Why did scientists compare megalodons and great white sharks? SCIENCE

(8c8p49) A 115g Frisbee is thrown from a point 1.00 m above the ground with a speed of 12.00 m/s. When it has reached a height o

IgorLugansk [536]

Answer:

The work done on the Frisbee is 1.36 J.

Explanation:

Given that,

Mass of Frisbee, m = 115 g = 0.115 kg

Initial speed of Frisbee, u = 12 m/s at a point 1 m above the ground

Final speed of Frisbee , v = 10.9674 m/s when it has reached a height of 2.00 m. Let W is the work done on the Frisbee by its weight. According to work energy theorem, the work done is equal to the change in its kinetic energy. So,

$W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times0.115\times\left((10.9674)^{2}-(12)^{2})\right)\\\\W=-1.36\ J$

So, the work done on the Frisbee is 1.36 J. Hence, this is the required solution.

3 0

3 years ago

Read 2 more answers

A seagull flies at a velocity of 9.00 m/s straight into the wind.

RideAnS [48]

a)If it takes the bird 18.0 minutes to fly 6 km away from the earth, the wind's speed will be 4 m/s.

b) The bird would need 7 minutes and 42 seconds to fly back 6 kilometers if he turned around and flew with the wind.

c)Compared to the 133.33 seconds it would take without the wind, the overall round-trip time is affected by the wind.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

A seagull flies at a velocity, $\rm v_{SA} = 9 \ m/sec$

The time the bird takes,t=18.0 min

The distance traveled relative to the earth = 6.00 km

a)

The seagull's relative velocity with reference to the ground as;

$\rm v_{sg} = \frac{6.00 \times 10^3 \ m }{(20 min) \times \frac{60 s }{1 \ min}} \\\\ v_{sg}= 5.00 \ m/sec$

Air velocity with reference to the ground is;

$\rm v_{AG}= v_{SG}-v_{SA} \\\\ v_{AG} = 5.00 \ m/sec - 9.00 \ m/sec \\\\ v_{AG} = -4.00 \ m/sec$

b)

If the bird turns around and flies with the wind, The time will he take to return 6.00 km is;

$\rm v_{SG}=v_{SA}+v_{AG} \\\\ v_{SG}=-900 \ m/sec +(-4.00 \ m/sec) \\\\ v_{SG}= -13.00 \ m/sec$

The time the bird takes;

$\rm t = \frac{x_{SG}}{v_{SG}} \\\\ t = \frac{6.00 \times 10^3 \ m }{13.00 \ m/sec } \\\\ t = 462 m/sec \\\\ t = 7 \ min \ and \ 42 \ sec$

c)\

The total round-trip time compared to what it would be with no wind. is;

$\rm t = 20 \ min( \frac{60 \ sec }{1 \ min} )+ 462 \ sec \\\\ t = 1200 \ sec +6 462 \ ec \\\\ t= 1662 \ sec$

The time for the round trip is;

$\rm t = \frac{12 \times 10^ 3 }{ 9 \ m/sec } \\\\ t = 1333.33 \ sec$

Hence the wind's speed, the time bird would need to fly back the total round-trip time will be 4 m/s, 7 minutes and 42 seconds and 1333.33 sec.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

4 0

2 years ago

50 points please help ASAP! !

Contact [7]

Answer:

The sum of the initial and final velocity is divided by 2 to find the average. The average velocity calculator uses the formula that shows the average velocity (v) equals the sum of the final velocity (v) and the initial velocity (u), divided by 2.

4 0

3 years ago

In order to open the clam it catches, a seagull will drop the clam repeatedly onto a hard surface from high in the air until the

Vedmedyk [2.9K]

Answer:

2.2 s

Explanation:

Hi!

Let's consider the origin of the coordinate system at the ground, and consider that the clam starts with zero velocity, the equation of motion of the clam is given by

$x(t) = 23.1 m - \frac{1}{2}(9.8 m/s^2) t^2$

We are looking for a time t for which x(t) = 0

$0 = 23.1 m - (4.9 m/s^2) t^2$

Solving for t:

$t = \sqrt{\frac{23.1}{4.9}} s = 2.17124 s$

Rounding at the first decimal:

t = 2.2 s

4 0

3 years ago

A hot-air balloon with a mass of 400 kilograms moves across the sky with 3,200 joules of kinetic energy. The velocity of the bal

kotegsom [21]

Answer:

4 m/s

Explanation:

KE =

Velocity of balloon will be 4 m/s.

!! Hope It Helps !!

4 0

3 years ago