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
GrogVix [38]
3 years ago
13

You hold a 0.125 kg glider A and a 0.500 kg glider B at rest on an air track with a compressed spring of negligible mass between

them. When you release the gliders, the spring pushes them apart so that they move in opposite directions. When glider A has moved 0.960 m to the left from its starting position, how far to the right from its starting position has glider B moved?
Physics
1 answer:
o-na [289]3 years ago
8 0

Answer:

Magnitude |VB| = 0.32 m/s

and

Direction is to the left

Explanation:

given data

glider A mass = 0.125 kg

glider B mass = 0.500 kg

glider A move = 0.960 m to the left

solution

we use here Conservation of momentum  that is

M1 × U1 + M2 × U2 =M1 × V1 + M2 × V2    ........................1

put here value and we get

0.125 × 0 + 0.500 × 0 = 0.125 × 0.960 + 0.375 × VB

VB = −0.32  m/s

Magnitude |VB| = 0.32 m/s

and

Direction is to the left

You might be interested in
Which graph accurately shows the relationship between kinetic energy and speed as speed increases?
mixer [17]

Answer:

B

Explanation:

kinetic energy (KE) is the energy possessed by moving bodies. It can be expressed as:

KE = \frac{1}{2}mv^{2}

Where: m is the mass of the object, and v its speed.

For example, a stone of mass 2kg was thrown and moves with a speed of 3 m/s. Determine the kinetic energy of the stone.

Thus,

KE =  \frac{1}{2} x 2 x 3^{2}

     = 9

KE = 9.0 Joules

Assume that the speed of the stone was 4 m/s, then its KE would be:

KE =  \frac{1}{2} x 2 x 4^{2}

     = 16

KE = 16.0 Joules

Therefore, it can be observed that as speed increases, the kinetic energy increases. Thus option B is appropriate.

3 0
3 years ago
In an RC circuit, what fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for
4vir4ik [10]

Answer:

The  fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is  

      k  = 0.903

Explanation:

From the question we are told that

     The time  constant  \tau  =  3

The potential across the capacitor can be mathematically represented as

     V  =  V_o  (1 -  e^{- \tau})

Where V_o is the voltage of the capacitor when it is fully charged

    So   at  \tau  =  3

     V  =  V_o  (1 -  e^{- 3})

     V  =  0.950213 V_o

   Generally energy stored in a capacitor is mathematically represented as

             E = \frac{1}{2 } * C  * V ^2

In this equation the energy stored is directly proportional to the the square of the potential across the capacitor

Now  since capacitance is  constant  at  \tau  =  3

        The  energy stored can be evaluated at as

         V^2 =  (0.950213 V_o )^2

       V^2 =  0.903  V_o ^2

Hence the fraction of the energy stored in an initially uncharged capacitor is  

      k  = 0.903

4 0
3 years ago
How are mixtures and pure substances related?
Elden [556K]
You are talking about make sure's and pearl substance I thought you was talking about mix in with something
4 0
3 years ago
A bar of length L = 8 ft and midpoint D is falling so that, when θ = 27°, ∣∣v→D∣∣=18.5 ft/s , and the vertical acceleration of p
777dan777 [17]

Answer:

alpha=53.56rad/s

a=5784rad/s^2

Explanation:

First of all, we have to compute the time in which point D has a velocity of v=23ft/s (v0=0ft/s)

v=v_0+at\\\\t=\frac{v}{a}=\frac{(23\frac{ft}{s})}{32.17\frac{ft}{s^2}}=0.71s

Now, we can calculate the angular acceleration  (w0=0rad/s)

\theta=\omega_0t +\frac{1}{2}\alpha t^2\\\alpha=\frac{2\theta}{t^2}

\alpha=\frac{27}{(0.71s)^2}=53.56\frac{rad}{s^2}

with this value we can compute the angular velocity

\omega=\omega_0+\alpha t\\\omega = (53.56\frac{rad}{s^2})(0.71s)=38.02\frac{rad}{s}

and the tangential velocity of point B, and then the acceleration of point B:

v_t=\omega r=(38.02\frac{rad}{s})(4)=152.11\frac{ft}{s}\\a_t=\frac{v_t^2}{r}=\frac{(152.11\frac{ft}{s})^2}{4ft}=5784\frac{rad}{s^2}

hope this helps!!

6 0
3 years ago
Read 2 more answers
If you lived by the sea, what effect of the moon would you see? Describe the effect.
dolphi86 [110]

Answer:

the changes in waves

Explanation:

the moon has its own gravitational pull thus making waves  and the rising tides

8 0
3 years ago
Other questions:
  • According to the _______ the amount of energy in the universe doesn't change.
    12·2 answers
  • A circuit has a resistance of 10ohmsband a current of 42 amps.caculate the voltage
    15·1 answer
  • A car speeds up from rest to +16 m/ s in 4s. calculate the acceleration
    15·2 answers
  • What is measurement​
    11·1 answer
  • Structures that trap light energy and perform photosynthesis
    13·1 answer
  • A wave with a frequency of 56 Hz has a wavelength of 27 meters. At what speed will this wave travel?
    8·1 answer
  • A 100 N force causes an object to accelerate at 2 m/s2. What is the mass of the<br> object?
    13·1 answer
  • A metal wire breaks when its tension reaches 100 newton. If the radius and length of the wire were both doubled then it would br
    5·1 answer
  • What part of the scientific method tests the hypothesis?
    6·1 answer
  • If a 5.00 kg box slides down a ramp inclined at 60.0° above the horizontal, what is the
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!