A 3.00-kg object has a velocity 1 6.00 i ^ 2 2.00 j ^2 m/s. (a) what is its kinetic energy at this moment? (b) what is the net work done on the object if its velocity changes to 1 8.00 i ^ 1 4.00 j ^

Answers

Answer 1
Answer: (a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:
v= √((6.00m/s)^2+(2.00 m/s)^2)=6.32 m/s
And so, the kinetic energy of the object is
K= (1)/(2)mv^2= (1)/(2)(3.00 kg)(6.32 m/s)^2=60 J

(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is
v= √((8.00 m/s)^2+(4.00 m/s)^2)=8.94 m/s
And so the new kinetic energy is
K= (1)/(2)mv^2= (1)/(2)(3.00 kg)(8.94 m/s)^2=120 J

So, the work done on the object is the variation of kinetic energy of the object:
W=\Delta K=120 J-60 J=60 J
Answer 2
Answer:

Final answer:

The initial kinetic energy of the 3.00-kg object traveling at a velocity of 2.00 m/s is 6.00 Joules. When the object's velocity changed to 4.47 m/s, its kinetic energy became 30.02 Joules. Hence, the net work done on the object is 24.02 Joules.

Explanation:

The kinetic energy of any object can be calculated using the formula KE = 0.5 * m * v^2, where m is the object's mass and v is its velocity. For the 3.00-kg object with a velocity of 6.00 i ^ 2 and 2.00 j ^2 m/s, its velocity magnitude would be the square root of (6.00^2 + 2.00^2), which is 2.00 m/s. Plugging the values into the formula, the kinetic energy (a) would be 0.5 * 3.00 * 2.00^2 = 6.00 Joules.

The net work done on an object (b) can be obtained by finding the change in kinetic energy when the object’s velocity changes to 8.00 I and 4.00 j. The final velocity's magnitude would be the square root of (8.00^2 + 4.00^2), which is 4.47 m/s. Hence, the final kinetic energy is 0.5 * 3.00 * 4.47^2 = 30.02 Joules. Therefore, the net work done equals the change in kinetic energy, which is 30.02 - 6.00 = 24.02 Joules.

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Answers

Answer:

Imp = 5626.488\,(kg\cdot m)/(s)

Explanation:

First, it is required to model the function that models the increasing force in the +x direction:

a =(781..25\,N)/((1.27\,s)^(2))

a = 484 (N)/(s^(2))

The equation is:

F_(x) = 484\,(N)/(s^(2))\cdot t^(2)

The impulse done by the engine is given by the following integral:

Imp=484\,(N)/(s^(2)) \int\limits^(3.50\,s)_(2\,s) {t^(2)} \, dt

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Imp = 5626.488\,(kg\cdot m)/(s)

What is the average kinetic energy of hydrogen atoms on the 5500°C surface of the sun?

Answers

Answer: The average kinetic energy of hydrogen atoms is 1.19562* 10^(-19)J

Explanation:

To calculate the average kinetic energy of the atom, we use the equation:

K=(3)/(2)kT

where,

K = average kinetic energy

k = Boltzmann constant = 1.3807* 10^(-23)J/K

T = temperature = 5500^oC=[5500+273]K=5773K

Putting values in above equation, we get:

K=(3)/(2)* 1.3807* 10^(-23)J/K* 5773K\n\nK=1.19562* 10^(-19)J

Hence, the average kinetic energy of hydrogen atoms is 1.19562* 10^(-19)J

Which of the following is correct? *PLEASE HELP MEEEE
1 cm = 100 m
1 mm = 100 cm
100 mm = 1 cm
1 m = 100 cm

Answers

Answer:

The last one

1m = 100 cm

Explanation:

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A boy flies a kite with the string at a 30∘ angle to the horizontal. The tension in the string is 4.5 N. Part A Part complete How much work does the string do on the boy if the boy stands still?

Answers

Answer:

Work done is zero

Explanation:

given data

Angle of kite with horizontal =  30 degree

tension in the string =  4.5 N

WE KNOW THAT

Work =  force * distance

horizontal force =  Tcos\theta = 4.5*cos30 = 3.89 N

DISTANCE = 0 as boy stands still. therefore

work done = 3.89 *0 = 0

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Answers

Answer:210000N

Explanation:

Pressure=3x10^5pa

area=0.7m^2

Force = pressure x area

Force=3x10^5x0.7

Force=210000N

What frequency corresponds to a period of 4.31s.
T =1/f = 1/4.31s = 0.232hz correct?

Answers

Answer:correct

Explanation: Period T is the reciprocal of frequency (i.e T=1/f)

Frequency is the reciprocal of period (i.e F= 1/T)

Therefore if T=4.31s

Frequency F= 1/4.31s=0.232hz