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
noname [10]
2 years ago
6

A chair of mass 11.5 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force F

= 42.0 N that is directed at an angle of 38.0 below the horizontal and the chair slides along the floor.
*Use Newton's laws to calculate the normal force that the floor exerts on the chair.

Physics
2 answers:
Oksi-84 [34.3K]2 years ago
5 0

Answer:112.7 N

Explanation: mass of chair = 11.5kg, g= acceleration due to gravity = 9.8m/s^2

F = applied force = 42.0 N

The normal reaction is always acting upwards which is given as

R = mg = 11.5 × 9.8 = 112.7 N

stira [4]2 years ago
4 0

Answer: Normal force = 139N

Explanation:

Please find the attached file for the solution

You might be interested in
Sarah is sitting at the top of the slide in the school's playground. Which of the following changes will increase Sarah's kineti
Ray Of Light [21]
B; kinetic energy is anything in motion
7 0
3 years ago
A loop of current-carrying wire has a magnetic dipole moment of 5. 0 10–4 am2. if the dipole moment makes an angle of 57° with a
Digiron [165]

The potential energy will be 1.46*10^-4J.

To find the answer, we have to know about the torque acting on a current loop in a uniform magnetic field.

<h3>How to find the potential energy of the loop?</h3>
  • We have the expression for torque acting on a current loop in a uniform magnetic field as,

                         \tau=MBsin\theta

where; M is the magnetic dipole moment, B is the magnetic field , and theta is the angle between M and B.

  • As we know that, the torque is equal to force times the perpendicular distance. Thus, it is equivalent to the work done. This work is stored as the potential energy in the loop.
  • Thus, the potential energy will be,

            \tau=W=U=MBsin\theta=5*10^{-4}*0.35*sin57=1.46*10^{-4}J

Thus, we can conclude that, the potential energy will be 1.46*10^-4J.

Learn more about the torque here:

brainly.com/question/27949876

#SPJ4

7 0
1 year ago
How long will it take to travel 200 km traveling 100m/s?
Damm [24]
Convert: 1km = 1000 m 200000 m * 0.01 s/m = 2000 s or 33 minutes and 20 seconds.
7 0
3 years ago
Why are all machines not 100% efficient<br> A. Mass<br> B. Gravity <br> C. Friction <br> D. Distance
Anon25 [30]
All machines are not 100% efficient because of <span>C. Friction</span>
7 0
3 years ago
Read 2 more answers
Three point charges are placed on the y-axis: a charge q at y=a, a charge –2q at the origin, and a charge q at y= –a. Such an ar
den301095 [7]

Answer:

electric field   Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

Explanation:

The electric field is a vector, so it must be added as vectors, in this problem both the charges and the calculation point are on the same x-axis so we can work in a single dimension, remembering that the test charge is always positive whereby the direction of the field will depend on the load under analysis, if the field is positive, if the field is negative.

 a) Let's write the electric field for each charge and the total field

       E = k q /r

With k the Coulomb constant, q the charge and r the distance of the charge to the test point

       Et = E1 + E2 + E3

       E1 = k q / (x-a)²

       E2 = k (-2q) / x²  

       E3 = k q / (x + a)²

       Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

The direction of the field is along the x axis

b) To use a binomial expansion we must have an expression the form (1-x)⁻ⁿ  where x << 1, for this we take factor like x from all the equations

       Et = kq/ x² [1 / (1-a/x)² - 2 + 1 / (1+a/x)²]

We use binomial expansion

     (1+x)⁻² = 1 -nx + n (n-1) 2! x² +… x << 1

     (1-x)⁻² = 1 +nx + n (n-1) 2! x² + ...

They replace in the total field and leaving only the first terms

       

   Et =kq/x² [-2 +(1 +2 a/x + 2 (2-1)/2 (a/x)² +…) + (1 -2 a/x + 2(2-1) /2 (a/x)² +.) ]

   Et = kq/x² [a²/x² + a²/x²2] = kq /x² [2 a²/x²]

Et = k q 2a²/x⁴

point charge

Et = k q 1/x²

Dipole

E = k q a/x³

3 0
2 years ago
Other questions:
  • 7. É dado o movimento S = 100 + 8t, no (S.I) Determine:
    10·1 answer
  • Calculate (A⃗ ×B⃗ )⋅C⃗ for the three vectors A⃗ with magnitude A = 4.86 and angle θA = 23.5 ∘ measured in the sense from the +x
    13·1 answer
  • The change in state of a liquid to a gas​
    14·2 answers
  • Which astronomer spent 20 years plotting the positions of the planets
    13·1 answer
  • What is the potential difference across a 15Ω resistor that has a current of 3.0 A?
    10·2 answers
  • An air balloon is moving upward at a constant speed of 3 m/s. Suddenly a passenger realizes that she left her camera on the grou
    9·1 answer
  • . Calculate the kinetic energy of a 100.0-kg meteor approaching the Earth at a speed of 10.0 km/s. Remember that 1 km = 1000 m.
    12·2 answers
  • Quentin is playing baseball and is batting. He swings and hits the ball sending it flying to the outfield. Which is greater- the
    8·1 answer
  • A 10 kg wagon is accelerated by a constant force of 60 N from an initial velocity of 5.0 m/s to a final velocity of 11 m/s. What
    15·1 answer
  • At rest, hydrogen has a spectral line at 116 nm. if this line is observed at 107 nm for the star sirius, how fast is sirius movi
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!