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irina [24]
3 years ago
6

An enemy sub is approaching a us submarine at 25.0 km/hr that is waiting in ambush. the us submarine needs to know the exact pos

ition of the enemy sub in order to calibrate their torpedo. calculate the distance the enemy sub is when the time for a reflected sonar signal to return to the us sub is 3.5 seconds
Physics
1 answer:
Illusion [34]3 years ago
3 0
Speed of the Enemy sub .....> <span> 25.0 km/hr which is (25/3.6) m/s = 6.94 m/s

We use the expression Speed = Distance/ time to find the distance

D = Speed*Time = 6.94 m/s * 3.5  = 24.305 m

The enemy sub will be 24.305 m closer 
</span>
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g Which of the following wavefunctions are: a) square-integrable on the interval provided (1 point); b) valid wavefunction satis
Goshia [24]

Answer:

The answer is given in the attachment

Explanation:

5 0
3 years ago
1000 millicoulombs of charge passes through a point. The amount of current passing through the point is
Rzqust [24]

The amount of current passing through the point is 1 A

The amount of current passing through the point can be calculated using the formula below.

⇒ Formula:

  • Q = i/t......................... Equation 1

⇒ Where:

  • Q = Charge
  • i = current
  • t = time.

⇒ Make "i" the subject of the equation.

  • i = Qt....................... Equation 2

From the question,

⇒ Given:

  • Q = 1000 millicoulombs = 1 coulombs
  • t = 1 seconds. (Assuming the time is 1 seconds)

⇒ Substitute these values into equation 2

  • i = 1/1
  • i = 1 A.

Hence, The amount of current passing through the point is 1 A.

Learn more about charges here: brainly.com/question/4158552

3 0
2 years ago
wo parallel conducting plates are separated by 10.0 cm, and one of them is taken to be at zero volts. (a) What is the magnitude
melisa1 [442]

Answer:

<em> -18896.49 V/m</em>

<em></em>

Explanation:

Distance between the two plates = 10 cm = 10 x 10^{-2} m = 0.1 m

Also, one of the plates is taken as<em> zero volt.</em>

a. The potential strength between the zero volt plate, and 7.05 cm (0.0705 m) away is 393 V

b. The potential strength between the other plate, and 2.95 cm (0.0295 m) away is 393 V

<em>Potential field strength = -dV/dx</em>

where dV is voltage difference between these points,

dx is the difference in distance between these points

For the first case above,

potential field strength = -393/0.0705 = -5574.46 V/m

For the second case ,

potential field strength = -393/0.0295 = -13322.03 V/m

Magnitude of the field strength across the plates will be

-5574.46 + (-13322.03) = -5574.46 + 13322.03 =<em> -18896.49 V/m</em>

6 0
3 years ago
What is the to top-level consumers
Harrizon [31]

Level 1: Plants and algae make their own food and are called producers. Level 2: Herbivores eat plants and are called primary consumers.

3 0
4 years ago
Mt. Asama, Japan, is an active volcano. In 2009, an eruption threw solid volcanic rocks that landed 1 km horizontally from the c
Nata [24]

Answer:

a) 69.3 m/s

b) 18.84 s

Explanation:

Let the initial velocity = u

The vertical and horizontal components of the velocity is given by uᵧ and uₓ respectively

uᵧ = u sin 40° = 0.6428 u

uₓ = u cos 40° = 0.766 u

We're given that the horizontal distance travelled by the projectile rock (Range) = 1 km = 1000 m

The range of a projectile motion is given as

R = uₓt

where t = total time of flight

1000 = 0.766 ut

ut = 1305.5

The vertical distance travelled by the projectile rocks,

y = uᵧ t - (1/2)gt²

y = - 900 m (900 m below the crater's level)

-900 = 0.6428 ut - 4.9t²

Recall, ut = 1305.5

-900 = 0.6428(1305.5) - 4.9 t²

4.9t² = 839.1754 + 900

4.9t² = 1739.1754

t = 18.84 s

Recall again, ut = 1305.5

u = 1305.5/18.84 = 69.3 m/s

7 0
3 years ago
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