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MaRussiya [10]
1 year ago
11

an object is acted on by a drag force with a magnitude that is proportional to the speed. the object accelerates downward at 3.0

0 m/s2 when it is falling with a speed 20.0 m/s. what is the terminal speed of the object as it is falling?
Physics
1 answer:
Nutka1998 [239]1 year ago
7 0

The terminal speed of the object falling down is 66.67 m/s.

The terminal speed acquired by the body when,

Weight of the body = Drag force of the body

It is given,

Drag force is directly proportional to the speed,

So,

F = CV

Where F is drag force,

V is the speed,

C is the constant,

So, it can be written as C = F/V.

The weight of the body = mg

The weight of the body = 10m

M is the mass and g is the acceleration due to gravity,

The drag force when the speed is 20m/s.

Drag force = ma

a is the acceleration during the drag force which is given to be 3m/s²,

Drag force = 3m

Now we can write,

F₁/V₁ = F₂/V₂

F₁ is the drag force at 20m/s speed.

F₂ is the weight of the body and V₂ is the terminal speed,

Now, it can be written,

3m/20 = 10m/V₂

V₂ = 66.67 m/s.

So, the terminal speed is 66.67m/s.

To know more about terminal speed, visit,

brainly.com/question/14605362

#SPJ4

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Vector A with arrow, which is directed along an x axis, is to be added to vector B with arrow, which has a magnitude of 5.5 m. T
ohaa [14]

Answer:

Magnitude of vector A = 0.904

Explanation:

Vector A , which is directed along an x axis, that is

                   \vec{A}=x_A\hat{i}

Vector B , which has a magnitude of 5.5 m

                   \vec{B}=x_B\hat{i}+y_B\hat{j}

                   \sqrt{x_{B}^{2}+y_{B}^{2}}=5.5\\\\x_{B}^{2}+y_{B}^{2}=30.25

The sum is a third vector that is directed along the y axis, with a magnitude that is 6.0 times that of vector A                    \vec{A}+\vec{B}=6x_A\hat{j}\\\\x_A\hat{i}+x_B\hat{i}+y_B\hat{j}=6x_A\hat{j}

Comparing we will get

                  x_A=-x_B\\\\y_B=6x_A

Substituting in x_{B}^{2}+y_{B}^{2}=30.25

                  \left (-x_{A} \right )^{2}+\left (6x_{A} \right )^{2}=30.25\\\\37x_{A}^2=30.25\\\\x_{A}=0.904

So we have

    \vec{A}=0.904\hat{i}

Magnitude of vector A = 0.904

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Potential energy is defined by formula

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m = mass

g = acceleration due to gravity

h = height

Now here two different stones are located at same height

while mass of stone A is twice that of stone B

so here we can say potential energy of A is

U_a = (2m)gh

Similarly potential energy of B is

U_b = mgh

now if we take the ratio of two energy

\frac{U_a}{U_b} = 2

so we can say potential energy of stone A is two times the potential energy of B

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