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____ [38]
4 years ago
6

A 1.5-kilogram cart initially moves at 2.0 m/s. It is brought to rest by a constant net force in 0.3 seconds. What is the magnit

ude of the net force?
Physics
1 answer:
ser-zykov [4K]4 years ago
5 0
We know, F = m.a
Here, m = 1.5 Kg
a = v/t = 2/0.3 = 6.67 m/s

Substitute their values, 
F = 1.5 * 6.67
F = 10 N

In short, Your Answer would be 10 Newtons

Hope this helps!
You might be interested in
The following data were collected during a short race between two friends. Velocity (m/s) 0 0.5 1 1.5 2 2 4 6 2 0 Time (s) 0 2 4
scoundrel [369]

The characteristics of the kinematics allow to find the results for the questions about the movement of the body are:

a)  we have four sections;

  • 0 to 8 s The body is accelerating.
  • 8 to 10 s The body goes at a constant speed, the acceleration is zero.
  • 10 to 14 Body accelerating.
  • 14 to 18 Body slowing down.

b)  The acceleration is the first 8 s is:  a = 0.25 m / s²

c) The maximum acceleration is:    a = 1 m / s²

d) The displacement   is:  i) d₁ =  8m,     ii)  d_{total}= 16 m

e) maximum speed  is:      v = 6 m / s

Kinematics studies the movement of bodies by finding relationships between the position, speed and acceleration of bodies.

        v = v₀ + a t

        y = v₀ t + ½ a t²

where v and v₀ is the current and initial velocity, respectively, a is the acceleration and t is time.

In many circumstances graphs are made for their analysis, in a graph of speed versus time when we have a horizontal line the speed is constant, the acceleration is zero and in the case of a slope there is an acceleration, we have two cases:

  • Positive slope the body is accelerating and the speed is increasing.
  • Negative slope the body is stopping, the speed decreases.

Let's answer the different questions about the system.

a) in the attached we have a graph of the velocity versus time, each section corresponds to a change in the slope of the graph, we have four sections;

  • 0 to 8 s The body is accelerating.
  • 8 to 10 s The body goes at a constant speed, the acceleration is zero.
  • 10 to 14 Body accelerating.
  • 14 to 18 Body slowing down.

b) The acceleration is the first 8 s

          v = v₀ + a t

          a = \frac{v-v_o}{\Delta t}  

          a = \frac{2-0}{8-0}  

          a = 0.25 m / s²

c) The maximum acceleration is when the slope is maximum.

          a = \frac{6-2}{ 14-10}  

          a = 1 m / s²

Therefore the acceleration is maximum in the section between 10 and 14 s

d) The total displacement is the sum of the displacements of each section.

         d_{total } = d_1 +d_2 + d_3 +d_4  

We look for every displacement.

       d₁ = v₀ + ½ a₁ Δt²

       d₁ = 0 + ½ 0.25 8²

       d₁ = 8 m

In the second section the velocity is constant

         d₂ = v₂ Δt₂

         d₂ = 2 (10-8)

         d₂ = 4 m

The third section.

    d₃ = v₀ + ½ a t²

    d₃ = 2 + ½ 1 (14-10) ²

    d₃ = 10 m

The distance of the fourth section.

       

we look for acceleration

          a₄ = \frac{v-v_o}{\Delta t}  

          a₄ = \frac{0-6}{18-14}  

          a₄ = -1.5 m / s²

     

          d₄ = 6 + ½ (-1.5) (1814) ²

          d₄ = -6 m

The total displacement is;

          d_{total} = 8 + 4 + 10 -6

          d_{total} = 16 m

e) The maximum speed is the highest point in the graph of speed versus time that in the attachment we can see corresponds to

          v = 6 m / s

In conclusion using the characteristics of kinematics we can find the results for the questions about the motion of bodies are:

  a)  we have four sections;

  • 0 to 8 s The body is accelerating.
  • 8 to 10 s The body goes at a constant speed, the acceleration is zero.
  • 10 to 14 Body accelerating.
  • 14 to 18 Body slowing down.

b)  The acceleration is the first 8 s is:  a = 0.25 m / s²

c) The maximum acceleration is:    a = 1 m / s²

d) The displacement   is:  i) d₁ =  8m,     ii)  d_{total}= 16 m

e) maximum speed  is:      v = 6 m / s

Learn more about kinematics here: brainly.com/question/24783036

3 0
2 years ago
A motor has a rotor (with armature) of moment of inertia ????m . The rotor is attached to a gear box of gear ratio G > 1. The
Inessa [10]

Answer:

hello your question is incomplete attached below is the complete question

answer : The moment of inertial felt by someone ( J ) is greater that the moment of inertia felt by the motor  i.e. J > Jm

Explanation:

Gear ratio G > 1

a) Determine the moment of inertia felt by the motor

moment of inertia felt by Motor = moment of Inertia at the armature

b) Determine the moment of inertial felt by someone who is rotating the mass by hand

moment of inertia felt by someone is = J

The moment of inertial felt by someone ( J ) is greater that the moment of inertia felt by the motor

attached below is a detailed solution

3 0
3 years ago
A red laser beam goes from crown glass with refraction index n=1.3 to air (n=1) with an incident angle of 0.23 radians. What is
Klio2033 [76]

Answer:

   θ = 10.28º

Explanation:

To find the angle of refraction use the equation of refraction

        n₁ sin θ₁ = n₂ sin θ₂

where index 1 is for incident light and index 2 is for refracted light.

         sin θ₂ = n₁ / n₂ sin θ

let's calculate

         sin = 1 / 1.3 sin 0.23

         sin = 0.175

        θ= 0.17528 rad

let's reduce to degrees

       θ = 0.17528 rad (180ª / pi rad)

       θ = 10.28º

6 0
3 years ago
Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and
strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance R_e defined by the formula:

\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\

and when this is combined with the third resistor in series, the equivalent resistance (R''_e) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}

The problem states that the difference between the equivalent resistances in both circuits is given by:

R''_e=R_e+630 \,\Omega

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega

8 0
3 years ago
Where does most star formation occur in the milky way galaxy?
k0ka [10]

Answer:

Star formation occurs most rapidly in the spiral arms, where the density of interstellar matter is highest.

4 0
2 years ago
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