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Greeley [361]
2 years ago
11

A stone weighing 1.5 kilograms is resting on a rock at a height of 20 meters above the ground. The stone rolls down 10 meters an

d comes to rest on a patch of moss. The gravitational potential energy of the stone on the moss is joules. (Use PE = m × g × h, where g = 9.8 N/kg.)
Physics
2 answers:
Andrews [41]2 years ago
6 0
The potential energy of the moss at its position is 147 Joules. 
34kurt2 years ago
4 0
Given that the equation for the potential energy of an object is: PE = mgh, where, m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above the ground, simply substitute the given values to obtain the potential energy. In this case, m = 1.5 kg, g = 9.8 N/kg, and h = 10 m. The height is set to 10 m since this is the height of the moss above the ground after it fell. Thus, the potential energy of the moss at its position is 147 Joules. 
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Answer:

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Explanation:

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F=ma

With F the maximum force the truck applies to the SUV, m the mass of the SUV and a the acceleration of the SUV; solving for a:

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How many significant figures are in 0.0069
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Two significant figures, the 6 and the 9
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The resultant motion is given by pithagoras, since the two components (north and east) are perpendicular to each other.
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A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by
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a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The position of the car as a function of time t is given by

x(t)=\alpha t^2 - \beta t^3

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\alpha = 1.50 m/s^2

\beta = 0.05 m/s^3

The average velocity is given by the ratio between the displacement and the time taken:

v=\frac{\Delta x}{\Delta t}

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

The position at t = 2.00 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

So the displacement is

\Delta x = x(2)-x(0)=5.6-0=5.6 m

The time interval is

\Delta t = 2.0 s - 0 s = 2.0 s

And so, the average velocity in this interval is

v=\frac{5.6 m}{2.0 s}=2.8 m/s

b)

The position at t = 0 is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

While the position at t = 4.00 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

So the displacement is

\Delta x = x(4)-x(0)=20.8-0=20.8 m

The time interval is

\Delta t = 4.0 - 0 = 4.0 s

So the average velocity here is

v=\frac{20.8}{4.0}=5.2 m/s

c)

The position at t = 2 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

While the position at t = 4 s is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

So the displacement is

\Delta x = 20.8 - 5.6 = 15.2 m

While the time interval is

\Delta t = 4.0 - 2.0 = 2.0 s

So the average velocity is

v=\frac{15.2}{2.0}=7.6 m/s

Learn more about average velocity:

brainly.com/question/8893949

brainly.com/question/5063905

#LearnwithBrainly

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