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AveGali [126]
3 years ago
12

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?)
Physics
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

4 0
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:

brainly.com/question/6763771

brainly.com/question/6443626

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7 0
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.

5 0
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

3 0
3 years ago
Difference between static and kinetic friction in physics
KIM [24]
The force of static friction keeps a stationary object at rest. Once the force of static friction is overcome, the force of kinetic friction is what slows down a moving object.
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