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Sever21 [200]
4 years ago
5

Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time (when they are not slee

ping or eating) joyfully scampering about on the cage's floor. Bryce tracks his mice's health diligently and just now recorded their masses as 22.3 g, 17.9 g, 19.1 g, and 10.1 g. At this very instant, the x ‑ and y ‑components of the mice's velocities are, respectively, (0.349 m/s, −0.301 m/s), (−0.699 m/s, −0.815 m/s), (0.745 m/s, 0.975 m/s), and (−0.905 m/s, 0.717 m/s). Calculate the x ‑ and y ‑components of Bryce's mice's total momentum, px and py.
Physics
1 answer:
user100 [1]4 years ago
8 0

Answer:

I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s ,  I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s ,  I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s  and I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s

Explanation:

The impulse is equal to the variation of the moment, to apply this relationship to our case, we will assume that initially the mouse was at rest

    I = Δp = m v_{f} -m v₀

    I = m (v_{f}  -v₀)

Bold indicates vector quantities, let's calculate the momentum of each mouse in for the x and y axes

We recommend bringing all units to the SI system

Mouse 1.

It has a mass of 22.3 g = 22.3 10⁻³ kg, a final velocity of (v = 0.349 i ^ - 0.301 j ^) m / s with an initial velocity of zero

    Iₓ = m (v_{fx}  - v₀ₓ)

    Iₓ = 22.3 10⁻³ (0.349 -0)

    Iₓ = 7.78 10⁻³ J s

   I_{y} = m (v_{fy}  -v_{oy} )

   I_{y} = 22.3 10⁻³ (-0.301)

   I_{y} = -6.71 10⁻³ J s

   I₁ = (7.78 i ^ - 6.71 j ^) 10⁻³ J s

Mouse 2

Mass 17.9 g = 17.9 10⁻³ kg

Speed ​​(-0.699 i ^ - 0.815 j ^) m / s

    Iₓ = m (v_{fx}  - v₀ₓ)

    Iₓ = 17.9 10⁻³ (-0.699 -0)

    Iₓ = -12.5 10⁻³ J s

    I_{y} = 17.9 10⁻³ (-0.815 - 0)

    I_{y} = -14.6 10⁻³ J s

   I₂ = (-12.5 i ^ -14.6 j ^) 10⁻³ J s

Mouse 3

Mass 19.1 g = 19.1 10⁻³ kg

Speed ​​(0.745i ^ + 0.975 j ^) m / s

    Iₓ = 19.1 10⁻³ (0.745 -0)

    Iₓ = 14.2 10⁻³ J s

    I_{y} = 19.1 10⁻³(0.975 -0)

    I_{y} = 18.6 10⁻³ J s

    I₃ = (19.1i ^ + 18.6 j ^) 10⁻³ J s

Mouse 4

Mass 10.1 g = 10.1 10⁻³ kg

Speed ​​(-0.905i ^ + 0.717j ^) m / s

    Iₓ = 10.1 10⁻³ (-0.905 -0)

    Iₓ = -9.14 10⁻³ J s

    I_{y} = 10.1 10⁻³ (0.717 -0)

    I_{y} = 7.24 10⁻³ J s

   I₄ = (-9.14i ^ + 7.24 j ^) 10⁻³ J s

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Answer:

1.94\cdot 10^{20} N

Explanation:

The magnitude of the gravitational force between two objects is given by the equation:

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

The gravitational force is always attractive.

In this problem, we have:

m_1 = 5.98\cdot 10^{24}kg is the mass of the Earth

m_2 = 7.34\cdot 10^{22} kg is the mass of the Moon

r=3.88\cdot 10^8 m is the separation between the Earth and the Moon

Therefore, the gravitational force between them is

F=(6.67\cdot 10^{-11})\frac{(5.98\cdot 10^{24})(7.34\cdot 10^{22})}{(3.88\cdot 10^8)^2}=1.94\cdot 10^{20} N

6 0
3 years ago
Can the velocity of a body revese the direction when acceleration is constant?
TEA [102]

Answer:

Yes, the velocity of the object can reverse direction when its acceleration is constant. For example consider that the velocity of any object at any time t is given as: ... At At t = 0 sec, the magnitude of velocity is 2m/s and is moving in the forward direction i.e.v (t) = -2.

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Lila is a track and field athlete. She must complete four laps around a circular track. The track itself is a 400 meter track an
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Answer:

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Explanation:

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Given

Distance = 400m

Time = 6minutes = 6*60 = 360 secs

Substitute the given parameter into the formula;

Speed = 400/360

Speed = 1.1m/s

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Velocity = Displacement/Time

Velocity = 0/3600

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Answer:

Displacement: 6.71 m, Direction: 63.4 degrees north of east

Explanation:

In the attached image we can aprecciate each one of the movements of the parade. Let's say that the parade started from the origin (point (0,0)) then it moves to the east 4 blocks it means now the parade is located at point (4,0).

Then the parade went to the south three blocks, so it moves to the coordinate (4,-3). After this the parade went to the west one block so the new coordinate point is (3, -3).

And finally the movement of the 0 parade was 9 blocks to the north. It means the final point is now (0,9) - (3,-3) = (3,6)

And the displacement will be defined by the folliwing vector operation:

A (0,0) = oi + 0j\\F (3,6) = 3i + 6 j\\Displacement vector = (3-0)i + (6-0)j = 3i + 6j

We know that the magnitude of the displacement vector is defined by the phytagoras theorem

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