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2 years ago

Would you expect to observe every supernova in our own Galaxy? Why or why not?

Physics

1 answer:

Taya2010 [7]2 years ago

3 0

Explanation:

Only few supernova are observed in our galaxy -

Type II supernovae ( i.e. the explosions of the massive stars ) occurred in the Milky Way, and they might be hidden by the intervening dust if they are located in the more distant parts of our Galaxy .

Type Ia supernovae , which need a white dwarf star in the binary star system , are brighter than the type II supernovae , but some of them could also happen in the older parts of Galaxy which are hidden due to the buildup of the dust and gas .

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At which point or points does the pendulum have the greatest kinetic energy?

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At the lowest point of its motion, kinetic energy is maximum and potential energy is minimum. This is where the velocity is a maximum. At the highest point of its motion, kinetic energy is minimum (i.e. zero) and potential energy is maximum.

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If HST has a tangential speed of 7,750 m/s, how long is HST’s orbital period? The radius of Earth is 6.38 × 106 m. s

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the orbital period is 5170 s

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3 years ago

An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of th

jeyben [28]

Answer:

the shooting angle ia 18.4º

Explanation:

For resolution of this exercise we use projectile launch expressions, let's see the scope

R = Vo² sin (2θ) / g

sin 2θ = g R / Vo²

sin 2θ = 9.8 75/35²

2θ = sin⁻¹ (0.6)

θ = 18.4º

To know how for the arrow the tree branch we calculate the height of the arrow at this point

X2 = 75/2 = 37.5 m

We calculate the time to reach this point since the speed is constant on the X axis

X = Vox t

t2 = X2 / Vox = X2 / (Vo cosθ)

t2 = 37.5 / (35 cos 18.4)

t2 = 1.13 s

With this time we calculate the height at this point

Y = Voy t - ½ g t²

Y = 35 sin 18.4 1.13 - ½ 9.8 1,13²

Y = 6.23 m

With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch

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3 years ago

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A scientist bred fruit flies in two separate containers with different food sources for many generations. When she put the fruit

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Fruit flies prefer mates adapted to the same food source.

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A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest fro

svetoff [14.1K]

Answer:

The equation of motion is $x(t)=-$ $\frac{1}{3} cos4\sqrt{6t}$

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

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2 years ago