Physics, 31.07.2019 15:30 itsseand23

A0.55-kg ball, attached to the end of a horizontal cord, is revolved in a circle of radius 1.3 m on a frictionless horizontal surface. if the cord will break when the tension in it exceeds 75 n, what is the maximum speed the ball can have?

Answers: 1

Show answers

Answers

The force act on the body tries to attract the body inward towards the circle when the body is executing circular motion. The maximum value of speed will be 13.3 m/sec.

What is centripetal force?

The force operating on an object in curvilinear motion directed toward the axis of rotation or center of curvature is known as centripetal force.

Newton is the unit of centripetal force.

The centripetal force is always perpendicular to the direction of displacement of the item. The centripetal force of an item traveling on a circular route always works towards the center of the circle.

Due to rotation, the tension is act in the chord which is equal to the centripetal force act the ball.

$\rm T = F_c= \frac{mv^2}{r}$

Tension in the cord = T= 75 N

Centripetal force = F_c

Mass of the ball= m= 0.55 Kg

Linear speed =v= ?

The radius of the circle.= r = 1.3 m

T is the maximum tension before the cord breaks. So according to the condition the obtained velocity will be ;

$\rm v = \sqrt{\frac{Tr}{m} } \\\\ \rm v = \sqrt{\frac{75\times1.3}{0.55} }\\\\ \rm v = 13.3 \;m/sec$

Hence the maximum value of speed will be 13.3 m/sec.

To learn more about the centripetal force refer to the link;

link

Answer from: bm42400

The tension of the cord is the centripetal force that keeps the ball in circular motion:
$T=F_c = m \frac{v^2}{r}$
where T is the tension of the cord, F_c

is the centripetal force, m=0.55 Kg

is the mass of the ball, v its speed and r=1.3 m

is the radius of the circle.

The maximum allowed tension is T=75 N, before the cord breaks. Using this value inside the formula, we can find which is the maximum allowed value fot the speed v:
$v= \sqrt{ \frac{T r}{m} }= \sqrt{ \frac{(75 N)(1.3 m)}{(0.55 kg)} }=13.3 m/s$

Answer from: Quest

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

$a = 32 ft/s^2$

so as we know that

$a = \frac{v^2}{r}$

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

$a_c = 32 ft/s^2$

$a_t = 0$

now radius of curvature is given as

$r = \frac{v^2}{a_c}$

$r = \frac{150^2}{32}$

$r = 703.125ft$

so the radius of curvature just after it is dropped is 703.125 ft

Answer from: Quest

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle

Another question on Physics

Physics, 22.06.2019 00:30

What is the acceleration of a spaceship that is initially at rest and then speeds up to 220 m/s in 20 seconds

Answers: 1

Answer

Physics, 22.06.2019 12:10

Awater slide of length l has a vertical drop of h. abby's mass is m. an average friction force of magnitude f opposes her motion. she starts down the slide at initial speed vi. use work-energy ideas to develop an expression for her speed at the bottom of the slide. then evaluate your result using unit analysis and limiting case analysis. express your answer in terms of the variables h, m, l, vi, f and appropriate constants. vf v f

Answers: 2

Answer

Physics, 22.06.2019 15:20

A20 cm long spring is attached to a wall. the spring stretches to a length of 22 cm when you pull on it with a force of 100 n. what is the spring constant

Answers: 3

Answer

Physics, 22.06.2019 16:30

Acalorimeter uses the principle that thermal energy flows from hotter material to colder material until both reach the same

Answers: 1

Answer

You know the right answer?

A0.55-kg ball, attached to the end of a horizontal cord, is revolved in a circle of radius 1.3 m on...

Questions