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miss Akunina [59]
3 years ago
15

A submarine sends a signal and receives an echo after 5 seconds. Calculate the speed of sound if distance of object from submari

ne
in 3625m.
Physics
2 answers:
quester [9]3 years ago
8 0
Distance of d object frm submarine = 3625m
time taken = 5/2

then speed = 3625/5/2. (speed = distance /time)
=> 1450m/s


seldon.
Reil [10]3 years ago
3 0

speed = distance \div time
Since the sound travels from the submarine to the object AND back, it actually travelled 3625x2=7250m.

7250 \div 5 = 1450
Speed of sound: 1450m/s
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When the net force of opposite forces is zero , the forces are
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The answer is balanced
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3 years ago
An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
lys-0071 [83]

Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

3 0
3 years ago
The image of an object formed by plane mirror is​
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Answer of your question is in this photo

8 0
2 years ago
In a closed system, a cart with a mass of 1.5 kg is rolling to the right at 1.4 m/s, while another cart of mass 1.0 kg is rollin
bija089 [108]

Answer: The correct option is (c.).

Explanation:

Mass of the cart A= 1.5 kg

Velocity of Cart A = 1.4 m/s towards right

Mass of the cart B = 1.0 kg

Velocity of Cart B = 1.4 m/s towards left

Momentum (P)= Mass × Velocity

P_A=1.5 kg\times 1.4 m/s=2.1 kg m/s

P_B=1.0 kg\times (-1.4m/s)=-1.4 kg m/s

(Negative sign means velocity of the cart is in opposite direction of that of the cart A)

Total Momentum =P_A+P_B=2.10 kg m/s-1.40 kg m/s=0.70 kg m/s

Hence, the correct option is (c.).

7 0
3 years ago
Read 2 more answers
Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the
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Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

q =75*10^-6 C

A = 0.25

eo = 8.85*10^-12

Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

8 0
3 years ago
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