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Snezhnost [94]
4 years ago
7

Why is there no drag in space?????? Help me please!!! Thanks!!!

Physics
2 answers:
dolphi86 [110]4 years ago
8 0
If there is no fluid, there is no drag. Drag is generated by the difference in velocity between the solid object and the fluid. 
astraxan [27]4 years ago
7 0
It is beacuse of fluid If there is no fluid, there is no drag. Drag is generated by the difference in velocity between the solid object and the fluid. If this statement is correct then how can there be drag in space if there is no air?
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Suppose that you are standing on a train accelerating at 0.20g. What minimum coefficient of static friction must exist between y
Ilia_Sergeevich [38]
Acceleration = (0.2 x g) = 1.96m/sec^2. 
<span>Accelerating force on 1kg. = (ma) = 1.96N. </span>
<span>1kg. has a weight (normal force) of 9.8N. </span>
<span>Coefficient µ = 1.96/9.8 = 0.2 minimum. </span>

<span>Coefficient is a ratio, so holds true for any value of mass to find accelerating force acting. </span>
<span>e.g. 75kg = (75 x g) = 735N. </span>
<span>Accelerating force = (735 x 0.2) = 147N</span>
5 0
4 years ago
Read 2 more answers
A flat loop of wire consisting of a single turn of cross-sectional area 7.10 cm2 is perpendicular to a magnetic field that incre
MAVERICK [17]
<span>From Ohm's law. V = IR. Where R = 2 Ohms. To calculate the induced current I; We need to calculate the electromotive force or voltage, V. From Faraday's law induced EMF = (The rate of change of magnetic flux density x Area)/ (changein time). Or EMF = BA/t. Where B = Bf - Bi. And BA = Bf* A - Bi* A. Bf = 2.00 and Bi = 0.500 and t = 0.93s and the area, A = 7.1 cm^2 is 0.000071 m^2. 2 So Emf = 2.00 (0.000071) - 0.500(0.000071) /(0.93) = 1.0654 * 10^(-4)/ 0.93 = 1.1415 * 10^(-4). Substituting into ohms law, we have, I = (1.1415 * 10^(-4)) / 2 = 0.57075 * 10^(-4)</span>
3 0
3 years ago
A beam of light is emitted 8.6 cm beneath the surface of a liquid and strikes the air surface 7.2 cm from the point directly abo
Makovka662 [10]

Answer:

n = 1.56

Explanation:

The total reflection attempts occurs when a light beam passes from a medium with a higher index to a medium with a lower nest, at an angle where it occurs we can find them by the refractive relationship

             n₁ sin θ₁ = n₂ sin θ₂

             n1 = n2 / sin θ₁

For this relationship to be fulfilled, the liquid index must be greater than the air index divided by the sine of the critical angle

Let's use trigonometry to find angle

               tan θ = y / x

                θ = tan⁻¹ 7.2 / 8.6

                θ = 39.94º

         

                n₁ = 1 / sin 39.94

                n = 1.56

This is the refractive index of the liquid

4 0
3 years ago
What would the reaction force of a person be that stands completely still with a mass of 100 kg?
Anastaziya [24]

-981N would be the reaction force of a person be that stands completely still with a mass of 100 kg.

An opposing force to an action force is referred to as a reaction force. The response force that results from surface engagement and adhesion while sliding is known as friction. Usually, the activities of applied forces result in reaction forces and reaction moments.

As,

F = ma

where, F = force

           m = mass of object

           a = acceleration due to gravity

Plugging in the values we get,

F = 100kg × 9.81m/s

F = 981 N

Since, reaction force acts opposite to action force so the correct answer is -981N.

Learn more about force here;

brainly.com/question/13191643

#SPJ4

4 0
2 years ago
A car is traveling at a constant speed along the road ABCDE shown in the drawing. Section Ab and DE are straight. Rank the accel
vladimir2022 [97]

Answer:

AB = DE <CD <BC

Explanation:

This is an exercise in kinetics, the accelerations defined as the change in velocity over the time interval, therefore the accelerations of a vector.

Because the acceleration is a vector, it has two parts, the modulus that the numerical value of the magnitude and the direction, a change in any of them implies the existence of a relationship.

Let's apply these reasoning to our problem.

AB Path  

this path is straight and as they indicate that the constant speed the acceleration is zero

DE path

This path is straight and since the velocity is constant the zero steps

BC path

This path is a curve and the velocity modulus is constant, but its directional changes therefore there is an acceleration called centripetal, given by the expression

         a_{c} = v² / r

where r is the radius of the curve and the direction of acceleration is towards the center of the curve

CD path

This path is a curve and it also has centripetal acceleration, as can be seen in the drawing, the radius of the curve is greater than in section BC, therefore the acceleration is less

              a_{BC} > a_{CD}

In  summary lower accelerations are

 AB = DE <CD <BC

3 0
3 years ago
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