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zzz [600]
3 years ago
5

According to the videos seen in Modules 8, please write a short paragraph answering the following questions: a. What are some ex

amples of on-the-spot feedback? b. What are some examples of delayed feedback?
Physics
1 answer:
Sindrei [870]3 years ago
8 0

Answer:

.

Explanation:

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7)An object is thrown straight up with an initial velocity of 3 m/s. What is its acceleration at its maximum height?
ser-zykov [4K]
C.
It is a motion with uniform acceleration, meaning that the acceleration will not change.
The object is thrown upwards with a positive velocity. This shows that the upward direction is positive. The object will decelerate due to gravity at a magnitude of 9.81 m/s2. Therefore, the acceleration is -9.81 m/s2.
Note that even though the velocity of the object is momentarily 0 m/s at maximum height, there is still a constant acceleration.
This allows the object first decelerate upwards, then change direction at max height, and finally accelerate downwards. So in this case, the acceleration is always negative and unchanged.
7 0
3 years ago
Consider a point on a bicycle wheel as the wheel makes exactly four complete revolutions
Slav-nsk [51]
It makes a cycloid pattern in parametrics if that's what you're asking~
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3 years ago
WILL GIVE BRAINLIEST
max2010maxim [7]
D, the plant takes Solar/Light energy and converts it into chemical energy.
5 0
4 years ago
Read 2 more answers
Like the filters falling through the air, a car on the freeway represents an object with a high Reynolds number traveling throug
Goshia [24]

Answer:

ΔF=125.22 %

Explanation:

We know that drag force on the car given as

F_D=\dfrac{1}{2}\rho C_DA v^2

C_D=Drag coefficient

A=Projected area

v=Velocity

ρ=Density

All other quantity are constant so we can say that drag force and velocity can be given as

\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}

Now by putting the values

\dfrac{F_D_1}{F_D_2}=\dfrac{v_1^2}{v_2^2}

\dfrac{F_D_1}{F_D_2}=\dfrac{50^2}{75^2}

\dfrac{F_D_1}{F_D_2}=0.444

Percentage Change in the drag force

\Delta F=\dfrac{F_D_2-F_D_1}{F_D_1}\times 100

\Delta F=\dfrac{F_D_2-0.444F_D_2}{0.444F_D_2}\times 100

\Delta F=\dfrac{1-0.444}{0.444}\times 100

ΔF=125.22 %

Therefore force will increase by 125.22  %.

3 0
3 years ago
In Ampere's law, ∮????⃗∙????????⃗=????0???? the direction of the integration around the path:
kupik [55]

Answer:

C) must be such as to follow the magnetic field lines.

Explanation:

Ampere's circuital law helps us to calculate magnetic field due to a current carrying conductor. Magnetic field due to a current forms closed loop around the current . If a  net current of value I creates a magnetic field B around it , the line integral of magnetic field around a closed path becomes equal to μ₀ times the net current . It is Ampere's circuital law . There may be more than one current passing through the area enclosed by closed curve . In that case we will take net current by adding or subtracting them according to their direction.

It is expressed as follows

∫ B.dl = μ₀ I . Here integration is carried over closed path . It may not be circular in shape. The limit of this integration must follow magnetic field lines.

the term ∫ B.dl is called line integral of magnetic field.

3 0
4 years ago
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