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Masja [62]
3 years ago
5

An oceanic depth-sounding vessel surveys the ocean bottom with ultrasonic waves that travel at 1530 m/s in seawater. The time de

lay of the echo to the ocean floor and back is 6 s. ?
Physics
1 answer:
Brrunno [24]3 years ago
5 0

Answer:

d = 4590 m

Explanation:

given,

Speed of ultrasonic wave, v = 1530 m/s

time of the echo, t = 6 s

Let d be the depth of the ocean

now,

total distance travel by the ultrasonic wave = 2 d

we know,

distance = speed x t

2 d = 1530 x 6

d = 1530 x 3

d = 4590 m

Hence, the depth of the ocean floor is equal to d = 4590 m

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What is the equation: If F=10 N, a=5 m/s², m=?
Romashka-Z-Leto [24]

Answer:

2 kg

Explanation:

Remember:

F = m * a       re-arrange to

F/a   = m      substitute in the given values

10 / 5   =   2 kg

8 0
2 years ago
Determine the direction of the force that will act on the charge in each of the following situations. A negative charge moving t
wlad13 [49]

Answer:

a) DOWN direction,  b)  directed INTO THE SCREEN, c)    F = 0

Explanation:

The direction of the force is

for electric force

           F = q E

where we assume a positive test charge, for which the force has the direction of the electric field.

For a magnetic field

in this case the direction of the force is given by the right hand rule.

For a positive test charge, the thumb points in the direction of velocity, the other fingers extended in the direction of the magnetic field, and the palm gives the direction of force for a positive charge.

           F = q v x B

Let us apply these considerations to our case.

a) negative charge moving to the left

in a magnetic field points away from the screen

In this case the thumb goes to the left, the fingers extended outwards and the palm points upwards, but since the charge is negative the force has a DOWN direction.

b) negative charge moves to the left

in electric field it points off the screen.

The outside is in the direction of the electric field and since the charge is negative, the force is directed INTO THE SCREEN

c) positive charge moves down

in magnetic field points up

in this case the velocity and the field have the same direction so the vector product of them is zero

       F = q v  B sin 0

       F = 0

6 0
3 years ago
What is an ellipse?
Rom4ik [11]

Answer:

i think it's C thx correct me if wrong

6 0
3 years ago
Before beginning a long trip on a hot day, a driver inflates anautomobile tire to a gauge pressure of 2.70 atm at 300 K. At the
anygoal [31]

Answer:

The value is the temperature of the air inside the tire T_{2} = 340.54 K

% of the original mass of air in the tire should be released 99.706 %

Explanation:

Initial gauge pressure = 2.7 atm

Absolute pressure at inlet P_{1} = 2.7 + 1 = 3.7 atm

Absolute pressure at outlet P_{2} = 3.2 + 1 = 4.2 atm

Temperature at inlet T_{1} = 300 K

(a) Volume of the system is constant so  pressure is directly proportional to the temperature.

\frac{T_{2} }{T_{1} } = \frac{P_{2} }{P_{1} }

\frac{T_{2} }{300}  = \frac{4.2}{3.7}

T_{2} = 340.54 K

This is the value is the temperature of the air inside the tire

(b). Since volume of the tyre is constant & pressure reaches the original value.

From ideal gas equation P V = m R T

Since P , V & R is constant. So

m T = constant

m_{1}  T_{1} =  m_{2}  T_{2}

\frac{m_{2} }{m_{1}  } = \frac{T_{1} }{T_{2} }

\frac{m_{2} }{m_{1}  } = \frac{300}{354.54}

\frac{m_{2} }{m_{1}  } =0.00294

value of  the original mass of air in the tire should be released is  \frac{m_{2} - m_{1}}{m_{1}}

\frac{0.00294-1}{1}

⇒ -0.99706

% of the original mass of air in the tire should be released 99.706 %.

8 0
3 years ago
What was created 300 years ago by scientist carolus linnaeus
Norma-Jean [14]
<span>the classification system</span>
6 0
3 years ago
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