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
blagie [28]
4 years ago
7

A net force of magnitude 36 N gives an object of mass m1 an acceleration of 6.0 m/s^2. The same net force gives m1 and another o

bject of mass m2 fastened together an acceleration of 2.0 m/s^2. What acceleration will m2 experience if the same net force acts on it alone?
Physics
1 answer:
Jet001 [13]4 years ago
3 0
This is a beautiful problem to test whether a student actually understands
Newton's 2nd law of motion . . . Force = (mass) x (acceleration). 

That simple law is all you need to solve this problem, but you need to
use it a few times.

m₁ alone:
                     Force = (mass) x (acceleration)

                       36 N = ( m₁ ) x  (6 m/s²)

                         m₁ = (36 N) / (6 m/s²)   

                         m₁  =  6 kilograms .

m₁ and m₂ glued together:
                     Force = (mass) x (acceleration)

                       36 N = (6 kg + m₂) x (2 m/s²)

                     6 kg + m₂  =  (36 N) / (2 m/s²)  =  18 kilograms

                               m₂  =  12 kilograms .

m₂ alone:
                     Force = (mass) x (acceleration)

                       36 N = (12 kg) x (acceleration)

                     Acceleration = (36 N) / (12 kg)

                     Acceleration  =   3 m/s²


You might be interested in
How does light energy work
OLEGan [10]

Answer:

Light consists of photons, which are produced when an object's atoms heat up. Light travels in waves and is the only form of energy visible to the human eye.

Explanation:

6 0
3 years ago
Read 2 more answers
What does it mean when work is positive?
solniwko [45]

Answer:

O The environment did work on an object

Explanation:

8 0
3 years ago
Read 2 more answers
A projectile is launched at some angle to the horizontal with some initial speed vi, and air resistance is negligible.(a) Is the
Paul [167]

Answer:

Explanation:

When a projectile is launched at some angle to the horizontal with some speed vi , and air resistance is negligible , it is definitely a freely falling body .

It is so because it is free to accelerate towards the earth with acceleration of g . Air has no resistance , hence no force is acting on it except the gravitational force . Hence it is a freely falling body .

b )

The acceleration in the vertical direction is due to force exerted by the earth that is gravitational force on it  . Hence its  acceleration is equal to g in vertically downward direction .

c )

It has zero acceleration in horizontal direction . It is so because no force is acting on it in horizontal direction . So no acceleration will be present in horizontal direction . It will move in horizontal direction with constant speed of vi cos θ where θ is the angle vi make with the horizontal .

8 0
3 years ago
5) Old-style dimmers consisted of a variable resistor so you could dial in more resistance to lower the current and, hence, dim
NeX [460]

Answer:

The power of the bulb is reduced to a quarter

Explanation:

The power in an electrical circuit is

         P = V I = I² R

If the dimmer cuts the current in half

        I = I₀ / 2

The resistance of the light bulb after hot is little affected by small changes in the current

We substitute in the equation

       P = I₂ R

       P = (I₀ /2)² R

      P = ¼ I₀² R

      P = ¼ P₀

The power of the bulb is reduced to a quarter

4 0
4 years ago
A 1350 kg uniform boom is supported by a cable. The length of the boom is l. The cable is connected 1/4 the
olchik [2.2K]

Answer:

Tension= 21,900N

Components of Normal force

Fnx= 17900N

Fny= 22700N

FN= 28900N

Explanation:

Tension in the cable is calculated by:

Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium

FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)

Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)

Ftorque= 2/3FBcostheta+ 4/3FWcostheta

Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°

Ftorque= 21900N

b) components of Normal force

Efx=FNx-FTcos(90-theta)=0 static equilibrium

Fnx=21900cos(90-55)=17900N

Fy=FNy+ FTsin(90-theta)-FB-FW=0

FNy= -FTsin(90-55)+FB+FW

FNy= -21900sin(35)+(1350+2250)×9.81=22700N

The Normal force

FN=sqrt(17900^2+22700^2)

FN= 28.900N

4 0
4 years ago
Other questions:
  • If a cart of 8 kg mass has a force of 16 newtons exerted on it what is its acceleration?
    7·1 answer
  • Why can we see a red shirt
    15·2 answers
  • A car is traveling on a straight, level road under wintry conditions. Seeing a patch of ice ahead of her, the driver of the car
    15·1 answer
  • Heredity includes all of the following except
    8·2 answers
  • A woman lifts her 100-newton child up one meter and carries her for a distance of 50 meters to the child's bedroom. How much wor
    15·2 answers
  • Any object that is given any initial velocity and which follows a path due to gravitational force acting on it and by the fricti
    6·1 answer
  • The doppler effect is when objects are not moving?<br><br> True <br><br> False
    9·1 answer
  • The pulley system has a mechanical advantage of 5. Determine how much input force is required to achieve 1000 N of lifting force
    6·1 answer
  • A measurement of a change<br> in distance over time
    8·1 answer
  • The number of wavelengths that pass a fixed point on a wave is called ______?
    7·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!