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Gnoma [55]
3 years ago
10

A ball is whirled on the end of a string in a horizontal circle of radius R at speed v. By which one of the following means can

the centripetal acceleration of the ball be increased by a factor of 1.5 (or 3/2)?
Physics
1 answer:
Vera_Pavlovna [14]3 years ago
4 0

Explanation:

When an object moves in a circular path, it will have circular acceleration. Its magnitude of acceleration is given by :  

a=\omega^2R

Since, \omega=\dfrac{2\pi }{T}R

a=(\dfrac{2\pi}{T})^2R

T is the time period

R is the radius of the circular path

To increase the centripetal acceleration bu a factor of 1.5 or 3/2, radius of circle must be increase by a factor of 6 and T is increased by a factor of 2 such that,

R'=6R and T'=2T

So,

a'=(\dfrac{2\pi}{T'})^2R'

a'=(\dfrac{2\pi}{(2T)})^2(6R)

a'=\dfrac{6}{4}(\dfrac{2\pi}{T})^2R

a'=\dfrac{3}{2}(\dfrac{2\pi}{T})^2R

Hence, this is the required solution.

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A. Why are elements placed in the same group on the periodic table?
jekas [21]

Answer:

A) Elements are placed in teh same group because they have similar properties. B) Every element in a family of elements has similar atomic numbers as the others in its family. C) Alkali Metals

Explanation:

5 0
3 years ago
6. A .25 kg arrow with a velocity of 12 m/s to the west strikes and pierces the center of a 6.8 kg target. a. What is the final
Alenkasestr [34]

Answer:

(a) the final velocity of the combined mass is 9.43 m/s

(b) the decrease in kinetic energy during the collision is 386.1 J

Explanation:

Given;

mass of arrow, m₁ = 25 kg

initial velocity of arrow, u₁ = 12 m/s

mass of target, m₂ = 6.8 kg

initial velocity of the target, u₂ = 0

Part (a)

From the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = v(m₁+m₂)

where;

v is the final velocity of the combined mass

25 x 12 + 0 = v(25 + 6.8)

300 = v(31.8)

v = 300/31.8

v = 9.43 m/s

Part(b)

Kinetic Energy, K.E = ¹/₂mv²

Initial kinetic energy =  ¹/₂m₁u₁² + ¹/₂m₂u₂²  = ¹/₂ x 25 x (12)² + 0 = 1800 J

Final kinetic energy = ¹/₂m₁v² + ¹/₂m₂v² = ¹/₂v²(m₁ + m₂)

                                                               = ¹/₂ x (9.43)²(25+6.8)

                                                               = 1413.91 J

Decrease in kinetic energy = Initial K.E - Final K.E

Decrease in kinetic energy = 1800J - 1413.91 J = 386.1 J

                               

4 0
3 years ago
For work to be accomplished we must have?
KiRa [710]

Answer:

a force and a movement in the same direction as the force.

Explanation:

4 0
3 years ago
A spaceship far from all other objects uses its impulse power system to attain a speed of 104 m/s. The crew then shuts off the p
asambeis [7]

Answer:

Velocity remains the same at 104 m/s

Explanation:

According to Newton's 1st law of motion, an object subjected to no force or net force equal 0 would maintain its velocity. In our case the crew shuts off the power, spaceship is in space and far from all other objects (so no gravity whatsoever) would have no force acting on it. Therefore its velocity would stay the same at 104 m/s

4 0
3 years ago
a 5 charge is locataed 1.25 m to the left of a -3 charge. What is the magnitude and direction of the electrostatic force on the
Alika [10]

Answer:

The force is 86.5×10^9 N towards the negative charge (to the right)

Explanation:

The electrostatic force on the charges is given by Coulomb's law;

F= Kq1q2/r^2

This an inverse square law.

F= electrostatic force on the charges

K= constant of Coulomb's law

q1 and q2= magnitude of the charges

Since K= 9.0×10^9Nm^2C^2

F= 9.0×10^9 × 5 × 3/(1.25)^2 = 135×10^9/1.56

F= 86.5×10^9 N

The force is 86.5×10^9 N towards the negative charge.

5 0
3 years ago
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