Is a vector necessarily changed if it is rotated through an angle ?Explain

Ayuda por favor cual es el objetivo del ultimate?

postnew [5]

Answer: i d k

Explanation:

3 0

3 years ago

regrine falcons frequently grab prey birds from the air. Sometimes they strike at high enough speeds that the force of the impac

solmaris [256]

Answers:

a) 30 m/s

b) 480 N

Explanation:

The rest of the question is written below:

a. What is the final speed of the falcon and pigeon?

b. What is the average force on the pigeon during the impact?

<h3>a) Final speed</h3>

This part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum $p_{i}$ before the collision must be equal to the final momentum $p_{f}$ after the collision:

$p_{i}=p_{f}$ (1)

Being:

$p_{i}=MV_{i}+mU_{i}$

$p_{f}=(M+m) V$

Where:

$M=480 g \frac{1 kg}{1000 g}=0.48 kg$ the mas of the peregrine falcon

$V_{i}=45 m/s$ the initial speed of the falcon

$m=240 g \frac{1 kg}{1000 g}=0.24 kg$ is the mass of the pigeon

$U_{i}=0 m/s$ the initial speed of the pigeon (at rest)

$V$ the final speed of the system falcon-pigeon

Then:

$MV_{i}+mU_{i}=(M+m) V$ (2)

Finding $V$ :

$V=\frac{MV_{i}}{M+m}$ (3)

$V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg}$ (4)

$V=30 m/s$ (5) This is the final speed

<h3>b) Force on the pigeon</h3>

In this part we will use the following equation:

$F=\frac{\Delta p}{\Delta t}$ (6)

Where:

$F$ is the force exerted on the pigeon

$\Delta t=0.015 s$ is the time

$\Delta p$ is the pigeon's change in momentum

Then:

$\Delta p=p_{f}-p_{i}=mV-mU_{i}$ (7)

$\Delta p=mV$ (8) Since $U_{i}=0$

Substituting (8) in (6):

$F=\frac{mV}{\Delta t}$ (9)

$F=\frac{(0.24 kg)(30 m/s)}{0.015 s}$ (10)

Finally:

$F=480 N$

7 0

3 years ago

Stop lamps must be visible within____ to the rear

Charra [1.4K]

The answer is 300 feet. The stop lamp or lamps on the rear of a vehicle must show a red light that is set in motion upon application of the service or foot brake and, in a vehicle manufactured or assembled on or after January 1, 1964, must be visible from a distance of not less than 300 feet to the rear in normal sunlight. Take note, if the vehicle is manufactured or assembled January 1, 1964, the stop lamp or lamps must be visible from a distance of not less than 100 feet. Also, the stop lamp may be combined with one or more other rear lamps.

3 0

3 years ago

Read 2 more answers

A glass rod and a steel rod are of equal length at 0C. At 100C they differ in length by

NeX [460]

The given lengths at 0 °C are 2.5 m

Let l₀ be the given lengths of the glass and steel rods at 0 °C. Let l and l' be the lengths of the glass and steel rods at 100 °C respectively.

From our expression for linear expansivity,

l = l₀ + l₀αΔθ where α = linear expansivity of glass = 0.000008/°C and Δθ = temperature change = θ - θ' where θ = 100 °C and θ' = 0 °C. So, Δθ = 100 °C - 0 °C = 100 °C.

Also,

l' = l₀ + l₀α'Δθ where α' = linear expansivity of steel = 0.000012/°C and Δθ = temperature change = θ - θ' where θ = 100 °C and θ' = 0 °C. So, Δθ = 100 °C - 0 °C = 100 °C.

Since the difference in their lengths at 100 °C = 0.001 m, we have that

l - l' = l₀ + l₀αΔθ - (l₀ + l₀α'Δθ)

l - l' = l₀ + l₀αΔθ - l₀ - l₀α'Δθ)

l - l' = l₀αΔθ - l₀α'Δθ

l - l' = l₀(α- α')Δθ

Making l₀ subject of the formula, we have

l₀ = (l - l')/[(α- α')Δθ]

Substituting the values of the variables into the equation, we have

l₀ = (l - l')/[(α- α')Δθ]

l₀ = 0.001 m/[(0.000008/°C - 0.000012/°C)100 °C.]

l₀ = 0.001 m/[(-0.000004/°C)100 °C.]

l₀ = 0.001 m/-0.0004

l₀ = -2.5 m

Neglecting the negative sign,

l₀ = 2.5 m

So, the given lengths at 0 °C are 2.5 m

Learn more about linear expansion here:

brainly.com/question/14089545

6 0

2 years ago

In the model of the hydrogen atom due to Niels Bohr, the electron moves around the proton at a speed of 3.3 × 106 m/s in a circl

irga5000 [103]

Answer:

$1.5048\times 10^{-23}\ Am^2$

Explanation:

q = Charge of proton = $1.6\times 10^{-19}\ C$

r = Radius of circle = $5.7\times 10^{-11}\ m$

v = Velocity of proton = $3.3\times 10^6\ m/s$

Magnetic moment is given by

$M=\frac{1}{2}qrv\\\Rightarrow M=\frac{1}{2}1.6\times 10^{-19}\times 5.7\times 10^{-11}\times 3.3\times 10^6\\\Rightarrow M=1.5048\times 10^{-23}\ Am^2$

The magnetic moment associated with this motion is $1.5048\times 10^{-23}\ Am^2$

5 0

3 years ago

Is a vector necessarily changed if it is rotated through an angle ?​Explain

Is a vector necessarily changed if it is rotated through an angle ?Explain