My <br> stoked.elle <br> its a new acc

Canadian geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at

Wewaii [24]

Answer:

θ=19.877⁰

Explanation:

Given data

Velocity Va=34.0 km/h

Velocity Va=100 km/h

To find

Angle θ

Solution

We want the bird to fly with velocity Vb=100 km/h with an angle θ relative to the ground so that the bird fly due south relative to the ground.From figure which is attached we got

Sinθ=(Va/Vb)

Sinθ=(34.0/100)

θ=Sin⁻¹(34.0/100)

θ=19.877⁰

5 0

3 years ago

I need help on all of this

wel

dude is -2m/s, ...ettte is +2m/s

pos vel is when dudette, eg, is going in increasing x, in this case.

neg vel is when dudette, eg, is going in decreasing x, in this case. ie she turns round and runs

zero vel is zero speed. dudette standing still

positive vel neg pos top dia

no

yes

all yes looks like ...

4 0

2 years ago

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

Each of the following would cause an increase in blood pressure except __________.

kicyunya [14]

Answer:

an inhibitor of angiotensin II

Explanation:

Angiotensin, specifically angiotensin II binds to many receptors in the body to affect several systems. It can normally increase blood pressure by constricting the blood vessels but with the introduction of an inhibitor, it wouldn't bring about an increase in blood pressure.

8 0

3 years ago

Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more t

olga2289 [7]

The force result in stretching the spring 10.0 centimeters is 2.5N.

<h3>What is Hooke's law?</h3>

If a spring is stretched from its equilibrium position, then a force with magnitude proportional to the increase in length from the equilibrium length is pulling each end.

F = kx

where k is the proportionality constant called the spring constant or force constant.

Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more than 10.0 centimeters, however, it no longer follows that simple linear rule.

Let the spring constant be very low 0.04N/m

The force applied is

F = 10 cm / 0.04

F = 0.1 m / 0.04

F = 2.5 N

Thus, the force result in stretching the spring 10cm is 2.5 N.

Learn more about hooke's law.

brainly.com/question/13348278

#SPJ1

5 0

1 year ago

My stoked.elle its a new acc