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Scrat [10]
3 years ago
14

What is the terminal velocity of a 6.00-kg mass object in falling with a drag force with a magnitude that depends on speed, v, a

s Fdrag = (30.0 N·s/m)v?
Physics
1 answer:
3241004551 [841]3 years ago
6 0

When object reached the terminal speed then its acceleration is zero

So as per Newton's II law we can say

F_{net} = 0

now in that case we can say that net force is zero so here weight of the object is counter balanced by the drag force when it will reach at terminal speed

so we can write

mg - F_d = 0

so here we are given that

F_d = 30

6*9.8 - 30*v = 0

58.8 - 30 *v = 0

v = \frac{58.8}{30}

v = 1.96 m/s

so terminal speed will be nearly 2 m/s

You might be interested in
In which stage of life will the sun undergo the most change?.
Nat2105 [25]
Between the two asymptotic gigantic branches, the Sun changes the greatest in size, brightness, and temperature.
8 0
2 years ago
g 1. A mass undergoing simple harmonic motion along the x-axis has a period of T = 0.5 s and an amplitude of 25 mm. Its position
kondor19780726 [428]

Answer:

Explanation:

a )

Amplitude A = 14 mm , angular frequency ω = 2π / T

= 2π / .5

ω = 4π rad /s

φ₀ = initial phase

Putting the given values in the equation

14 = 25 cos(ωt + φ₀ )

14/25 = cosφ₀

φ₀ = 56 degree

x(t) = 25cos(4πt + 56° )

b )

maximum velocity = ω A

=  4π  x 25

100 x 3.14 mm /s

= 314 mm /s

At x = 0 ( equilibrium position or middle point , this velocity is achieved. )

maximun acceleration = ω² A

= 16π² x A

= 16 x 3.14² x 25

= 3943.84 mm / s²

3.9 m / s²

It occurs at x = A or at extreme position.

7 0
3 years ago
Two drag cars race. They line up at the starting line at rest. The winning car accelerates at a constant rate a and reaches the
Mila [183]

Answer:d=\frac{v^2}{8a}

Explanation:

Given

winning car accelerates with a and its final velocity is v

considering they both start from rest

time taken by winning car is

v=u+at

where u=initial velocity

a=acceleration

t=time

v=at

t=\frac{v}{a}

Now loosing car is accelerating with \frac{a}{4}

Distance traveled by loosing car in time t

s_1=ut+\frac{at^2}{2\cdot 4}

s_1=0+\frac{a}{8}\times (\frac{v}{a})^2

s_1=\frac{v^2}{8a}

Thus distance d traveled by loosing car is given by d=\frac{v^2}{8a}

5 0
4 years ago
in three situations, a briefly applied horizontal force changes the velocity of a hockey puck that slides over frictionless ice.
victus00 [196]

The work done on the Puck by the applied force from the most positive to the most negative is c, b, a respectively.

According to Newton's second law of motion, the force applied to an object is directly proportional to the product of mass and acceleration of the object.

F = ma

F= \frac{mv}{t}

The force applied to an object increases with increases in the velocity of the object.

In the given diagram, the resultant velocity of the puck is calculated as follows;

Figure a:

\Delta v = v_f -v_i\\\\\Delta v = 5 - 6 = - 1 \ m/s

Figure b:

v = \sqrt{4^2 + 3^2} \\\\v = 5 \ m/s

Figure c:

\Delta v = 4 - (-2)\\\\\Delta v = 6 \ m/s

Thus, the work done on the Puck by the applied force from the most positive to the most negative is c, b, a respectively.

Learn more here:brainly.com/question/19498865

4 0
3 years ago
What work is done by a system on its environment when an ideal gas is maintained at a constant pressure of 200 newtons/meter2 du
Crank

<u>Answer:</u>

Work done done by a system on its environment is W=40 Joules

<u>Explanation:</u>

<u>Given:</u>

Ideal gas is maintained at a constant pressure P=200 \frac{N}{m^{2}}

Change in Volume d v= -0.2 m^{3}

<u>To find:</u>  

Work done by a system during an Isobaric Process

<u>Solution:</u>  

In a Isobaric Process, constant pressure is maintained, P=200 \frac{N}{m^{2}}

According to the formula,

work done  \bold{W=P \times d v}

Substitute the values of pressure and change in volume in the above formula,

=200 \times -0.2

=40 \text { Joules }

<u>Result:</u>  

Work done done by a system W= -40 Joules

8 0
3 years ago
Read 2 more answers
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