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m_a_m_a [10]
3 years ago
11

One of the ways that scientists determine the distance between Earth and other objects in space is to send light waves toward an

object and measure how long it takes for the waves to bounce back. If a scientist sends light waves to four different objects in space, which of the following will take the longest to bounce the wave back to Earth?
a star in the Milky Way Galaxy


a star in the Andromeda Galaxy


a planet in Earth's Solar System


a planet in another star system within the Milky Way Galaxy
Physics
1 answer:
Ahat [919]3 years ago
7 0

Answer:

a star in andromeda

Explanation:

all of the other objects are in the milkyway (where we are) and the andromeda galaxy is 2 million light years away from us

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A 50.0-kg bungee jumper steps off a bridge with a light bungee cord tied to her and to the bridge. The unstretched length of the
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The the analysis for the free fall part should be done under the constant acceleration.

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20. A semiconductor is a
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D. crystalline solid that conducts current under certain conditions

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5 0
2 years ago
In a game of angry birds you launch a bird with an angle of 53 degrees to horizontal. Unfortunatly, its not a good shot and the
Alisiya [41]

Answer:

The maximum height covered is 3.25 m.

The horizontal distance covered is 9.81 m.

The total time in the air is 1.63 seconds.

Explanation:

The launch speed, u_0= 10 m/s.

Angle of launch with the horizontal, \theta = 53 ^{\circ}

So, the vertical component of the initial velocity,

u_0\sin\theta=10 \sin 53 ^{\circ}\cdots(i).

The horizontal component of the initial velocity,

u_0\cos\theta=10 \cos 53 ^{\circ}

Let, t be the time of flight, to the horizontal distance covered

D=10 \cos (53 ^{\circ})t\cdots(ii).

Not, applying the equation of motion in the vertical direction.

s= ut +\frac 1 2 at^2

Where s is the displacement in time t, u is the initial velocity and a is the acceleration.

In this case, u =10 \sin 53 ^{\circ} (from equation (i), s=0 (as the final height is same as the launch height) and a = -9.81 m/s^2 (negative sign is due to the downward direction).

\Rightarrow 0 = 10 (\sin 53 ^{\circ})t-\frac 1 2 (9.81)t^2

\Rightarrow t= \frac {2\times 10 (\sin 53 ^{\circ})}{9.81}=1.63 seconds.

So, the total time in the air is 1.63 seconds.

From equation (i),

Total horizontal distance covered is

D=10 \cos (53 ^{\circ})\times 1.63 = 9.81 m.

Now, for the maximum height, H, applying the equation of motion as

v^2=u^2+2as

Here, v is the final velocity and v=0 (at the maximum height), and h=H.

So, 0^2=(10 \sin 53 ^{\circ})^2-2(9.81)H

\Rightarrow H = \frac {(10 \sin 53 ^{\circ})^2}{2\times 9.81}

\Rightarrow H = 3.25 m.

Hence, the maximum height covered is 3.25 m.

8 0
3 years ago
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