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4 years ago

How much is the moon in the first quarter actually lit

Physics

2 answers:

grandymaker [24]4 years ago

8 0

Half of the moon is illuminated.

slamgirl [31]4 years ago

7 0

There's always 50% (half) of the moon lit by the sun. The thing that changes is how much of the lit part we can see.

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Which is an example of radiation?

Y_Kistochka [10]

A common example of radiation would have to be sunlight

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3 years ago

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A wildlife researcher is tracking a flock of geese. The geese fly 4.0km due west, then turn toward the north by 40 degrees and f

BaLLatris [955]

Answer:

(a). The distance from the initial position is 6.68 km.

(b). The magnitude of their displacement is 7.05 km.

Explanation:

Given that,

Geese fly 4.0 km due west, then turn to north by 40° and fly another 3.5 km.

(a). We need to calculate the distance from the initial position

Using formula of distance

$D=AB+BC\cos\theta$

Put the value into the formula

$D=4.0+3.5\cos40$

$D=6.68\ km$

(b). We need to calculate the magnitude of their displacement

Using formula of displacement

$AC=\sqrt{CD^2+AD^2}$

$AC=\sqrt{(3.5\sin40)^2+(6.68)^2}$

$AC=7.05\ Km$

Hence, (a). The distance from the initial position is 6.68 km.

(b). The magnitude of their displacement is 7.05 km.

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3 years ago

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Which exoplanet has the most Earth-like orbit?

Digiron [165]

Answer: Kepler-452b

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3 years ago

Calculate the amount of energy produced in joules by 100- watt light bulb lit for 2.5 hours.

9966 [12]

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4 years ago

A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+88 determines the height of the rock above

babymother [125]

1) 88 ft

2) 4.09 s

3) 1.38 s

4) 118.2 m

Explanation:

For an object thrown upward and subjected to free fall, the height of the object at any time t is given by the suvat equation:

$h(t) = h_0 + ut - \frac{1}{2}gt^2$ (1)

where

$h_0$ is the height at time t = 0

$u$ is the initial vertical velocity

$g=32 ft/s^2$ is the acceleration due to gravity

The function that describes the height of the rock above the surface at a time t in this problem is

$f(t)=-16t^2+44t+88$ (2)

By comparing the terms with same degree of eq(1) and eq(2), we observe that

$h_0 = 88 ft$

which means that the rock is at height h = 88 ft when t = 0: therefore, this means that the height of the bridge above the water is 88 feet.

The rock will hit the water when its height becomes zero, so when

$f(t)=0$

which means when

$0=-16t^2+44t+88$

First of all, we can simplify the equation by dividing each term by 4:

$0=-4t^2+11t+22$

This is a second-order equation, so we solve it using the usual formula and we find:

$t_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-11\pm \sqrt{(11)^2-4(-4)(22)}}{2(-4)}=\frac{-11\pm \sqrt{121+352}}{-8}=\frac{-11\pm 21.75}{-8}$

Which gives only one positive solution (we neglect the negative solution since it has no physical meaning):

t = 4.09 s

So, the rock hits the water after 4.09 seconds.

Here we want to find how many seconds after being thrown does the rock reach its maximum height above the water.

For an object in free fall motion, the vertical velocity is given by the expression

$v=u-gt$

where

u is the initial velocity

g is the acceleration due to gravity

t is the time

The object reaches its maximum height when its velocity changes direction, so when the vertical velocity is zero:

$v=0$

which means

$0=u-gt$

Here we have

$u=+44 ft/s$ (initial velocity)

$g=32 ft/s^2$ (acceleration due to gravity)

Solving for t, we find the time at which this occurs:

$t=\frac{u}{g}=\frac{44}{32}=1.38 s$

The maximum height of the rock can be calculated by evaluating f(t) at the time the rock reaches the maximum height, so when

t = 1.38 s

The expression that gives the height of the rock at time t is

$f(t)=-16t^2+44t+88$

Substituting t = 1.38 s, we find:

$f(1.38)=-16(1.38)^2 + 44(1.38)+88=118.2 m$

So, the maximum height reached by the rock during its motion is

$h_{max}=118.2 m$

Which means 118.2 m above the water.

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4 years ago