dot product of force and the distance. Mathematically, it is given by :

$W=Fd\ cos\theta$