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3 years ago

Hi i need help re-stating Newton's Second law. It needs to be at least two sentences.

Physics

2 answers:

jek_recluse [69]3 years ago

8 0

Answer:

The newton's second law states that the acceleration of an object is directly proportional to the force acting upon the object and inversely proportional to the mass

dmitriy555 [2]3 years ago

4 0

Answer:

The newton's second law states that the acceleration of an object is directly proportional to the force acting upon the object and inversely proportional to the mass

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Is there gravity in space? If so name one piece of evidence

Eddi Din [679]

There is gravity everywhere in space. In fact, it's how the moon orbits the Earth. Also, gravity in space assists the Earth to orbit the sun. These are caused by small areas of gravity. The term for this is called "microgravity".

I hope this helps!

8 0

4 years ago

What is the inverse square law and how does it relate to gravity?

Nesterboy [21]

Answer:

Inverse Square Law Newton proposed the Inverse Square Law. The effect of gravity (and also on forces such as sunlight) works like this. If say we have a half-mass Earth, it would produce a gravity of not half but a quarter (the square of 2).

7 0

4 years ago

How much force is needed to accelerate a 1,100 kg car at a rate of 1.5 m/s2?

Anna [14]

Assuming there is no force of friction...

F = ma
F = (1300kg)(1.5m/s^2)
F = 1950N
Just multiply mass by acceleration.
1300 x 1.5 = 1950N.

7 0

3 years ago

Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (As

Naddik [55]

Answer:

$\lambda = 1.667 nm$

$\theta = 0.8681^o$

Explanation:

From the question we are told that

The distance of separation is $d = 0.220 \ mm = 0.00022 \ m$

The is distance of the screen from the slit is $D = 2.60 \ m$

The distance between the central bright fringe and either of the adjacent bright $y = 1.97 cm = 1.97 *10^{-2}\ m$

Generally the condition for constructive interference is

$d sin \tha(\theta ) = n \lambda$

From the question we are told that small-angle approximation is valid here.

So $sin (\theta ) = \theta$

=> $d \theta = n \lambda$

=> $\theta = \frac{n * \lambda }{d }$

Here n is the order of maxima and the value is n = 1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright is mathematically represented as

$y = D * sin (\theta )$

From the question we are told that small-angle approximation is valid here.

$y = D * \theta$

=> $\theta = \frac{ y}{D}$

$\frac{n * \lambda }{d } = \frac{y}{D}$

$\lambda =\frac{d * y }{n * D}$

substituting values

$\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }$

$\lambda = 1.667 *10^{-6}$

$\lambda = 1.667 nm$

In the b part of the question we are considering the next set of bright fringe so n= 2

Hence

$dsin (\theta ) = n \lambda$

$\theta = sin^{-1}[\frac{ n * \lambda }{d} ]$

$\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ]$

$\theta = 0.8681^o$

7 0

4 years ago

A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how

GalinKa [24]

Answer: 473.640 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=48.1 m/s$ is the cannonball's initial velocity

$\theta=0$ because we are told the cannonball is shot horizontally

$t$ is the time since the cannonball is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=1.5m$ is the initial height of the cannonball

$y=0$ is the final height of the cannonball (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it $X_{max}$ and this occurs when $y=0$ .

So, firstly we will find the time with (2):

$0=1.5 m+48. 1 m/s sin(0\°) t-(4.9 m/s^{2})t^2$ (3)

Rearranging the equation:

$0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m$ (4)

$-(4.9 m/s^{2})t^2+(48. 1 m/s) t+1.5 m=0$ (5)

This is a <u>quadratic equation</u> (also called <u>equation of the second degree</u>) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula: