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3 years ago

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Say you want to make a sling by swinging a mass M of 1.9 kg in a horizontal circle of radius 0.042 m, using a string of length 0

.042 m. You wish the mass to have a kinetic energy of 15.0 Joules when released. How strong will the string need to be? (How much tension will there be in the string?)

1 answer:

padilas [110]3 years ago

3 0

Answer: T= 715 N

Explanation:

The only external force (neglecting gravity) acting on the swinging mass, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:

T = mv² / r

At the moment that the mass be released, it wil continue moving in a straight line at the same tangential speed that it had just an instant before, which is the same speed included in the centripetal force expression.

So the kinetic energy will be the following:

K = 1/2 m v² = 15. 0 J

Solving for v², and replacing in the expression for T:

T = 1.9 Kg (3.97)² m²/s² / 0.042 m = 715 N

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A ball is projected horizontally from the top of a cliff. At the same moment, a second identical ball is dropped from rest from

almond37 [142]

Answer:3

Explanation:

First ball is thrown with horizontal velocity while other ball is dropped from cliff such that both have zero vertical velocity. So both balls have to cover a distance equal to the height of cliff with same initial velocity.

time taken is given by $t=\sqrt{\frac{2h}{g}}$

where h=height of cliff

g=acceleration due to gravity

horizontal velocity to first ball will make the ball to travel more horizontal distance as compared to second ball.

Option 3 is correct

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3 years ago

A force of 350N is applied to a body. If the work done is 40kJ, what is the distance through which the body moved?

Studentka2010 [4]

The distance covered by the body is 114.3 m

Explanation:

The work done by a force exerted on an object is given by

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and of the displacement

For the object in this problem, we have

F = 350 N is the force applied

$W=40 kJ = 40,000 J$ is the work done

$\theta=0^{\circ}$ if we assume that the force is applied parallel to the motion of the object

Solving for d, we find the distance covered by the object:

$d=\frac{W}{F cos \theta}=\frac{40,000}{(350)(cos 0)}=114.3 m$

Learn more about work:

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3 years ago

Which list places the layers of the sun in the correct order from outermost to innermost?

Yuki888 [10]

Is D

is D because the inner layers are the Core, Radiative Zone and Convection Zone.

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3 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

a

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

b

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

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