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ch4aika [34]
4 years ago
10

A squirrel is trying to locate some nuts he buried for the winter. He moves 4.3 m to the right of a stone and dogs unsuccessfull

y. Then he moves 1.1 m to the left of his hole, changes his mind, and moves 6.3 m to the right of that position and digs a second hole. No luck. Then he moves 8.0 m to the left and digs again. He finds a nut at last. What is the squirrel's total displacement from his starting point
Physics
1 answer:
krek1111 [17]4 years ago
8 0

Answer:

The total displacement from the starting point is 1.5 m.

Explanation:

You need to sum and substract, depending on the movement (to the right, sum; to the left, substract).

First, it moves 4.3 m right and return 1.1 m. So the new distance from the starting point is 3.2 m.

Second, it moves 6.3 m right, so the new distance is 9.5 m.

Finally it moves 8 m to the left, so 9.5 m - 8 m= 1.5 m.

Summarizing, at the end the squirrel is 1.5 m from its starting point.

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A model rocket is launched straight upward with an initial speed of 52.0 m/s. It accelerates with a constant upward acceleration
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Explanation:

(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.

(b) We must separate the motion into two parts, when the rocket's engines is on  and when the rocket's engines is off.

First we must find the rocket speed when the engines stop:

v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}

This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:

v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m

So, the maximum height reached by the rocket is:

h=y_1+y_2\\h=160m+154.28m=314.28m

(c) In the first part we have:

v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s

And in the second part:

t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s

So,  the time it takes to reach the maximum height is:

t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s

(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:

v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}

t_4=\frac{v_f-v_0}{g}\\t_4=\frac{78.48\frac{m}{s}-0}{9.8\frac{m}{s^2}}\\t_4=8.01s

So, the total time is:

t=t_3+t_4\\t=8.60s+8.01s\\t=16.61s

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4 years ago
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Explanation:

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A crane raises a crate with a mass of 150 kg to a height of 20 m. Given that
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Answer:

\boxed {\boxed {\sf 29,400 \ Joules}}

Explanation:

Gravitational potential energy is the energy an object possesses due to its position. It is the product of mass, height, and acceleration due to gravity.

E_P= m \times g \times h

The object has a mass of 150 kilograms and is raised to a height of 20 meters. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.

  • m= 150 kg
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  • h= 20 m

Substitute the values into the formula.

E_p= 150 \ kg \times 9.8 \ m/s^2 \times 20 \ m

Multiply the three numbers and their units together.

E_p=1470 \ kg*m/s^2 \times 20 m

E_p=29400 \ kg*m^2/s^2

Convert the units.

1 kilogram meter square per second squared (1 kg *m²/s²) is equal to 1 Joule (J). Our answer of 29,400 kg*m²/s² is equal to 29,400 Joules.

E_p= 29,400 \ J

The crate has <u>29,400 Joules</u> of potential energy.

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A frictionless piston-cylinder device contains 4.5 kg of nitrogen at 110 kPa and 200 K. Nitrogen is now compressed slowly accord
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Answer:

427.392 kJ

Explanation:

m = Mass of gas = 4.5 kg

Initial temperature = 200 K

Final temperature = 360 K

R = Mass specific gas constant = 296.8 J/kgK

\gamma = Specific heat ratio = 1.5

Work done for a polytropic process is given by

W=\frac{mR\Delta T}{1-\gamma}\\\Rightarrow W=\frac{4.5\times 296.8(360-200)}{1-1.5}\\\Rightarrow W=-427392\ J\\\Rightarrow W=-427.392\ kJ

The work input during the process is -427.392 kJ

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3 years ago
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