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MakcuM [25]
3 years ago
6

Becky leaves home and rides a distance of 30km it took her 2.5 hours what is her speed

Physics
2 answers:
Sedaia [141]3 years ago
7 0
Her speed would be 12.5km p/s
ololo11 [35]3 years ago
7 0
12.5km. I hope this helps.
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A constant amount of charge passes through a conductor. How is current affected if the same amount of charge passes in less time
baherus [9]

Answer:

B. The current increases.

Explanation:

As we know that rate of flow of charge through the conductor is known as electric current

So we have

i = \frac{q}{t}

here we know that charge Q flowing through the conductor is constant while the time in which it passes through it is decreased

so we can say that the ratio of charge and time will increase

so here we have

i = increased

So correct answer will be

B. The current increases.

4 0
3 years ago
Which projectiles will be visibly affected by air resistance when they fall?
NemiM [27]
Any object that is launched as a projectile will lose speed and, as a result, altitude, as it travels through the air. The rate at which the object loses speed and altitude depends on the amount of force that way applied to it when it was launched. It is also dependent on the size and shape of the item. This is why something like, say, a football is much faster to fall to the ground than a bullet.
4 0
3 years ago
Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl
melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar

The boundary layer thickness is equal to:

\delta=\frac{4.91*1}{Re^{0.5} }  =\frac{4.91*1}{246507^{0.5} } =0.0098 ft

The shear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{1}  )(246507)^{0.5} =0.012

b) If the railing edge is 2 ft, the Reynold number is:

Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar

The boundary layer is equal to:

\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft

The sear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{2}  )(493015.6^{0.5} )=0.0082

c) The drag coefficient is equal to:

C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019

The friction drag is equal to:

F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf

7 0
3 years ago
What velocity must a 2.25 kg croquet mallet have in order to have the same momentum as 1.25 kg ball that has a momentum 6.25 kg•
Ksenya-84 [330]

abcdefghijklmnopqrstuvwxyz

3 0
3 years ago
Помогите плз!
Softa [21]

Все написано в скобках правильно

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