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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 nonconducting sphere has radius R = 2.81 cm and uniformly distributed charge q = +2.35 fC. Take the electric potential at the
Sladkaya [172]

Answer:

(a). The electric potential at 1.650 cm is -1.219\times10^{-4}\ V.

(b). The electric potential at 2.81 cm is -3.759\times10^{-4}\ V.

Explanation:

Given that,

Radius of sphere R=2.81 cm

Charge = +2.35 fC

Potential at center of sphere

V = 0

(a). We need to calculate the potential at a distance r = 1.60 cm

Using formula of potential difference

V_(r)-V_(0)=-\int_{0}^{r}{E(r)}dr

V_{r}-0=-\int_{0}^{r}{\dfrac{qr}{4\pi\epsilon_{0}R^3}}dr

V_{r}=-(\dfrac{qr^2}{8\pi\epsilon_{0}R^3})_{0}^{1.60\times10^{-2}}

V_{r}=-(\dfrac{2.35\times10^{-15}\times(1.60\times10^{-2})^2}{8\times\pi\times8.85\times10^{-12}\times(2.81\times10^{-2})^3})

V_{r}=-0.00012190\ V

V_{r}=-1.219\times10^{-4}\ V

The electric potential at 1.650 cm is -1.219\times10^{-4}\ V.

(b). We need to calculate the potential at a distance r = R

Using formula of  potential difference

V_{R}=-\dfrac{2.35\times10^{-15}}{8\pi\times8.85\times10^{-12}\times2.81\times10^{-2}}

V_{R}=-0.0003759\ V

V_{R}=-3.759\times10^{-4}\ V

The electric potential at 2.81 cm is -3.759\times10^{-4}\ V.

Hence, This is the required solution.

5 0
3 years ago
Aisha is sitting on frictionless ice and holding two heavy ski boots. Aisha weighs 637 N, and each boot has a mass of 4.50 kg. A
Studentka2010 [4]

Answer:

-0.73 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum.

In fact, in absence of external forces (the ice is frictionless, so no friction), the total momentum of Aisha + two boots is conserved.

At the beginning, their total momentum is zero, since they are at rest:

p_i = 0 (1)

After, their total momentum is:

p_f = Mv + 2mv' (2)

where:

M is Aisha's mass

v is Aisha's velocity relative to the ground

m = 4.50 kg is the mass of each boot

v' is the boot's velocity relative to the ground

We can find:

M=\frac{W}{g}=\frac{637 N}{9.8 N/kg}=65 kg is Aisha's mass (where W = 637 N was her weight)

v' can be rewritten as:

v'=v+6

because 6 m/s is the velocity of the boots relative to her, while v' is their velocity relative to the ground.

Substituting and combining (1) and (2) we find:

0=Mv+2m(v+6)\\0=Mv+2mv+12m\\v=\frac{-12m}{M+2m}=\frac{-12(4.50)}{65+2(4.50)}=-0.73 m/s

and the negative sign indicates that the direction is opposite to that of the boots.

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Answer: I think it’s D.

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