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GREYUIT [131]
2 years ago
9

If a fish experiences an acceleration of 4.9 m/s over a period of 1.84 a , what was the change in velocity

Physics
1 answer:
Veseljchak [2.6K]2 years ago
5 0

Explanation:

they'll dewwel jss q.v hewn red f quill DC noo

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A sound wave moving with a speed of 1500 m/s is sent from a submarine to the ocean floor. It reflects off the
Alekssandra [29.7K]

Answer:

the distance between the submarine and the ocean floor is 11,250 m

Explanation:

Given;

speed of the wave, v = 1500 m/s

time of motion of the wave, t = 15 s

The time taken to receive the echo is calculated as;

time \ of \ motion \ (t) = \frac{total \ distance }{speed \ of \ wave} = \frac{2d}{v}  \\\\2d = vt\\\\d = \frac{vt}{2} \\\\d = \frac{1500 \times 15}{2} \\\\d = 11,250 \ m

Therefore, the distance between the submarine and the ocean floor is 11,250 m

3 0
2 years ago
How much force was applied to an object if was moved 2 meters and the work done on the object was 8 joules?
serious [3.7K]

Answer:

The answer is 4N or B

Explanation:

Just the equation W = F x D.

We have W = 8 J and D = 2 m using algebra ....

8J/2m = F ... F = 4.

6 0
3 years ago
Read 2 more answers
The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz. What is the period of the crystal's motion?
Elan Coil [88]

Answer:

Time period, T=3.05\times 10^{-5}\ s

Explanation:

Given that,

The quartz crystal used in an electric watch vibrates with a frequency of 32,768 Hz, f = 32768 Hz

We need to find the period of the crystal's motion. The relationship between the frequency and the time period is given by :

T=\dfrac{1}{f}

T is the time period of the crystal's motion.

Time period is given by :

T=\dfrac{1}{32768}

T=3.05\times 10^{-5}\ s

So, the time period of the crystal's motion is 3.05\times 10^{-5}\ s. Hence, this is the required solution.

8 0
3 years ago
A 1451 kg car is traveling at 48.0 km/h. How much kinetic energy does it possess? K.E. =
jarptica [38.1K]

           Kinetic energy  =  (1/2) (mass) (speed)²

BUT . . . in order to use this equation just the way it's written,
the speed has to be in meters per second.  So we'll have to
make that conversion.

        KE  =  (1/2) · (1,451 kg) · (48 km/hr)² · (1000 m/km)² · (1 hr/3,600 sec)²

               =  (725.5) · (48 · 1000 · 1 / 3,600)²  (kg) · (km·m·hr / hr·km·sec)²

               =  (725.5) · ( 40/3 )²  ·  ( kg·m² / sec²)

               =    128,978  joules  (rounded)
8 0
2 years ago
The owner of a company that manufactures drinking cups decides it would be impressive to build an inground swimming pool that is
garri49 [273]

Answer:

The depth is 5.15 m.

Explanation:

Lets take the depth of the pool = h m

The atmospheric pressure ,P = 101235 N/m²

The area of the top = A m²

The area of the bottom =  a m²

Given that A= 1.5 a

The force on the top of the pool = P A

The total pressure on the bottom = P + ρ g h

ρ =Density of the  water = 1000 kg/m³

The total pressure at the bottom of the pool =  (P + ρ g h) a

The bottom and the top force is same

(P + ρ g h) a = P A

P a +ρ g h a =  P A

ρ g h a =  P A - P a

h=\dfrac{P ( A-a)}{\rho g a}

h=\dfrac{P ( 1.5 a-a)}{\rho g a}

h=\dfrac{P ( 1.5- 1)}{\rho g}

h=\dfrac{101235 ( 1.5- 1)}{1000\times 9.81}\ m

h=5.15 m

The depth is 5.15 m.

7 0
3 years ago
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