1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
jeyben [28]
3 years ago
9

A Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia In one set of experimen

ts, they studied the maximum speed that quolls could run around a curved path without slipping. One quoll was running at 2.4 m/s around a curve with a radius of 1.6 m when it started to slip.
What was the coefficient of static friction between the quoll's feet and the ground in this trial?
Physics
1 answer:
balu736 [363]3 years ago
7 0

Answer:

Coefficient of static friction = 0.37

Explanation:

At the point the the quoll slides, quoll attains its maximum velocity.

So Ne = (mv^2)/r ....equa 1

And N =mg....equ 2

Where N vertical force of qoull acting on the surface, e = coefficient of friction, m=mass, g=9.8m/s^2, r =radius =1.6m, v= max velocity of quill = 2.4m/s

Sub equ 2 into equ 1

Mge= (mv^2)/r ...equa3

Simplfy equ3

e = v^2/(gr)...equ 4

Sub figures above

e = 5.76/(9.8*1.6)

e = 0.37

You might be interested in
What type of bond results from the side‑on overlap of orbitals?
Serga [27]

Answer:

A pi bond

Explanation:

A pi bond is a type of covalent bond that results from the formation of a molecular orbital by the side-to-side overlap of atomic orbitals along a plane perpendicular to a line connecting the nuclei of the atoms.

4 0
3 years ago
An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis
Ivan

(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

W=\Delta K (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

W=\Delta K + E_{lost}

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane (\Delta K) and part is lost because of the air resistance (E_{lost}).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N

Now we can calculate the acceleration of the plane, by using Newton's second law:

a=\frac{F}{m}=\frac{2.70\cdot 10^4 N}{1.60\cdot 10^4 kg}=1.69 m/s^2

where m is the mass of the plane.

Finally, we can calculate the final speed of the plane by using the equation:

v^2- u^2 = 2aS

where

v=? is the final velocity

u=66.0 m/s is the initial velocity

a=1.69 m/s^2 is the acceleration

S=5.00 \cdot 10^2 m is the distance travelled

Solving for v, we find

v=\sqrt{u^2+2aS}=\sqrt{(66.0 m/s)^2+2(1.69 m/s^2)(5.00\cdot 10^2 m)}=77.8 m/s

8 0
3 years ago
A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest in 1
attashe74 [19]

The average act on her during the deceleration is 4.47 meters per second.                                                                

<u>Explanation</u>:

<u>Given</u>:

youngster mass m = 50.0 kg

She steps off a 1.00 m high platform that is s = 1 meter

She comes to rest in the 10-meter second

<u>To Find</u>:

The average force and momentum

<u>Formulas</u>:

p = m * v

F * Δ t = Δ p

vf^2= vi^2+2as

<u>Solution</u>:

a = 9.8 m/s

vi = 0

vf^2= 0+2(9.8)(1)

vf^2 = 19.6

vf = 4.47 m/s .

Therefore the average force is 4.47 m/s.                                          

                                 

5 0
3 years ago
1. When in the past have you pushed your personal limits? Give at least one
Yuri [45]

Answer:

Umm that's a personal question. All u have to do is say when have u pushed your personal limits....... Ummm one for me is when i had to try out for a select soccer and that is past my comfort zone.  

Explanation:

4 0
3 years ago
If the truck has a mass of 2,000 kilograms , what's its momentum?(v=35 m/s)
katen-ka-za [31]

Answer:

\boxed {\boxed {\sf  70,000 \ kg*m/s}}

Explanation:

Momentum is the product of mass and velocity.

p=m*v

The mass of the truck is 2,000 kilograms and the velocity is 35 meters per second.

m= 2000 \ kg \\v= 35 \ m/s

Substitute the values into the formula and multiply.

p= 2000 \ kg * 35 \ m/s \\p= 70,000 \ kg*m/s

The truck's momentum is <u>70,000 kilograms meters per second.</u>

8 0
3 years ago
Other questions:
  • What is the exact time of a full earth rotation?
    5·2 answers
  • Obi Wan hears the destruction of a planet and all of its people through 'the force'. These sounds are only in his head and are n
    8·2 answers
  • What is the average acceleration during the time interval 0 seconds to 10 seconds?
    12·1 answer
  • Which of the following is not found in living things? Which of the following is not found in living things?
    14·2 answers
  • Photovoltaic technology is most commonly used to generate electricity using what type of power?
    8·1 answer
  • How does the electric force between two charged objects change when the
    7·1 answer
  • An asteroid has acquired a net negative charge of 149 C from being bombarded by the solar wind over the years, and is currently
    7·1 answer
  • 100 POINTS 100 POINTS 100 POINTS!!!!!<br> HELP PLEASE I DON'T KNOW WHAT TO DO!!!!!
    13·2 answers
  • Một đĩa tròn có bán kính 20cm đang đứng yên, được quanh nhanh dần đều sau 10s thì đạt vận tốc góc 180 vòng/phút. Hãy xác định: a
    6·1 answer
  • Runs 4 m North and then runs 9 m West. Calculate Distance and Displacement.
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!