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2 years ago

If the momentum of a system is to be conserved, which must be true of the net external force acting on the system?

Physics

1 answer:

kari74 [83]2 years ago

7 0

Answer:

D. zero

Explanation:

For momentum of an isolated or closed system to be conserved (initial momentum must equal final momentum), the net external force acting on the system must be zero.

There is always external forces acting on a system, for this system’s momentum to remain constant, all the external forces acting on the system must cancel out, so that the net external force on the system is zero.

$F_{ext} = 0$

Therefore, the correct option is "D"

D. zero

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Why is it important to consider the experimental error in all the empirical results presented?

irinina [24]

Answer:

It is very important because scientists, especially the ones with empirical experiments and results, are prone to error and the empirical data is in need to be under strict observation done not only by many scientists but also by expermiented ones. This guards everybody to change the parameters suddenly which can affect the real results of an experiment

Explanation:

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3 years ago

Read 2 more answers

Lord Beckett and members of the EIT Co. spot the Black Pearl in the distance making its way towards land. As a

KATRIN_1 [288]

Answer:

Explanation:

The cannonball goes a horizontal distance of 275 m . It travels a vertical distance of 100 m

Time taken to cover vertical distance = t ,

Initial velocity u = 0

distance s = 100 m

acceleration a = 9.8 m /s²

s = ut + 1/2 g t²

100 = .5 x 9.8 x t²

t = 4.51 s

During this time it travels horizontally also uniformly so

horizontal velocity Vx = horizontal displacement / time

= 275 / 4.51 = 60.97 m /s

Vertical velocity Vy

Vy = u + gt

= 0 + 9.8 x 4.51

= 44.2 m /s

Resultant velocity

V = √ ( 44.2² + 60.97² )

= √ ( 1953.64 + 3717.34 )

= 75.3 m /s

Angle with horizontal Ф

TanФ = Vy / Vx

= 44.2 / 60.97

= .725

Ф = 36⁰ .

6 0

2 years ago

a cylindrical jar is 10cm long and has a cross sectional area of 36cm. if it is completely filled with a fluid of relative densi

ki77a [65]

Answer:

The mass of the fluid is 72 g.

Explanation:

The following data were obtained from the question:

Height (h) = 10 cm

Area of cross section (A) = 36cm²

Relative density = 0.2

Mass =..?

Next, we shall determine the volume of the cylinder. This can be achieved by doing the following:

Volume = Area x Height

Volume = 36 x 10

Volume = 360 cm³

Next, we shall determine the density of the liquid.

This can be obtained as follow:

Relative density = density of substance/density of water.

Relative density = 0.2

Density of water = 1 g/cm³

Density of fluid =...?

Relative density = density of substance/density of water.

0.2 = density of fluid / 1 g/cm³

Cross multiply

Density of fluid = 0.2 x 1 g/cm³

Density of fluid = 0.2 g/cm³

Finally, we shall determine the mass of fluid as follow:

Volume = 360 cm³

Density of fluid = 0.2 g/cm³

Mass of fluid =...?

Density = mass /volume.

0.2 g/cm³ = mass of fluid /360 cm³

Cross multiply

Mass of fluid = 0.2 g/cm³ x 360 cm³

Mass of fluid = 72 g

Therefore, the mass of the fluid in the jar is 72 g.

6 0

3 years ago

"The ________ method can determine whether a string contains a value that can be converted to a specific data type before it is

dem82 [27]

Answer:

The method to determine whether a string contains a value that can be converted to a specific data type before it is converted to that data type

Explanation:

The Java string includes () method to check whether a particular sequence of character is the part of given sub string or not.

One string contain another string in Java or not.

The indexof() to check the string and substring in java

Hence java string include () method to check whether a string contains a value that can be converted to a specific data type before it is converted to that data type

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3 years ago

Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the ce

jasenka [17]

Answer:

$\lambda= 506.25 nm$

Explanation:

Diffraction is observed when a wave is distorted by an obstacle whose dimensions are comparable to the wavelength. The simplest case corresponds to the Fraunhofer diffraction, in which the obstacle is a long, narrow slit, so we can ignore the effects of extremes.

This is a simple case, in which we can use the Fraunhofer single slit diffraction equation:

$y=\frac{m \lambda D}{a}$

Where:

$y=Displacement\hspace{3}from\hspace{3} the\hspace{3} centerline \hspace{3}for \hspace{3}minimum\hspace{3} intensity =1.35mm\\\lambda=Light\hspace{3} wavelength \\D=Distance\hspace{3}between\hspace{3}the\hspace{3}screen\hspace{3}and\hspace{3}the\hspace{3}slit=2m\\a=width\hspace{3}of\hspace{3}the\hspace{3}slit=0.750mm\\m=Order\hspace{3}number=1$

Solving for λ:

$\lambda=\frac{y*a}{mD}$

Replacing the data provided by the problem:

$\lambda=\frac{(1.35\times 10^{-3})*(0.750\times 10^{-3})}{1*2} =5.0625\times 10^{-7}m =506.25nm$

7 0

3 years ago