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
faust18 [17]
3 years ago
11

Lucy is cruising through space in her new spaceship. As she coasts along, a tiny spacebug drifts into her path and bounces off t

he window. Consider several statements concerning this scenario. Evaluate each statement according to the law of momentum conservation and match it to the appropriate category.
Physics
1 answer:
Elden [556K]3 years ago
4 0

Your question is not complete, please let me assume this to be your complete question:

Lucy is cruising through space in her new spaceship. As she coasts along, a spacebug drifts into her path and bounces off the window. Consider several statements concerning this scenario. Evaluate each statement according to the law of momentum conservation and match it to the appropriate category.

Below are several statements concerning this scenario. Evaluate each statement and decide if it is true, false, or undetermined by the principle of momentum conservation.

i) The total change in momentum for this interaction is zero.

ii) The change in the space bug's momentum is greater than the change in the spaceship's momentum.

iii) If the space bug had stuck to the spaceship instead of bouncing off, momentum would not have been conserved for this interaction.

Answer:

I is true

II is false

III is undetermined

Explanation:

STATEMENT I : This statement is TRUE because because they was no change in velocity and mass of the spaceship and spacebug, after the collision, as the bug bounced outside the window and the spaceship retains it's velocity. Therefore the total momentum in the system before the collision, is equal to the total momentum of the system after the collision.

Where;

momentum = Mass × velocity

M1V1 = M2V2

Therefore;

M2V2 - M1V1 = 0

STATEMENT II : This statement is FALSE, because the change in the momentum of the spaceship and bug are equal, as the spaceship and bug remains in constant motion after collision. The collision did not have any effect in their velocity nor mass.

Ps = Pb

Ps = Momentum of spaceship

Pb = Momentum of spacebug

STATEMENT III : The statement is UNDETERMINE, because it will depend on the momentum we are considering. If the spaceship is still in constant motion, that means moment of the spaceship is conserved, while that of the spacebug is not conserved.

You might be interested in
A 50 kg bumper car with a 40 kg child and it is at rest when a 60 kg child in her own bumper car slams into it the collision las
IrinaK [193]

Answer:

 F = 99 v₂₀

v₂₀ = 1 m / s,        F = 99 N

Explanation:

In this exercise it is asked to find the force during the collision, for this we use the relationship between the momentum and the momentum of car 1

            I = Δp

            F t = p_f- p₀

            F t = m (v_f -v₀)                        (1)

We must find the final speed of car 1, for this we define a system formed by the two cars, in this case the forces during the collision are internal and the moment is conserved

initial instant. Before the crash

        p₀ = 0 + m₂ v₂₀

         

final instant. After the crash

        p_f = m₁ v₁ + m₂ v_{2f}

the moment is preserved

        p₀ = p_f

        m₂ v₂₀ = m₁ v_{1f} + m₂ v_{2f}           (2)

        m₂ (v₂₀ - v_2f}) = m₁ v_{1f}

as the collision is elastic the kinetic energy is also conserved

        K₀ = K_f

        ½ m₂ v₂₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²

        m₂ (v₂₀² -v_{2f}²) = m₁ v_{1f}²

let's write our system of equations, using

         a² - b² = (a + b) (a-b)

         m₂ (v₂₀ - v_{2f}) = m₁ v_{1f}

         m₂ (v₂₀ -v_{2f}) (v₂₀ + v_{2f}) = m₁ v_{1f}²

to solve we divide the equations

       v₂₀ + v_{2f} = v_{1f}

with this we substitute in equation 2 and find the speed of each car, in this case we need the speed of car 1

         m₂ v₂₀ = m₁ v_{1f} + m₂ (v_{1f}-v₂₀)

         2m₂ v₂₀ = (m₁ + m₂) v_{1f}

          v_{1f} = \frac{2m_2}{m_1+m_2}  v_{2o}

We substitute in the drive ratio of car 1

            F t = m (v_f -v₀)

            F = m₁ (\frac{2m_2}{m_1+m_2}  v_{2o} - 0) / t

            F = \frac{2m_1 m_2 }{m_1+m_2}   \   \frac{v_{2o}}{t}

the mass of each car is the mass of the car plus the mass of the boy

           m₁ = 50 +40 = 90 kg

           m₂ = 50 +60 = 110 kg

     

time is t = 1

         

we substitute the values

           F = \frac{ 2\  90 \ 110}{90+110}  \ \frac{v_{2o}}{1}2 90 100/90 + 110 vo2 / 1

           F = 99 v₂₀

The value of the initial velocity of car 2 is not indicated in the problem, if this velocity is known it can be included and the force value is obtained, suppose that the initial velocity v₂₀ = 1 m / s

           F = 99 N

4 0
3 years ago
Most people would consider a comfortable room temperature to be
Alla [95]

B

A  is extremly hot

and c is -330.07 degrees farenheit

4 0
3 years ago
Read 2 more answers
If the weight of displaced liquid and the weight of an object are the same, the object will:
liberstina [14]

Answer:

c

Explanation:

I think but if I'm wrong I'm sorry

8 0
3 years ago
Read 2 more answers
¡¡¡AYUDA CON ESTOS EJERCICIOS DE FÍSICA!!!
asambeis [7]

Answer:

(a) 8 V, (b) 144000 V, (c) 2 x 10^(-8) C

Explanation:

(a) charge, q = 5 μC , Work, W = 40 x 10-^(-6) J

The electric potential is given by

W = q V

40\times10^{-6}=5 \times10^{-6}\times V\\\\V = 8 V

(b)

charge, q = 8 x 10^(-6) C, distance, r = 50 cm = 0.5 m

Let the potential is V.

V =\frac{k q}{r}\\\\V =\frac{9\times 10^{9}\times 8\times 10^{-6}}{0.5}\\\\V =144000 V

(c)

Work, W = 8 x 10^(-5) J, Potential difference, V = 4000 V

Let the charge is q.

W= q V

8\times10^{-5}= q\times 4000\\\\q =2\times 10^{-8} C

3 0
3 years ago
A brick sits on the top of a hill with a gravitational potential energy of 245 J. To determine the gravitational potential of th
Bogdan [553]

Answer:

The mass of the object, its acceleration due to gravity and the distance between the top of the hill and the ground level.

Explanation:

gravitational potential energy is the energy possessed by a body under influence of gravitational force by virtue of its position.

In order to determine the gravitational potential energy of the brick, we must know the mass (m) of the brick, its acceleration due to gravity (g) since it is acting under the influence of gravitational force and the distance between the top of the hill and the ground level. (The height).

Potential energy of a body is calculated as mass × acceleration due to gravity × height.

5 0
3 years ago
Other questions:
  • 1kg slab of concrete loses 12,000j of heat when it cools from 30 Celsius to 26 Celsius. Determine the specific heat capacity of
    10·1 answer
  • The temperature of the soda
    8·2 answers
  • A 5 kg block is being pulled to the right by a rope tied to it. the block is accelerating at 2 m/s2 to the right. how much force
    8·2 answers
  • You are riding in a train that is traveling at a speed of 120km /h how long will it take to travel 950 km
    13·1 answer
  • How is circular motion different from variable motion?
    9·1 answer
  • How do we gather most of the information we get about the universe around us?
    13·1 answer
  • Question 3 of 5
    14·1 answer
  • What affects the strength of electric and magnetic forces?
    9·1 answer
  • Superman strikes a golf ball on the ground at a 38 degree angle above the horizontal at 147 m/s. What is the maximum height the
    13·1 answer
  • When ripples pass though a gap in a barrier what is observed
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!