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
Anton [14]
4 years ago
15

A dog travels 18 meters south across the backyard in 11 seconds. What is the dog's speed?

Physics
1 answer:
o-na [289]4 years ago
4 0
The dog’s speed is
A) 0.61 m/s
You might be interested in
Which two statements best determine whether an engineering design is
pshichka [43]

Answer: A and B

Explanation:

just took the apex test

7 0
3 years ago
Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10
Mars2501 [29]

Answer:

a) a=5.7551 \times g

b) d=20.5539\times g

Explanation:

Given:

  • speed of rocket initially, v_i=0\ m.s^{-1}
  • top speed of rocket after acceleration, v=282\ m.s^{-1}
  • time taken to get to the top speed, t_i=5\ m.s^{-1}
  • final speed of the rocket, v_f=0\ m.s^{-1}
  • time taken to get to the final speed after reaching the top speed, t_f=1.4\ s

Now the acceleration:

a=\frac{v-v_i}{t_i}

a=\frac{282-0}{5}

a=56.4\ m.s^{-2}

Now as a fraction of gravity:

a=\frac{56.4}{9.8}\times g

a=5.7551 \times g

Now, the deceleration:

d=\frac{0-282}{1.4}

d=201.4285\ m.s^{-2}

Now as a fraction of gravity:

d=\frac{201.4285}{9.8}\times g

d=20.5539\times g

6 0
3 years ago
Dwayne ‘The Rock’ Johnson needs to escape from the fourth floor of a burning building (in a movie). He ties a rope around his wa
ZanzabumX [31]

Answer:

Final Speed of Dwayne 'The Rock' Johnson = 15.812 m/s

Explanation:

Let's start out with finding the force acting downwards because of the mass of 'The Rock':

Dwayne 'The Rock' Johnson: 118kg x 9.81m/s = 1157.58 N

Now the problem also states that the kinetic friction of the desk in this problem is 370 N

Since the pulley is smooth, the weight of Dwayne Johnson being transferred fully, and pulls the desk with a force of 1157.58 N. The frictional force of the desk is resisting this motion by a force of 370 N. Subtracting both forces we get the resultant force on the desk to be: 1157.58 - 370 = 787.58 N

Now lets use F = ma to calculate for the acceleration of the desk:

787.58 = 63 x acceleration

acceleration = 12.501 m/s

Finally, we can use the motion equation:

v^2 - u^2 = 2*a*s

here u = 0 m/s (since initial speed of the desk is 0)

a = 12.501 m/s

and s = 10 m

Solving this we get:

v^2 - 0 = 2 * 12.501 * 10

v = 15.812 m/s

Since the desk and Mr. Dwayne Johnson are connected by a taught rope, they are travelling at the same speed. Thus, Dwayne also travels at            15.812 m/s when the desk reaches the window.

5 0
3 years ago
I start going 10 m/sIf I slow down at 2 m/s/s how far will I go before I stop
klasskru [66]

Answer:

25 m

Explanation:

Given:

v₀ = 10 m/s

v = 0 m/s

a = -2 m/s²

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (10 m/s)² + 2 (-2 m/s²) Δx

Δx = 25 m

5 0
4 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
Other questions:
  • What happens to the chemical structure of water when it changes state?
    12·1 answer
  • Jumping up before the elevator hits. After the cable snaps and the safety system fails, an elevator cab free-falls from a height
    11·1 answer
  • Can someone help me on this question
    5·1 answer
  • Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g
    11·1 answer
  • A 2 kg block is pushed by an external force against a spring with spring constant 131 N/m until the spring is compressed by 2.1
    11·1 answer
  • What If? The coil and applied magnetic field remain the same, but the circuit providing the current in the coil is now changed.
    13·1 answer
  • A rock weighing 20 N (mass = 2 kg) is swung in a horizontal circle of radius 2 m at a constant speed of 6 m/s. What is the centr
    5·1 answer
  • Your body exerts the same amount of gravitational force on the Moon as the Moon exerts on your body. True or, false?
    11·2 answers
  • Watch the video and make a ninja star. Once you have made a ninja star, 5 Point Ninja Star. Explain the energy transfer in the n
    5·2 answers
  • The charge of an electron is 1.6x10^'' C. How many electrons does it take to make 1 C of charge?
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!