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Lynna [10]
3 years ago
14

The submarine emits a pulse of sound to detect other objects in the sea. The sped of sound in sea water is 1500m/s. An echo is r

eceived with a time delay of 0.50s after the original sound is emitted.
Calculate the distance between submarine and the other object
Physics
1 answer:
r-ruslan [8.4K]3 years ago
3 0

Answer:

d = 375 m

Explanation:

The speed of sound is constant in any medium, therefore we can use the uniform motion relationships

          v = x / t

          x = v t

In this case it indicates that the time since the sound is emitted and received is t = 0.50 s, in this time the sound traveled a round trip distance

           x = 2d

          2d = v t

          d = v t/2

     

let's calculate

          d = 1500 0.5 / 2

          d = 375 m

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A car of mass m goes around a banked curve of radius r with speed v. If the road is frictionless due to ice, the car can still n
Sholpan [36]

Answer:

horizontal component of normal force is equal to the centripetal force on the car

Explanation:

As the car is moving with uniform speed in circle then the force required to move in the circle is towards the center of the circle

This force is due to friction force when car is moving in circle with uniform speed

Now it is given that car is moving on the ice surface such that the friction force is zero now

so here we can say that centripetal force is due to component of the normal force which is due to banked road

Now we have

N sin\theta = \frac{mv^2}{R}

N cos\theta = mg

so we have

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3 years ago
What are the strengths and limitations of the doppler and transit methods? What kind of planets are easiest to detect with each
Arturiano [62]

\huge\mathfrak\red{✔Answer:-}

Strength: able to detect planets in a wide range of orbits, as long as orbits aren't face on

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3 years ago
What is the strength of the electric field 0.1 mmmm below the center of the bottom surface of the plate
Aleksandr [31]

Complete Question

A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Answer:

The  values is  E =248.2 \  N/C

Explanation:

From the question we are told that

   The  diameter is  d =  12 \  cm  =  0.12 \ m

    The charge  is  Q =  4.4 nC  =  4.4 *10^{-9} \  C

    The  distance from the center  is  k =  0.1 \ mm   =  1*10^{-4} \  m

Generally the radius is mathematically represented as

        r =  \frac{d}{2}

=>     r =  \frac{0.12}{2}

=>       r =  0.06 \  m

Generally electric field is mathematically represented as  

       E =  \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 +  k^2 } } ]

substituting values  

      E =  \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 +  (1.0*10^{-4})^2 } } ]

     E =248.2 \  N/C

4 0
3 years ago
Iron + Hydrochloric acid --> ?
marusya05 [52]
<h3>Iron - Fe</h3>

<h3>Hydrochloric Acid- HCl</h3>

<h2><u>Solution</u></h2>

\bold{Fe  +2HCl \rightarrow FeCl _{}{ \tiny2}  + H{ \tiny{2}}}

\therefore \bold{\fbox{{Balanced}}}

Iron + Hydrochloric Acid \rightarrow Ferrous Chloride + Hydrogen

<h2>Hope This Helps You ❤️</h2>
8 0
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avanturin [10]

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7 0
3 years ago
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