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
kondor19780726 [428]
3 years ago
6

How many chirps will a snowy tree cricket give is 21 seconds at a temperature of 22 degree celcius

Physics
1 answer:
Studentka2010 [4]3 years ago
7 0
There is an established correlation between the number of chirps of a cricket and temperature. The function is
ln C = -6713.2 / T + 25.868

Where c is the number of chirps per min and T is the temperature in K
We are given
T = 22 C + 273 = 295K

Solving for C
ln C = -6713.2 / 295 + 25.868
C = 22.45 chirps/min

In 21 seconds, the number of chips is
22.45 chirps / min (1 min/60 s) (21 s) = 7.86 chirps <span />
You might be interested in
An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the radius were changed to 1000 m, but the sp
laila [671]

Answer:

The centripetal acceleration changed by a factor of 0.5

Explanation:

Given;

first radius of the horizontal circle, r₁ = 500 m

speed of the airplane, v = 150 m/s

second radius of the airplane, r₂ = 1000 m

Centripetal acceleration is given as;

a = \frac{v^2}{r}

At constant speed, we will have;

v^2 =ar\\\\v = \sqrt{ar}\\\\at \ constant\ v;\\\sqrt{a_1r_1} = \sqrt{a_2r_2}\\\\a_1r_1 = a_2r_2\\\\a_2 = \frac{a_1r_1}{r_2} \\\\a_2 = \frac{a_1*500}{1000}\\\\a_2 = \frac{a_1}{2} \\\\a_2 = \frac{1}{2} a_1

a₂ = 0.5a₁

Therefore, the centripetal acceleration changed by a factor of 0.5

7 0
3 years ago
An engine absorbs 1.69 kJ from a hot reservoir at 277°C and expels 1.25 kJ to a cold reservoir at 27°C in each cycle.
Anna35 [415]

Answer

Given,

Energy absorbed, Q_H = 1.69\ kJ

Energy expels,Q_C =  1.25\ kJ

Temperature of cold reservoir, T = 27°C

a) Efficiency of engine

 \eta = \dfrac{Q_H - Q_C}{Q_H}\times 100

 \eta = \dfrac{1.69 - 1.25}{1.69}\times 100

\eta =26.03 %

b) Work done by the engine

 W = Q_H- Q_C

 W =1.69 - 1.25

 W = 0.44\ kJ

c) Power output

     t = 0.296 s

   P = \dfrac{W}{t}

   P = \dfrac{0.44}{0.296}

   P = 1.486\ kW

8 0
3 years ago
The sum of two component vectors is referred to as the
blondinia [14]
Resultant is the correct answer!
8 0
3 years ago
Help it’s multiple choice 11 through 15 please!
riadik2000 [5.3K]

1. • Here, force of gravity on the block = 20 N.

• Therefore, the normal force will also be the same, i.e., 20 N [According to Newton's Third Law, on every action, there is an equal and opposite reaction]

• The coefficient

u_{k} = 0.4

• Force of friction =

u_{k} \times  \: normal \:  \:  \: force \\  = 0.4 \times 20N \\  = 8N

• Hence, the force of sliding friction between the block and the ground is 8 N.

• So, it is option c. 8 N

2. The answer is option d. continue in the same direction with no change in speed.

We know, force = mass × acceleration. When force is 0, then acceleration will also be 0 since mass cannot be 0. So, there will be no change in speed.

3. It is option b. force that is required to give a one kilogram object the acceleration of 1 m/s^2.

Newton is the SI unit of force. As mentioned earlier, force = mass × acceleration. The SI unit of mass and acceleration is Kg and m/s^2 respectively.

So, 1 N = 1 Kg × 1 m/s^2.

4. It is d. not zero.

Acceleration is the change in speed. So, if the force is zero, then acceleration will not occur.

5. Force = 2 N

Acceleration of the object A = 2 m/s^2.

Acceleration of the object B = 1 m/s^2.

Therefore, mass of the object A = 2 N ÷ 2 m/s^2 = 1 Kg

And, mass of the object B = 2 N ÷ 1 m/s^2 = 2 Kg

So, the mass of object B is greater than that of object A.

Hence, the answer is option c. Object B has more mass.

Hope you could get an idea from here.

Doubt clarification - use comment section.

3 0
2 years ago
14.A 90 kg quarterback gets tackled by being hit by a 120 kg lineman backwards
quester [9]

The acceleration of the quarterback and the lineman is 5.55m/s² and 4.16m/s² respectively in the same direction.

As, we know, the 120 Kg lineman is moving with a force of 500N.

His net acceleration will be in the same direction as his motion.

It is already known that, If M is the mass of the body and a is the acceleration of the body, then the force F on the body can be calculated by using the formula,

F = Ma.

The weight of the quarterback is 90 Kg. He is being hit by a force of 500N.

So, the acceleration can be calculated using the formula,

500N = 90kg x a

a = 5.55 m/s².

Now, the weight if the lineman is 120kg, the force applied by him is 500N.

So, from the formula, his acceleration A will be,

500N = 120Kg x A

A = 4.16 m/s².

both of them will have acceleration in the same direction,

To know more about Force, visit,

brainly.com/question/25239010

#SPJ9

8 0
1 year ago
Other questions:
  • An 80-kg astronaut becomes separated from his spaceship. He is 15.0 m away from it and at rest relative to it. In an effort to g
    6·1 answer
  • What is special about a DC circuit?
    6·2 answers
  • What is an example of chemical potential energy in humans?
    15·1 answer
  • A satellite in the shape of a solid sphere of mass 1,900 kg and radius 4.6 m is spinning about an axis through its center of mas
    13·1 answer
  • How is ultraviolet light and infrared light alike? How are they different?
    12·1 answer
  • Need help on this please
    14·2 answers
  • ¿Cómo obtener razonadamente la primera ley de newton a partir de la segunda?
    6·1 answer
  • Which of the following is not a compound <br> A. Gold <br> B. Water<br> C. Salt<br> D. Sugar
    13·1 answer
  • Why can't 41 be represented in a binary table?
    15·1 answer
  • A merry-go-round rotates at the rate of
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!