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Zolol [24]
2 years ago
8

Give at least two reasons today’s astronomers are so interested in the discovery of additional Earth-approaching asteroids.

Physics
1 answer:
blondinia [14]2 years ago
4 0

Answer:

1. Potential hazard

2. Mining opportunity

Explanation:

The two reason, why astronomers are so interested in the discovery of additional Earth-approaching asteroids:

1. Potential hazard: We have proof that the dinosaurs got extinct because of an asteroid/comet strike on Earth. Also we have seen the effects of the Tunguska event and Chelyabinsk tragedy. These are enough to show us that asteroids can be very dangerous and wipe out the life from Earth.

2. Mining Opportunity: We have discovered a lot of asteroids which contains a lot of metal and precious elements. There can be a possibility of mining such asteroids in the future and reducing the burden on Earth.

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Time=speed/acceleration
Gravitaional Acceleration=9.8 m/s^2
Speed=24.5 m/s
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Stochastic effects are the effects that are caused by chance. Cancer is one of the main stochastic effects.

So, the correct option is (b) "the severity of stochastic effects, such as cancer".

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In a certain region of space, a uniform electric field has a magnitude of 4.30 x 104 n/c and points in the positive x direction.
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Two wires are stretched between two fixed supports and have the same length. One wire A there is a second-harmonic standing wave
lina2011 [118]

(a) Greater

The frequency of the nth-harmonic on a string is an integer multiple of the fundamental frequency, f_1:

f_n = n f_1

So we have:

- On wire A, the second-harmonic has frequency of f_2 = 660 Hz, so the fundamental frequency is:

f_1 = \frac{f_2}{2}=\frac{660 Hz}{2}=330 Hz

- On wire B, the third-harmonic has frequency of f_3 = 660 Hz, so the fundamental frequency is

f_1 = \frac{f_3}{3}=\frac{660 Hz}{3}=220 Hz

So, the fundamental frequency of wire A is greater than the fundamental frequency of wire B.

(b) f_1 = \frac{v}{2L}

For standing waves on a string, the fundamental frequency is given by the formula:

f_1 = \frac{v}{2L}

where

v is the speed at which the waves travel back and forth on the wire

L is the length of the string

(c) Greater speed on wire A

We can solve the formula of the fundamental frequency for v, the speed of the wave:

v=2Lf_1

We know that the two wires have same length L. For wire A, f_1 = 330 Hz, while for wave B, f_B = 220 Hz, so we can write the ratio between the speeds of the waves in the two wires:

\frac{v_A}{v_B}=\frac{2L(330 Hz)}{2L(220 Hz)}=\frac{3}{2}

So, the waves travel faster on wire A.

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2 years ago
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Answer:

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Explanation:

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