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3 years ago

Rotational dynamics physics question please help.

Physics

1 answer:

soldier1979 [14.2K]3 years ago

8 0

Looking at the force-time graph, wouldn't the force be integral fdt between 0 and 10s, a sort of "smoothed out pulse" ? And it looks like a familiar bell shaped curve. doesn't that produce a turning effect/torque and isn't there something about a circular analogue to F=ma in newtonian linear mechs ???
Looks like a difficult question for 5 points

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How do fossils provide evidence that evolution has taken place

zzz [600]

<span>Fossils provide solid evidence that organisms from the past are not the same as those found today; they show a progression of evolution. Scientists calculate the age of fossils and categorize them to determine when the organisms lived relative to each other. Hope this helps</span>

7 0

3 years ago

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

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;

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

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

What can increase the gravitational potential energy of a object

love history [14]

Answer:

Increase in mass and height

Explanation:

The gravitational potential energy of an object can increase if the mass and height of object is increased.

Gravitational potential energy is the energy due to the position of a body.

It is expressed as:

Gravitational potential energy = mass x acceleration due gravity x height

Increasing mass and height will cause an increase in gravitational potential energy.

6 0

3 years ago

If a 54 kg sprinter can accelerate from a standing start to a speed of 10 m/s in 3 s, what average power is generated?

lora16 [44]

Answer:

Power, P = 600 watts

Explanation:

It is given that,

Mass of sprinter, m = 54 kg

Speed, v = 10 m/s

Time taken, t = 3 s

We need to find the average power generated. The work done divided by time taken is called power generated by the sprinter i.e.

$P=\dfrac{W}{t}$

Work done is equal to the change in kinetic energy of the sprinter.

$P=\dfrac{\dfrac{1}{2}mv^2}{t}$

$P=\dfrac{\dfrac{1}{2}\times 54\ kg\times (10\ m/s)^2}{3\ s}$

P = 900 watts

So, the average power generated by the sprinter is 900 watts. Hence, this is the required solution.

3 0

3 years ago

What is the momentum of a 1.5 * 10 ^ 3 - kil kilogram car as it travels at 30. meters per second due east for 60. seconds ?

serg [7]

Answer:

45000kgm/s due east

Explanation:

Given parameters:

Mass of the car = 1.5 x 10³kg

Velocity = 30m/s

Time taken = 60s

Unknown:

Momentum = ?

Solution:

Momentum is the quantity of motion a body possess,

Momentum = mass x velocity

So;

Momentum = 1.5 x 10³ x 30 = 45000kgm/s due east

6 0

3 years ago