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KatRina [158]
3 years ago
10

What are all the moon phases

Physics
2 answers:
kicyunya [14]3 years ago
8 0
New Moon
Waxing Crescent
First Quarter
<span>Waxing Gibbous
</span>Full Moon
<span>Waning Gibbous
</span>Last Quarter
Waning Crescent

;)
photoshop1234 [79]3 years ago
5 0
New moon, waxing crescent, first quarter, waxing gibbous, full moon, waning gibbous, third quarter (or last quarter) , waning crescent .

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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
A Or b <br> Question in picture
Helen [10]

Answer:

the answer to the question is indeed B

5 0
3 years ago
Read 2 more answers
A 500kg car is driven forward with a thrust force of 1500N. Air resistance and friction acts against the motion of the car with
Wittaler [7]
2m/s^2, this is because F=ma, meaning a is also equal to F/m. The car applies 1500N in one direction and outside sources apply a total of -500N, meaning the 500kg car is moving forward with a total of 1000N of force. Taking the total 1000N and dividing it by 500kg gives you and acceleration of 2m/s^2. Hope this helps!
8 0
2 years ago
The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in va
bonufazy [111]

Answer:

10000 V

0.00225988700565 m²

8\times 10^{-12}\ F

Explanation:

E = Electric field = 4\times 10^6\ V/m

d = Gap = 2.5 mm

Q = Charge = 80 nC

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

Potential difference is given by

V=Ed\\\Rightarrow V=4\times 10^6\times 2.5\times 10^{-3}\\\Rightarrow V=10000\ V

The potential difference between the plates is 10000 V

Area is given by

A=\dfrac{Q}{\epsilon_0E}\\\Rightarrow A=\dfrac{80\times 10^{-9}}{8.85\times 10^{-12}\times 4\times 10^6}\\\Rightarrow A=0.00225988700565\ m^2

The area of the plate is 0.00225988700565 m²

Capacitance is given by

C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.85\times 10^{-12}\times 0.00225988700565}{2.5\times 10^{-3}}\\\Rightarrow C=8\times 10^{-12}\ F

The capacitance is 8\times 10^{-12}\ F

4 0
3 years ago
Write two main group of animals​
Nesterboy [21]

Answer:

Animals are classifield as two main groups in the animal kingdom : Vertebrates and invertebrates .

Explanation:

keep it up....

7 0
3 years ago
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