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Ad libitum [116K]
3 years ago
6

two horses are pulling a 325 kg wagon, initially at rest. The horses exert 250 N and 178 N forward forces, respectively. Ignorin

g friction, how fast is the wagon moving 3.50 s later?
Physics
1 answer:
Dovator [93]3 years ago
7 0

Answer:

AFter 3.5 s, the wagon is moving at:   4.62\,\,\frac{m}{s}

Explanation:

Let's start by finding first the net force on the wagon, and from there the wagon's acceleration (using Newton's 2nd Law):

Net force = 250 N + 178 N = 428 N

Therefore, the acceleration from Newton's 2nd Law is:

F=m\,*\,a\\a = \frac{F}{m} \\a= \frac{428}{325}\, \frac{m}{s^2} \\a\approx 1.32 \,\,\frac{m}{s^2}

So now we apply this acceleration to the kinematic expression for velocity in an object moving under constant acceleration:

v_f=v_i+a\,*\,t\\v_f=0+1.32\,*\,3.5\,\,\frac{m}{s} \\v_f=4.62\,\,\frac{m}{s}

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2 years ago
A 210 Ohm resistor uses 9.28 W of
IgorLugansk [536]
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Current=I

\boxed{\sf P=I^2R}

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2 years ago
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<span> For any body to move in a circle it requires the centripetal force (mv^2)/r. In this case a ball is moving in a vertical circle swung by a mass less cord. At the top of its arc if we draw its free body diagram and equate the forces in radial direction to the centripetal force we get it as T +mg =(mv^2)/r T is tension in cord m is mass of ball r is length of cord (radius of the vertical circle) To get the minimum value of velocity the LHS should be minimum. This is possible when T = 0. So minimum speed of ball v at top =sqrtr(rg)=sqrt(1.1*9.81) = 3.285 m/s In the second case the speed of ball at top = (2*3.285) =6.57 m/s Let us take the lowest point of the vertical circle as reference for potential energy and apllying the conservation of energy equation between top & bottom we get velocity at bottom as 9.3m/s. Now by drawing the free body diagram of the ball at the bottom and equating the net radial force to the centripetal force T-mg=(mv^2)/r We get tension in cord T=13.27 N</span>
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Answer:

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