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
klio [65]
4 years ago
11

A security guard walks at a steady pace traveling 170 m in one trip around the perimeter of a building.

Physics
1 answer:
WARRIOR [948]4 years ago
6 0
Speed= distance/time
Speed= 170/230
Speed=0.74 m/s
You might be interested in
Which phase of cell division is shown
ddd [48]
The last one , telophase
6 0
3 years ago
Last question! Please help me. You don't have to give the answer, just tell me how to start the equation. BUT if you want to sha
tensa zangetsu [6.8K]

Explanation :

It is given that :

speed of sound waves, v=331\ m/s

frequency of sound waves, \nu=15\ Hz

The relation between speed of sound wave and wavelength is

v=\nu \lambda

\lambda=\dfrac{v}{\nu}

\lambda=\dfrac{331\ m/s}{15\ Hz}

\lambda=22.06\ m

4 0
4 years ago
In a mercury barometer at atmospheric pressure, the height of the column of mercury in a glass tube is 760 mm. If another mercur
Archy [21]

Answer:

c. equal to 760 mm

Explanation:

We are told that another mercury barometer is used that has a tube of larger diameter. This means a larger area and the weight of the liquid in the tube will have increased since volume = area × height. Also, due to the larger area, the net upward force on the mercury will have also increased by the same amount because force = area × pressure. Therefore, as long as the pressure remains the same, the height of the mercury will also remain the same.

Thus the height of the mercury = 760 mm

7 0
3 years ago
What name is applied to sound vibrations above audio range?<br><br>​
asambeis [7]
Audio frequency

An audio frequency or audible frequency is a periodic vibration whose frequency is in the band audible to the average human, the human hearing range.
4 0
3 years ago
-What can you say about the snowboarder’s kinetic energy as he moves?
Damm [24]

Answer:

  His kinetic energy increases, potential energy decreases

  The sum of kinetic and potential energy is a constant at any instant before he comes to rest.

Explanation:

  Snowboarder is starting from a height and moving to the down direction. As he moves down his velocity increases, we know that kinetic energy is given by the expression \frac{1}{2} mv^2, so as he moves his kinetic energy increases.

  When the snowboarder is starting his potential energy is maximum(Potential energy = mgh), as he comes down his potential energy decreases.

  Based on this we can conclude that the sum of potential energy and kinetic energy is a constant at any instant for a snowboarder before he comes to rest.

                             mgh+\frac{1}{2} mv^2= Constant

 

7 0
4 years ago
Other questions:
  • A 15 kg box is accelerated from 15m/s to 30 m/s in 36 seconds. What is the net force exerted on the box?
    14·1 answer
  • find the resistance of a resistor connected to a 3v voltmeter and a 3A ammeter, resistance box along with cells of EMF 3V​
    11·1 answer
  • Elements, or single atoms, are built out of:
    11·2 answers
  • . MONEY The Wayside Hotel charges its guests $1 plus $0.80 per minute for long distance calls. Across the street, the Blue Sky H
    6·1 answer
  • If a gas has a volume of 1.25 L and a pressure of 1.75 atm, what will the pressure be if the volume is changed to 3.15 L?
    9·1 answer
  • What is the law of conservation of matter
    14·2 answers
  • A circuit containing an inductor and a capacitor in series is designed to have a resonant frequency of 4511 Hz. If the inductor
    9·1 answer
  • E=<br> (500.0lm)<br> 4 (100)
    7·1 answer
  • Which is the best explanation for his results?
    5·2 answers
  • What happen if we found the way to another universe? and how can we send people there ?​
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!