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
Dominik [7]
3 years ago
7

What will happen if a car experiences a 300 N force to the right from the engine and a separate 150 N force due to friction and

air resistance to the left? Show your work to support your position.
Physics
1 answer:
gayaneshka [121]3 years ago
5 0

Force applied on the car due to engine is given as

F_1 = 300 N towards right

Also there is a force on the car towards left due to air drag

F_2 = 150 N towards left

now the net force on the car will be given as

\vec F_{net} = \vec F_1 + \vec F_2

now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.

So we can say

F_{net} = F_1 - F_2

F_{net} = 300 - 150

F_[net} = 150 N

So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.

You might be interested in
Calculate the index of refraction for a medium in which the speed of light is 2.1x 108 m/s. The speed of light in vacuum is 3x10
strojnjashka [21]

Answer:

n = 1.42

Explanation:

The refractive index for a medium is given by the ratio of the speed of light in vacuum to the speed of light in a medium.

n=\dfrac{c}{v}\\\\n=\dfrac{3\times 10^8}{2.1\times 10^8}\\\\n = 1.42

So, the refractive index of the medium is 1.42.

5 0
3 years ago
How does temperature influence a chemical change?
Serga [27]
Your weight #sorryfortheweight
8 0
3 years ago
Which measurements are equal to 321 decimeters? Check all that apply.
S_A_V [24]
32,100 Millimeters
3,210 Centimeters
3.21 Decameters

Hope It Helps
5 0
3 years ago
Read 2 more answers
A rigid, nonconducting tank with a volume of 4 m3 is divided into two unequal parts by a thin membrane. One side of the membrane
kondor19780726 [428]

The final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

To find the answer, we need to know about the thermodynamic processes.

<h3>How to find the final temperature of the gas?</h3>
  • Any processes which produce change in the thermodynamic coordinates of a system is called thermodynamic processes.
  • In the question, it is given that, the tank is rigid and non-conducting, thus, dQ=0.
  • The membrane is raptured without applying any external force, thus, dW=0.
  • We have the first law of thermodynamic expression as,

                                dU=dQ-dW

  • Here it is zero.

                                  dU=0,

  • As we know that,

                             dU=C_pdT=0\\\\thus,  dT=0\\\\or , T=constant\\\\i.e, T_1=T_2

  • Thus, the final temperature of the system will be equal to the initial temperature,

                          T_1=T_2=100^0C=373K

<h3>How much work is done?</h3>
  • We found that the process is isothermal,
  • Thus, the work done will be,

                               W=RT*ln(\frac{V_2}{V_1} )=373R*ln(\frac{4}{\frac{4}{3} })\\ \\W=409.8R J

Where, R is the universal gas constant.

<h3>What is a reversible process?</h3>
  • Any process which can be made to proceed in the reverse direction is called reversible process.
  • During which, the system passes through exactly the same states as in the direct process.

Thus, we can conclude that, the final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

Learn more about thermodynamic processes here:

brainly.com/question/28067625

#SPJ1

7 0
2 years ago
Read 2 more answers
An open pipe, 0.29 m long, vibrates in the second overtone with a frequency of 1,227 Hz. In this situation, the fundamental freq
Andru [333]

Answer:

f = 409 Hz

Explanation:

We have,

Length of the open organ pipe, l = 0.29 m

Frequency of vibration of second overtone, f_2 = 1227 Hz

It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :

f_2=\dfrac{3v}{2l}

v is speed of sound

Let f is the fundamental frequency. It is given by :

f=\dfrac{v}{2l}

The relation between f and f₂ can be written as :

f_2=3f\\\\f=\dfrac{f_2}{3}\\\\f=\dfrac{1227}{3}\\\\f=409\ Hz

So, the fundamental frequency of the pipe is 409 Hz.              

6 0
3 years ago
Other questions:
  • What happens that a light ray if it incident on a reflective surface along the normal
    13·1 answer
  • Which of the following phrases best describes the term "magnetic flux"?
    11·1 answer
  • How much do you know about the water cycle and the impact that humans have on the water cycle?
    8·1 answer
  • What do astronomers mean by light pollution?
    6·1 answer
  • Why are lanthanum and actinium at the bottom of the periodic table?
    6·1 answer
  • “I am sure I will get a good grade on this project. I’ve worked really hard; my teacher will recognize my efforts!” This stateme
    5·2 answers
  • The question and answer options are in the photo!
    5·1 answer
  • a car accelerates at a constant rate from 15 m/s to 25 m/s while it travels a distance of 125 m. How long does it take to achiev
    11·1 answer
  • Plz help me thank you
    6·1 answer
  • 4. A 20-kg box sits on an incline of 30° from the horizontal. The coefficient of friction between the box and the incline is o.3
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!