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UkoKoshka [18]
3 years ago
14

A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.

Physics
1 answer:
fredd [130]3 years ago
6 0

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

F=\dfrac{mv^2}{r}-mg

F=m(\dfrac{v^2}{r}-g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

F=\dfrac{mv^2}{r}+mg

F=m(\dfrac{v^2}{r}+g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}

T=\dfrac{m}{r}(g(r-h)+v^2)

Since, v^2=u^2-2gh

T=\dfrac{m}{r}(u^2-3gh+gr)

Hence, this is the required solution.

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When a body of mass 0.25 kg is attached to a vertical massless spring, it is extended 5.0 cm from its unstretched length of 4.0
lora16 [44]

Answer:

d=0.165m

Explanation:

Given

m=0.25kg,x_{1}=5cm*\frac{1m}{100cm}=0.05m,x_{2}=4cm*\frac{1m}{100cm}=0.04m,v=2\frac{rev}{s}

The tension of the spring is

F_{k}=K*x_{1}=m*g

K=\frac{m*g}{x_{1}}

K=\frac{0.25kg*9.8m/s^2}{0.05m}=49N/m

The force in the spring is equal to centripetal force so

F_{c}=\frac{m*v^2}{r}

v=w*r=2\pi*r

But Fc is also

Fc=KxΔr

F_{c}=K*(r-x_{2})

Replacing

m*4\pi^2*r=K*(r-x_{2})

0.25kg*4\pi^2*r=49*(r-0.04m)

r=0.205m

total distance is

d=0.205-0.04=0.165m

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A football punker attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in the a
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v_x = \frac{67}{4.5}

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