When a tennis ball is spun around in a circle on a string and the string breaks the tennis ballwill be pulled in a curved path away from the center because of Centrifugal force
True or false


Answer 1



whenever the string breaks, the ball will follow the straight line tangential path


No, the ball will not follow a curved path after the string breaks. Since, the the direction of velocity is tangential to each point of the circular motion. Therefore, it changes at every point. This produces an acceleration in the circle called centripetal acceleration. There is also a tangential component of acceleration acting on the ball during this motion.

So, whenever the string breaks, the ball will follow the straight line tangential path. Hence, the given statement is false.

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1. If a net force of 412 N is required to accelerate an object at 5.82 m/s2, what must theobject's mass be?



The mass of the object is approximately 70.79 kilograms


We use Newton's second law to solve this problem. This law states that the net force on an object equals the product of its mass times the acceleration:


Therefore, for this case, since the net force on the object and its acceleration are given, we can use the equation above to solve for the unknown mass:

F_(net)=m\,a\n412\,N=m\,(5.82\,(m)/(s^2) )\nm=(412\,N)/(5.82\,(m)/(s^2) ) \nm=70.79\.kg

A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector r with arrow = (2.00 m)i hat − (3.00 m)j + (2.00 m)k, the force is F with arrow = Fxi hat + (7.00 N)j − (5.80 N)k and the corresponding torque about the origin is vector tau = (3.40 N · m)i hat + (2.80 N · m)j + (0.800 N · m)k. Determine Fx.





explanation is in attachment

A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above the horizontal. The horizontal distance to the net is 7.0 m, and the net is 1.0 m high. Does the ball clear the net?



ball clears the net


v_(o) = initial speed of launch of the ball = 20 ms^{-1}

\theta = angle of launch = 5 deg

Consider the motion of the ball along the horizontal direction

v_(ox) = initial velocity = v_(o) Cos\theta = 20 Cos5 = 19.92 ms^(-1)

t = time of travel

X = horizontal displacement of the ball to reach the net = 7 m

Since there is no acceleration along the horizontal direction, we have

X = v_(ox) t \n7 = v_(ox) t\nt = (7)/(v_(ox))       Eq-1

Consider the motion of the ball along the vertical direction

v_(oy) = initial velocity = v_(o) Sin\theta = 20 Sin5 = 1.74 ms^(-1)

t = time of travel

Y_(o) = Initial position of the ball at the time of launch = 2 m

Y = Final position of the ball at time "t"

a_(y) = acceleration in down direction = - 9.8 ms⁻²

Along the vertical direction , position at any time is given as

Y = Y_(o) + v_(oy) t + (0.5) a_(y) t^(2)\nY = 2 + (20 Sin5) ((7)/(20 Cos5)) + (0.5) (- 9.8) ((7)/(20 Cos5))^(2)\nY = 2.00758 m\n

Since Y > 1 m

hence the ball clears the net

At takeoff, a commercial jet has a speed of 72 m/s. Its tires have a diameter of 0.89 m. Part (a) At how many rev/min are the tires rotating? Part (b) What is the centripetal acceleration at the edge of the tire in m/s^2?



a) Revolutions per minute = 2.33

b) Centripetal acceleration = 11649.44 m/s²


a) Angular velocity is the ratio of linear velocity and radius.

Here linear velocity = 72 m/s

Radius, r  = 0.89 x 0. 5 = 0.445 m

Angular velocity

         \omega =(72)/(0.445)=161.8rad/s


         f=(2\pi)/(\omega)=(2* \pi)/(161.8)=0.0388rev/s=2.33rev/min

Revolutions per minute = 2.33

b) Centripetal acceleration


  Here linear velocity = 72 m/s

  Radius, r  = 0.445 m



Centripetal acceleration = 11649.44m/s²

A force of 140 140 newtons is required to hold a spring that has been stretched from its natural length of 40 cm to a length of 60 cm. Find the work done in stretching the spring from 60 cm to 65 cm. First, setup an integral and find a a, b b, and f ( x ) f(x) which would compute the amount of work done.



The work done in stretching the spring is 0.875 J.


Given that,

Force = 140 N

Natural length = 60-40 = 20 cm

Stretch length of the spring = 65-60 = 5 cm

We need to calculate the spring constant

Using formula of Hooke's law

F= kx




We need to calculate the work done



On integration



W=0.875\ J

Hence, The work done in stretching the spring is 0.875 J.

How many hours does earth take to complete one rotation?



24 hours take earth to complete rotation