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IRINA_888 [86]
3 years ago
9

Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time–when they\'re not sleep

ing or eating–joyfully scampering about on the cage\'s floor. Bryce tracks his mice\'s health diligently and just now recorded their masses as 0.0225 kg, 0.0223 kg, 0.0197 kg, and 0.0127 kg. At this very instant the x and y components of the mice\'s velocities are, respectively, (0.869 m/s, -0.283 m/s), (-0.883 m/s, -0.253 m/s), (0.345 m/s, 0.803 m/s), and (-0.555 m/s, 0.205 m/s). Calculate the x and y components of Bryce\'s mice\'s total momentum, px and py.
Physics
1 answer:
Anit [1.1K]3 years ago
7 0

Answer:

mice total momentum (-0.000250, 0.00639) Kg m

Explanation:

To calculate the moment of the mice we must multiply their mass by their velocities, remember that the moment is a vector quantity, so we use the components of velocity

mouse 1

  m1 = 0.0225 Kg

  V1 = (0.869, -0.283) m / s

 

  Px = m Vx

  Px1 = 0.0225 0.869

  Px1 = 0.01955 Kg m

  Py = m Vy

  Py1 = 0.0225 (-0.283)

  Py1 = -0.006368 Kg m

  P1 = (0.0196, -0.00637) Kg m

Mouse 2

 m2 = 0.0223 Kg

 Px2 = 0.0223 (-0.883) = -0.0196 Kg m

 Py2 = 0.0223 (-0.253) = -0.00564 Kg m

 P2 = (-0.0196, -0.00564) Kg m

Mouse 3

 m3 = 0.0197

 Px3 = 0.0197 0.345 = 0.00680 Kg m

 Py3 = 0.0197 0.803 = 0.0158 Kg m

 P3 = (0.00680, 0.0158) Kg m

Mouse4

  m4 = 0.0127 Kg

  Px4 = 0.0127 (-0.555) = -0.00705 Kg m

  Py4 = 0.0127 0.205 = 0.00260 Kg m

  P4 = (-0.00705, 0.00260) Kg m

To find the total momentum we must add each component of the individual moments

   Px = Px1 + Px2 + Px3 + Px4  

   Py = py1 + Py2 + Py3 + Py4

   Px = 0.0196 -0.0196 +0.00680 -0.00705

   Px = -0,000250 Kg m

   Py = -0.00637 -0.00564 +0.0158 +0.00260

   Py = 0.00639 Kg m

   P = (-0.000250, 0.00639) Kg m

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A horizontal disk with a radius of 10 cm rotates about a vertical axis through its center. The disk starts from rest at t = 0 an
Tom [10]

Answer:

0.69s

Explanation:

10 cm = 0.1 m

Let t be the time that radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude. At that time we have the angular velocity would be

\omega = \alpha t = 2.1 t

And so the radial acceleration is

a_r = \omega^2 r = (2.1t)^2 r = 2.1^2 t^2 * 0.1= 0.441 t^2 m/s^2

The tangential acceleration is always the same since angular acceleration is constant:

a_t = \alpha * r = 2.1 * 0.1 = 0.21 m/s^2

For these 2 quantities to be the same

a_r = a_t

0.441 t^2 = 0.21

t^2 = 0.21/0.441 = 0.4762

t = \sqrt{0.4762} = 0.69 s

6 0
3 years ago
What is the wavenumber of the stretching vibrational mode for the CO molecule, given that the force constant of the bond is 680
Gnesinka [82]

Answer:

1.10134 * 10⁻⁹m⁻¹

Explanation:

K = 680Nm⁻¹

μ = ?

μ = (m₁ + m₂) / m₁m₂

compound = CO

C = 12.0 g/mol = 0.012kg/mol

O = 16.0g/mol = 0.016kg/mol

μ = (m₁ + m₂) / m₁m₂

μ = (0.012 + 0.016) / (0.012*0.016) = 145.83

v = 1/2πc * √(k/μ)

ν = 1/ 2*3.142* 3.0*10⁸ * √(630/145.83)

v = 5.30*10⁻¹⁰ * 2.078

v = 1.10134*10⁻⁹m⁻¹

8 0
3 years ago
A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force
Tamiku [17]

Answer:

(a) 91 kg (2 s.f.)    (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

                                                   s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}

     Subsequently,

                                                  F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

                                                    v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}

     According to Newton's First Law which states objects will remain at rest

     or in uniform motion (moving at constant velocity) unless acted upon by

     an external force. Hence, the block of ice by the end of the first 5

     seconds, experiences no acceleration (a = 0) but travels with a constant

     velocity of 4.4 m \ s^{-1}.

                                                    s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}

      Therefore, the ice block traveled 22 m in the next 5 seconds after the

      worker stops pushing it.

4 0
2 years ago
Just need help with 1 and 2 please :D i’m having a bit of trouble :/
dexar [7]
1. Traveling by car means you have specific roads to follow. You won’t be able to go straight to Banning high from POLAHS. The 8.4km will be defined as distance. Traveling by helicopter you don’t have roads to follow that means you can fly directly to banning high. 6.8km will be defined as displacement.

2. A) 400m
B)0m
C)d=1/2(vi+vf)t
400=1/2(0+vf)92
8.7m/s
D) 0m/s
E) Not sure but instantaneous velocity refer to velocity at a given point. Average velocity is just the average. Usually instantaneous velocity won’t be same as the average velocity.
Plz like if it helped.
7 0
3 years ago
The owners want the speed at the bottom of the first hill doubled. How much higher must the first hill be built?
Sonja [21]
Neglecting friction and air resistance, the first hill must be built 4 times higher than it is now.
3 0
3 years ago
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