When Little Albert was conditioned to be afraid of the white mouse, he also became afraid of stuffed animals,

A 0.150 kg ball on the end of a 1.10 m long cord (negligible mass)is swung in a vertical circle..

Aneli [31]

<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>

3 0

3 years ago

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What is Potentiometer

Katena32 [7]

Answer:

an instrument for measuring an electromotive force by balancing it against the potential difference produced by passing a known current through a known variable resistance.

3 0

3 years ago

A small 12.00 g plastic ball is suspended by a string in a uniform, horizontal electric field. If the ball is in equilibrium whe

notsponge [240]

Answer:

$Q = \frac{0.068}{E}$

where E = electric field intensity

Explanation:

As we know that plastic ball is suspended by a string which makes 30 degree angle with the vertical

So here force due to electrostatic force on the charged ball is in horizontal direction along the direction of electric field

while weight of the ball is vertically downwards

so here we have

$QE = F_x$

$mg = F_y$

since string makes 30 degree angle with the vertical so we will have

$tan\theta = \frac{F_x}{F_y}$

$tan30 = \frac{QE}{mg}$

$Q = \frac{mg}{E}tan30$

$Q = \frac{0.012\times 9.81}{E} tan30$

$Q = \frac{0.068}{E}$

where E = electric field intensity

5 0

3 years ago

A certain white dwarf star was once an average star like our Sun. But now it is in the last stage of its evolution and is the si

solmaris [256]

Answer:

4.384 * 10^13

Explanation:

Given the expression :

[(6.67 * 10^-11) * (1.99 * 10^30)] ÷ [(1.74*10^3)*(1.74*10^3)]

Applying the laws of indices

[(6.67 * 1.99) *10^(-11 + 30)] ÷ [(1.74 * 1.74) * 10^3+3]

13.2733 * 10^19 ÷ 3.0276 * 10^6

(13.2733 / 3.0276) * 10^(19 - 6)

4.3840996 * 10^13

= 4.384 * 10^13

6 0

3 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$