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3 years ago

2. An airplane traveling north at 220. meters per second encounters a 50.0-meters-per-second crosswind

Physics

1 answer:

Alex777 [14]3 years ago

5 0

The resultant speed of the plane is (3) 226 m/s

Why?

We can calculate the resultant speed of the plane by using the Pythagorean Theorem since both speeds are perpendicular (forming a right triangle).

So, calculating we have:

$ResultantSpeed=\sqrt{VerticalSpeed^{2}+HorizontalSpeed^{2}}\\\\ResultantSpeed=\sqrt{(220\frac{m}{s})^{2}+50\frac{m}{s})^{2}$

$ResulntantSpeed=\sqrt{48400\frac{m^{2} }{s^{2} }+2500\frac{m^{2} }{s^{2} } } \\\\ResultantSpeed=\sqrt{50900\frac{m^{2} }{s^{2} }}=226\frac{m}{s}$

Hence, we have that the resultant speed of the plane is (3) 226 m/s

Have a nice day!

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A mass is attached to a vertical spring, which then goes into oscillation. At the high point of the oscillation, the spring is i

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Answer:

0.34 sec

Explanation:

Low point of spring ( length of stretched spring ) = 5.8 cm

midpoint of spring = 5.8 / 2 = 2.9 cm

Determine the oscillation period

at equilibrum condition

Kx = Mg

g= 9.8 m/s^2

x = 2.9 * 10^-2 m

k / m = 9.8 / ( 2.9 * 10^-2 ) = 337.93

note : w = $\sqrt{k/m}$ = $\sqrt{337.93}$ = 18.38 rad/sec

Period of oscillation = $2\pi / w$

= 0.34 sec

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3 years ago

A 1000 kg car moving a 10 m/s collides with a stationary 2000 kg truck. The two vehicles interlock as a result of the collision.

IgorLugansk [536]

Answer:

v₃ = 3.33 [m/s]

Explanation:

This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.

In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.

$(m_{1}*v_{1})+(m_{2}*v_{2})=(m_{1}+m_{2})*v_{3}$

where:

m₁ = mass of the car = 1000 [kg]

v₁ = velocity of the car = 10 [m/s]

m₂ = mass of the truck = 2000 [kg]

v₂ = velocity of the truck = 0 (stationary)

v₃ = velocity of the two vehicles after the collision [m/s].

Now replacing:

$(1000*10)+(2000*0)=(1000+2000)*v_{3}\\v_{3}=3.33[m/s]$

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3 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

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4 years ago

A bullet of mass m=26g is fired into a wooden block of mass M=4.7kg. The block is attached to a string of length 1.5m. The bulle

vodka [1.7K]

Answer:

u = 449 m/s

Explanation:

Given,

Mass of the bullet, m = 26 g

Mass of the wooden block,M = 4.7 Kg

height of the block,h = 0.31 m

initial speed of the block, u = ?

Using conservation of energy

$(M+ m)gh = \dfrac{1}{2}(M+m)v^2$

$gh = \dfrac{1}{2}v^2$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.81\times 0.31}$

v = 2.47 m/s

Now, using conservation of momentum to calculate the speed of the bullet.

m u + M u' = (M+m)v

m u = (M+m)v

0.026 x u = (4.7+0.026) x 2.47

u = 449 m/s

Hence, the speed of the bullet is equal to 449 m/s.

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3 years ago

What is the difference between transverse and longitudinal waves? Transverse waves always have greater frequencies than do longi

Anastasy [175]

Answer:

Explanation:

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3 years ago