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Korvikt [17]
2 years ago
12

Kon'nichiwa~please help me with this question!!​

Physics
1 answer:
andrezito [222]2 years ago
4 0

Let's check the relationship

\\ \rm\hookrightarrow g=\dfrac{GM}{r^2}

\\ \rm\hookrightarrow g\propto G

So

  • Raindrops will fall faster . .
  • Also walking on ground would become more difficult as g increases.

Option C is wrong by now .Let's check D once

\\ \rm\hookrightarrow T\propto \dfrac{1}{\sqrt{g}}

  • So time period of simple pendulum would decrease.

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To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d
Nimfa-mama [501]

Complete Question

The compete question is shown on the first uploaded question

Answer:

The speed is  v  =  350 \  m/s  

Explanation:

From the question we are told that

   The  distance of separation is  d =  4.00 m  

  The distance of the listener to the center between the speakers is  I =  5.00 m

  The change in the distance of the speaker is by k  =  60 cm  =  0.6 \  m

    The frequency of both speakers is f =  700 \  Hz

Generally the distance of the listener to the first speaker is mathematically represented as

       L_1  =  \sqrt{l^2 + [\frac{d}{2} ]^2}

       L_1  =  \sqrt{5^2 + [\frac{4}{2} ]^2}

        L_1  =   5.39 \  m

Generally the distance of the listener to second speaker at its new position is  

          L_2  =  \sqrt{l^2 + [\frac{d}{2} ]^2 + k}

       L_2  =  \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}

        L_2  =   5.64  \  m  

Generally the path difference between the speakers is mathematically represented as

        pD  = L_2 - L_1  =  \frac{n  *  \lambda}{2}

Here \lambda is the wavelength which is mathematically represented as

         \lambda =  \frac{v}{f}

=>    L_2 - L_1  =  \frac{n  *  \frac{v}{f}}{2}

=>    L_2 - L_1  =  \frac{n  *  v}{2f}  

=>    L_2 - L_1  =  \frac{n  *  v}{2f}  

Here n is the order of the maxima with  value of  n =  1  this because we are considering two adjacent waves

=>    5.64 - 5.39   =  \frac{1  *  v}{2*700}      

=>    v  =  350 \  m/s  

7 0
3 years ago
Poor measurement practices are likely to lead to data that are.....
zhannawk [14.2K]
Inconsistent. You should take three readings at least.
8 0
3 years ago
Read 2 more answers
A newly discovered particle, the SPARTYON, has a mass 945 times that of an electron. If a SPARTYON at rest absorbs an anti-SPART
Gennadij [26K]

Answer:

\nu =1166\times 10^{20}Hz  

Explanation:

We have given the rest mass of SPARTYON = 945 times of mass of electron

We know that mass of electron =9.11\times 10^{-31}kg

So mass of SPARTYON =945\times 9.11\times 10^{-31}=8608.95\times 10^{-31}kg

Speed of light c=3\times 10^8m/sec

According to Einstein equation energy is given by

E=mc^2=8608.95\times 10^{-31}\times (3\times 10^{8})^2=77480.55\times 10^{-15}j

Now according to planks's rule

Energy is given by

E=h\nu , here h is plank's constant h=6.6\times 10^{-34}

So 77480.55\times 10^{-15}=6.6\times 10^{-34}\nu

\nu =1166\times 10^{20}Hz  

3 0
3 years ago
the brightness of stars as they appear from earth is measured by _______ magnitude. a. light b. apparent c. relative d. absolute
xxTIMURxx [149]
How bright a star appears from Earth is the star's apparent magnitude.
8 0
3 years ago
What are one of the main uses of this device ?<br><br> anwser A
Inessa [10]

One of the main uses of this device is to test a beam of charged particles

What is cathode ray tube?

The cathode ray tube determines the charge flowing in a gas. When an electric field is set up using metal electrodes, the cathode ray tends to bend towards the positive electrode.

This concludes that a charge and the electrodes present helps determine the charge of the beam of charged particle.

Thus, one of the main uses of this device is to test a beam of charged particles.

Learn more about cathode ray tube.

brainly.com/question/1581574

#SPJ1

7 0
1 year ago
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