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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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Can somebody help me with this quiz please
mezya [45]
Sure what do u need help with
4 0
2 years ago
A(n) 131 g ball is dropped from a height
larisa [96]

Answer:

26.59 N/m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 131 g

Extention (e) = 4.82755 cm

Acceleration due to gravity (g) = 9.8 m/s²

Spring constant (K) =?

Next, we shall convert 131 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

131 g = 131 g × 1 Kg / 1000 g

131 g = 0.131 Kg

Thus, 131 g is equivalent to 0.131 Kg.

Next, we shall the force exerted by the ball on the spring. This can be obtained as follow:

Mass (m) = 0.131 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = ma

F = 0.131 × 9.8

F = 1.2838 N

Next, we shall convert 4.82755 cm to metre (m)

This can be obtained as follow:

100 cm = 1 m

Therefore,

4.82755 cm = 4.82755 cm × 1 m / 100 cm

4.82755 cm = 0.0482755 m

Thus, 4.82755 cm is equivalent to 0.0482755 m

Finally, we shall determine the spring constant as follow:

Force (F) = 1.2838 N

Extention (e) = 0.0482755 m

Spring constant (K) =?

F = Ke

1.2838 = K × 0.0482755

Divide both side by 0.0482755

K = 1.2838 / 0.0482755

K = 26.59 N/m

Thus the spring constant is 26.59 N/m

7 0
3 years ago
A sinusoidally oscillating current I ( t ) with an amplitude of 9.55 A and a frequency of 359 cycles per second is carried by a
UNO [17]

Answer:

P_{avg} = 6.283*10^{-9} \ W

Explanation:

Given that;

I₀ = 9.55 A

f = 359 cycles/s

b = 72.2 cm

c = 32.5 cm

a = 80.2 cm

Using the formula;

\phi = \frac{\mu_o Ic }{2 \pi} In (\frac{b+a}{b})

where;

E= \frac{d \phi}{dt}

E = \frac{\mu_o}{2 \pi}c In (\frac{b+a}{a}) I_o \omega cos \omega t

E_{rms} =   \frac { {\frac{\mu_o \ c}{2 \pi} In (\frac{b+a}{a}) I_o (2 \pi f)}}{\sqrt{2}}

Replacing our values into above equation; we have:

E_{rms} =   \frac { {\frac{4 \pi*10^{-7}*0.325}{2 \pi} In (\frac{72.2+80.2}{80.2}) *9.55 (2 \pi *359)}}{\sqrt{2}}

E_{rms} =   \frac {8.98909588*10^{-4} }{\sqrt{2}}

E_{rms} =   6.356*10^{-4} \ V

Then the P_{avg is calculated as:

P_{avg} = \frac{E^2}{R}

P_{avg} = \frac{(6.356*10^{-4})^2}{64.3}

P_{avg} = 6.283*10^{-9} \ W

6 0
3 years ago
If the pendulum took longer to complete one oscillation, how would the graph change?
Sindrei [870]

Answer:

took longer to complete one oscillation, that means its PERIOD increased, and the distance between the peaks of the graph would be longer.

line would be less. the period of oscillation would have any effect on the graph

7 0
2 years ago
Read 2 more answers
According to the principal of wave refraction, when light from the atmosphere enters a medium of greater density at an oblique a
Hoochie [10]

Answer:

d. velocity slows down and angle changes

Explanation:

6 0
2 years ago
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