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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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Aliun [14]

Answer:

The value is      A   = 39315 \  m^2

Explanation:

From the question we are told that

    The velocity which the rover is suppose to land with is  v  =  1 \ m/s

    The  mass of the rover and the parachute is  m  =  2270 \ kg

     The  drag coefficient is  C__{D}}  =  0.5

      The atmospheric density of Earth  is  \rho =  1.2 \  kg/m^3

     The acceleration due to gravity in Mars is  g_m  =  3.689 \  m/s^2

     

Generally the Mars  atmosphere density is mathematically represented as

          \rho_m  =  0.71 *  \rho

=>        \rho_m  =  0.71 *  1.2

=>        \rho_m  = 0.852 \  kg/m^3

Generally the drag force on the rover and the parachute  is mathematically represented as

          F__{D}} =  m  *  g_{m}

=>       F__{D}} =  2270   *  3.689  

=>       F__{D}} =  8374 \ N  

Gnerally this drag force is mathematically represented as

         F__{D}} =   C__{D}} *  A *  \frac{\rho_m * v^2 }{2}

Here A is the frontal area

So  

         A   =  \frac{2 *  F__D }{ C__D}  *  \rho_m  * v^2   }

=>       A   =  \frac{2 * 8374 }{ 0.5 *  0.852    *  1 ^2   }

=>       A   = 39315 \  m^2

8 0
3 years ago
A piece of styrofoam has a charge of 0.002 mC and is placed 0.5 m from a grain of salt with a charge of 0.03 nC. How much electr
aleksklad [387]

Answer:

2.16×10⁻⁶ N

Explanation:

Applying,

F = kqq'/r² (coulomb's Law)....................... Equation 1

Where F = electrostatic force, k = coulomb's constant, q = charge on the styrofoam, q' = charge on the grain of salt, r = distance between the charges.

From the question,

Given: q = 0.002 mC = 2.0×10⁻⁶ C, q' = 0.03 nC = 3.0×10⁻¹¹ C, r = 0.5 m

Constant: k = 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

F = (2.0×10⁻⁶)(3.0×10⁻¹¹)(8.99×10⁹)/0.5²

F = 2.16×10⁻⁶ N

5 0
3 years ago
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece
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Answer:

10.01 cm

Explanation:

Given that,

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.

The average propagation speed for sound in body tissue is 1540 m/s.

We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,

v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m

or

d = 10.01 cm

So, the reflection will occur at 10.01 cm.

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3 years ago
6. One minute after takeoff, a rocket carrying the space shuttle into outer space reaches a speed of 447 m/s.
Sidana [21]

Answer:

acceleration of the rocket is given as

a = 7.45 m/s^2

Explanation:

As we know that rocket starts from rest and then reach to final speed of 447 m/s after t = 1 min

so we have

v_i = 0

v_f = 447 m/s

t = 1 min = 60 s

so we have

a = \frac{v_f - v_i}{t}

a = \frac{447 - 0}{60}

a = 7.45 m/s^2

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3 years ago
When are you most aware of your motion in a moving vehicle: when it is moving steadily in a straight line or when it is accelera
jenyasd209 [6]

Answer:

A) Accelerating. You would not be aware of the motion if you did not look outside the car

Explanation:

Since human cannot sense the motion (unless visually), but the inertia force caused due to the acceleration of the motion. So if you are in a car with constant velocity, there's no acceleration and no inertia force, you would not be able to sense the motion at all unless you look outside.

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3 years ago
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