The law of conservation of mass states that ______________________________

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

user100 [1]

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

8 0

3 years ago

You have a battery marked " 6.00 V 6.00 V ." When you draw a current of 0.383 A 0.383 A from it, the potential difference betwee

Archy [21]

Answer:

V = 4.81 V

Explanation:

As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.
We can calculate this loss, applying Ohm's law to the internal resistance, as follows:

$V_{rint} = I* r_{int}$

The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

$V = V_{b} - V_{rint} = 5.03 V = 6.0 V - 0.383 A* r_{int}$

We can solve for rint, as follows:

$r_{int} = \frac{V_{b}-V}{I} =\frac{6.0V-5.03V}{0.383A} = 2.53 \Omega$

When the circuit draws from battery a current I of 0.469A, we can find the potential difference between the terminals of the battery, as follows:

$V = V_{b} - V_{rint} = 6.0 V - 0.469 A* 2.53 \Omega= 6.0 V - 1.19 V = 4.81 V$

As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.

5 0

2 years ago

Brainliest if correct

Bad White [126]

Answer:

D: Increase the distance between the objects.

E: Decrease the mass of one of the objects.

6 0

1 year ago

Pls tell the answer asap

Rzqust [24]

The maximum force acts between B and C as the graph is steepest showing maximum deceleration

5 0

2 years ago

A particular heat engine has a mechanical power output of 4.00 kW and an efficiency of 26.0%. The engine expels 8.55 103 J of ex

Ivahew [28]

To develop the problem we will start by finding the energy taken by each cycle through the efficiency of the motor and the exhausted energy. Later the work will be found for the conservation of energy in which this is equivalent to the difference between the two calculated energy values. Finally the estimated time will be calculated with the work and the power given,

$\text{Efficiency of the heat engine} = \eta = 26\% = 0.26$