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dedylja [7]
3 years ago
10

Blue to gray in color and reaching up to 10 feet (3.05 meters) in length and 220 pounds (99.8 kilograms) in weight, the sailfish

is considered the fastest marine creature. These streamlined beasts have been clocked at 68 miles per hour (30.4 meters per second). How far does the sailfish travel in 2 minutes?
Physics
1 answer:
vagabundo [1.1K]3 years ago
6 0

Answer:

the distance traveled by the fish is 3648 m

Explanation:

In general, animals have a small period of acceleration, which we will despise after which they travel at a constant speed so we can use the kinematic equations in uniform motion

   

We reduce the units to System SI

      t = 2 min (60s / 1 min) = 120 s

Calculate

       V = x / t

       x= V t

       x = 30.4 120

       x = 3648 m

 

This is the distance traveled by the fish

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Answer the attached question.<br>No Spam!<br>⚔️⚔️⚔️​
RoseWind [281]
<h3>Question:5</h3>

The given data can be written in following way in coordinates axes.

(0,0), (2,2), (4,4), (6,4), (8,4), (10,6),(12,4), (14, 2).

a) Average velocity for first 4 seconds

Average velocity = Total Displacement/ Time taken

= (4-0)/(4 - 0)

=4/4

= 1/1

<h3> =1 m/s</h3>

b) Average velocity for 4 to 8 seconds

Average velocity = Total Displacement/ Time taken

= (4 - 4)/(8-4)

= 0/4

<h3> = 0</h3>

c) Average velocity for last 6 seconds

Last 6 seconds = from 8 to 14 seconds

Average velocity = Total Displacement/Time taken

= (2 - 4)/(14 - 8)

= -2/6

<h3> = -1/3 m/s</h3>

<h3>Question 6:</h3>

The given data can be written in following way in coordinates axes.

(0,0), (10,20), (20,20), (30,20), (40,0)

a) State the kind of motion from Os to 10s and from 30s to 40s

It is obvious from the graph that the velocity between Os to 10s has been increased from 0 m/s to 20m/s. Hence there is a uniform Acceleration in the body.

b) What is the velocity of the body after 10s and 40s ?

It is clear from the graph and table as well that,

<h3>Velocity after 10s is 20m/s</h3>

and

<h3>Velocity after 40s is 0 m/s</h3>

c) Calculate the distance covered by the body between 10s to 30s.

Distance covered by the body between 10s to 30s will be given by area of rectangle ABCD

Area of rectangle ABCD= (30 - 10) × (20 - O)

=20×20

<h3> = 400 m</h3>

6 0
2 years ago
Light shines through a single slit whose width is 5.7 x 10-4 m. A diffraction pattern is formed on a flat screen located 4.0 m a
lilavasa [31]

Answer:

\lambda = 570\ nm

Explanation:

Given,

Width of slit, W = 5.7 x 10⁻⁴ m

Distance between central bright fringe, L = 4 m

distance between central bright fringe and first dark fringe, y = 4 mm

Diffraction angle

tan \theta = \dfrac{y}{L}

tan \theta = \dfrac{4}{4\times 10^3}

\theta = 0.0572

Now.

W sin \theta = m \lambda

m = 1

5.7 \times 10^{-4} \times sin (0.0572) = 1 \times \lambda

\lambda = 569.99 \times 10^{-9}\ m

\lambda = 570\ nm

4 0
3 years ago
A 92kg astronaut and a 1200kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving
Harman [31]

Answer:

13.7m

Explanation:

Since there's no external force acting on the astronaut or the satellite, the momentum must be conserved before and after the push. Since both are at rest before, momentum is 0.

After the push

m_av_a + m_sv_s = 0

Where m_a = 92kg is the mass of the astronaut, m_s = 1200kg is the mass of the satellite, v_s = 0.14 m/s is the speed of the satellite. We can calculate the speed v_a of the astronaut:

v_a = \frac{-m_sv_s}{m_a} = \frac{-1200*0.14}{92} = -1.83 m/s

So the astronaut has a opposite direction with the satellite motion, which is further away from the shuttle. Since it takes 7.5 s for the astronaut to make contact with the shuttle, the distance would be

d = vt = 1.83 * 7.5 = 13.7 m

4 0
3 years ago
Acid compounds contain a.Oxygen B. Ion c. Hydrogen
valentina_108 [34]

Answer:

c hydrogen

Explanation:

8 0
3 years ago
Read 2 more answers
Two cars are traveling in the same direction down a highway at 65 miles per hour. What is the relative velocity of the second ca
Levart [38]

Answer:

5 hours

Explanation:

Let the required time be x hours. The time will be the same for both cars.

The cars will cover different distances because they are travelling at different speeds.

<em>D=S×T </em>

The distance travelled by the slower car = 50×x miles.

The distance travelled by the faster car = 58×x miles.

The two distances differ by 40 miles.

58x−50x=40

8x=40

x=5 hours

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

A second method:

The difference in the distances is 40 miles

The difference in the speeds is #8mph.

The time to make up the 40 miles= \frac{40}{8}=5 hours

8 0
2 years ago
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