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Vsevolod [243]
3 years ago
6

A box moves 1000 m horizontally as a force F 2000 N is applied downward. What is the work done on the box by F during the displa

cement?​

Physics
1 answer:
stiks02 [169]3 years ago
5 0

Answer:

D. OJ

Explanation:

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A balloon filled with helium gas has an average density of Q,-0.41 kg/m'. The density of the air is Qa-1.23 kg/m3. The volume of
Citrus2011 [14]

Answer:

a) (Qa*g*Vb)-(Qh*Vb*g)=(Qh*Vb*a)\\where \\g=gravity [m/s^2]\\a=acceleration [m/s^2]

b) a = 19.61[m/s^2]

Explanation:

The total mass of the balloon is:

massball=densityheli*volumeheli\\\\massball=0.41 [kg/m^3]*0.048[m^3]\\massball=0.01968[kg]\\\\

The buoyancy force acting on the balloon is:

Fb=densityair*gravity*volumeball\\Fb=1.23[kg/m^3]*9.81[m/s^2]*0.048[m^3]\\Fb=0.579[N]

Now we need to make a free body diagram where we can see the forces that are acting over the balloon and determinate the acceleration.

In the attached image we can see the free body diagram and the equation deducted by Newton's second law

6 0
3 years ago
As capacitor was discharging, what did you observe about q on its plates and the motion of charges in the external circuit?
liubo4ka [24]

As capacitor was discharging, The charge on the plate got reversed and the motion of charge is opposite to the flow of current.

The charging contemporary asymptotically processes 0 as the capacitor becomes charged up to the battery voltage.

The capacitor is completely charged when the voltage of the electricity supply is equal to that at the capacitor terminals. that is referred to as capacitor charging; and the charging segment is over when modern-day stops flowing thru the electrical circuit.

A capacitor can be slowly charged to the important voltage and then discharged quick to provide the power wanted. it's far even viable to charge several capacitors to a positive voltage and then discharge them in any such way as to get extra voltage out of the gadget than became installed.

Learn more about capacitor here:-brainly.com/question/14883923

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8 0
1 year ago
Unpolarizedlight of intensity I_0 is incident on three polarizingfilters. The axis of the first is vertical, that of the secondi
Marina86 [1]

To solve this problem it is necessary to apply the concepts related to the law of Malus which describe the intensity of light passing through a polarizer. Mathematically this law can be described as:

I = I_0 cos^2\theta

Where,

I_0 = Indicates the intensity of the light before passing through the polarizer

I = Resulting intensity

\theta= Indicates the angle between the axis of the analyzer and the polarization axis of the incident light

From the law of Malus when the light passes at a vertical angle through the first polarizer its intensity is reduced by half therefore

I_1= \frac{I_0}{2}

In the case of the second polarizer the angle is directly 60 degrees therefore

I_2 = I_1 cos^2\theta

I_2 = (\frac{I_0}{2} ) cos^2(60)

I_2 = 0.125I_0

In the case of the third polarizer, the angle is reflected on the perpendicular, therefore, its angle of index would be

\theta_3 = 90-60 = 30

Then,

I_3 = I_2 cos^2\theta_3

I_3 = 0.125I_0 cos^2 (30)

I_3 = 0.09375I_0

Then the intensity at the end of the polarized lenses will be equivalent to 0.09375 of the initial intensity.

5 0
3 years ago
Maria makes a graphic organizer to compare open, closed, and short circuits. She adds the labels shown.
Lina20 [59]
The Answer is Label 1:Y Label 2: W
7 0
3 years ago
Read 2 more answers
Drag each tile to the correct location.
scoray [572]

Answer:

Relative: meredith's birth, the planting of jasmine's tree, and the extinction of the stegosaurus

Absolute: harrison's birth, the building of jasmine house, and the extinction of the triceratops

Explanation: plato

7 0
2 years ago
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