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ivanzaharov [21]
3 years ago
9

Give some reasons for our knowledge of the solar system has increased considerably in the past few years. Support your response

with at least 3 reasons with details regarding concepts from the units learned in this course.
Physics
1 answer:
djverab [1.8K]3 years ago
3 0

Answer:

Improvement in observational, and exploratory technology

Rapid increase in knowledge

International collaboration

Explanation:

Our knowledge of the solar system has increased greatly in the past few years due to to some factors which are listed below.

<u>Improvement in observational, and exploratory technology</u>: In recent years, developments in technology has led to the invention of advanced observational instruments and probes, that are used to study the solar system. Also more exploratory units are now developed to go out into the solar system and gather useful data which is then further processed to yield more results about our solar system.

<u>Rapid increase in knowledge:</u> The past few years has seen an increased number of theories proposed to explain phenomena in the solar system. Some of these theories have been seen to be accurate under experimentation, leading to newer and fresher insights into our solar system. Also, new experiments and research are carried out, all these leading to an exponential growth in our knowledge of the solar system.

<u>International Collaboration</u>: The sharing of knowledge by scientists all over has led to a better, quick understanding of the solar system. Also, scientists from  different countries, working together on different experiment and data sharing regarding our solar system now allows our knowledge of the solar system to deepen faster.

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In a double-slit experiment, if the central diffraction peak contains 13 interference fringes, how many fringes are contained wi
Ronch [10]

Answer:

Explanation:

Width of central diffraction peak is given by the following expression

Width of central diffraction peak= 2 λ D/ d₁

where d₁ is width of slit and D is screen distance and λ is wave length.

Width of other fringes become half , that is each of  secondary diffraction fringe is equal to

λ D/ d₁

Width of central interference  peak is given by the following expression

Width of  each of  bright fringe =  λ D/ d₂

where d₂ is width of slit and D is screen distance and λ is wave length.

Now given that the central diffraction peak contains 13 interference fringes

so ( 2 λ D/ d₁)  /  λ  D/ d₂ = 13

then (  λ D/ d₁)  /  λ  D/ d₂ = 13 / 2

= 6.5

no of fringes  contained within each secondary diffraction peak = 6.5

6 0
4 years ago
Benjamin, a transport technician, was punched by a patient in the
Masja [62]
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6 0
3 years ago
A subway car moves at a constant speed of 10 m/s over a period of 10 s. What is the instantaneous speed halfway through this mot
andre [41]

Answer:  10 m/s

We're told the speed is constant, so it's not changing throughout the time period given to us. So throughout the entire interval, the speed is 10 m/s.

5 0
3 years ago
Help me with the question in the image I provided​
inna [77]

Answer: 10 s, 30 m/s , 150 m

Explanation:

Given

The speed of motorcyclist is  u_1=15\ m/s

The initial speed of a police motorcycle is u_2=0\ m/s

acceleration of police motorcycle is a=3\ m/s^2

Police will catch the motorcyclist when they traveled equal distances

distance traveled by motorcyclist in time t is

\Rightarrow s_1=15\times t

Distance traveled by Police in time t is

\Rightarrow s_2=u_2t+\frac{1}{2}at^2\\\Rightarrow s_2=0+0.5\times 3\times t^2

put s_1=s_2

\Rightarrow 15t=0.5\times 3\times t^2\\\\\Rightarrow t(1.5t-15)=0\\\\\Rightarrow t=\dfrac{15}{1.5}=10\ s

Police officer's speed at that time is

\Rightarrow v_2=u_2+at\\\Rightarrow v_2=0+3\times 10=30\ m/s

Distance traveled by each vehicle is

\Rightarrow s_1=s_2=15\times t=15\times 10=150\ m

8 0
3 years ago
What is the increase in pressure required to decrease volume of mercury by 0.001%
REY [17]

Answer:

Explanation:

Using Boyles law

Boyle's law states that, the volume of a given gas is inversely proportional to it's pressure, provided that temperature is constant

V ∝ 1 / P

V = k / P

VP = k

Then,

V_1 • P_1 = V_2 • P_2

So, if we want an increase in pressure that will decrease volume of mercury by 0.001%

Then, let initial volume be V_1 = V

New volume is V_2 = 0.001% of V

V_2 = 0.00001•V

Let initial pressure be P_1 = P

So,

Using the equation above

V_1•P_1 = V_2•P_2

V × P = 0.00001•V × P_2

Make P_2 subject of formula by dividing be 0.00001•V

P_2 = V × P / 0.00001 × V

Then,

P_2 = 100000 P

So, the new pressure has to be 10^5 times of the old pressure

Now, using bulk modulus

Bulk modulus of mercury=2.8x10¹⁰N/m²

bulk modulus = P/(-∆V/V)

-∆V = 0.001% of V

-∆V = 0.00001•V

-∆V = 10^-5•V

-∆V/V = 10^-5

Them,

Bulk modulus = P / (-∆V/V)

2.8 × 10^10 = P / 10^-5

P = 2.8 × 10^10 × 10^-5

P = 2.8 × 10^5 N/m²

3 0
3 years ago
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