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Ede4ka [16]
3 years ago
10

A mass on a string of unknown length oscillates as a pendulum with a period of 4.00 s. What is the period if a. The mass is doub

led? b. The string length is doubled? c. The string length is halved? d. The amplitude is halved? Parts a to d are independent questions, each referring to the initial situation.?

Physics
2 answers:
Blizzard [7]3 years ago
7 0

Answer:

a) T= 4.00s

b)T = 5.66s

c)T  = 2.83s

d)T = 4.00s

Explanation:

a)  The pendulum period is express as

T_0 = 2\pi \sqrt{\frac{L_0}{g} }  = 4.00s

The period does not depend on the mass and depend only on the length

so we have,

T = T_0 = 4.00s

b) for new length

L = 2L_0

so, we have

T_0 = 2\pi \sqrt{\frac{L_0}{g} }  = \sqrt{2T_0}  = 5.66s

c) For a new length

T_0 = 2\pi \sqrt{\frac{L_0}{g} }  = \frac{1}{\sqrt{2} } T_0= 2.83s

d) the period does not depend on the amplitude as long as there is simple harmonic motion

so, we have

T = 4.00s

Alex17521 [72]3 years ago
3 0

Explanation:

Below are attachments containing the solution.

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A 2.3 kg cart is rolling across a frictionless, horizontal track towards a 1.5 kg cart that is initially held at rest. The carts
Inga [223]

Answer:

total momentum = 8.42 kgm/s

velocity of the first cart is 3.660 m/s

Explanation:

Given data

mass m1 = 2.3 kg

mass m2 = 1.5 kg

final velocity V2 = 4.9 m/s

final velocity V3 = - 1.9 m/s

to find out

total momentum  and velocity of the first cart

solution

we know mass and final velocty

and initial velocity of second cart V1 = 0

so now we can calculate total momentum that is m1 v2 + m2 v2

total momentum =  2.3 ×4.9 + 1.5 ×(-1.9)

total momentum = 8.42 kgm/s

and

conservation of momentum  is

m1 V + m2 v1  = m1 v2  + m2 v3

put all value and find V

2.3 V + 1.5 ( 0) = 2.3 ( 4.9 ) + 1.5 ( -1.9)

V = 8.42 / 2.3

V = 3.660 m/s

so velocity of the first cart is 3.660 m/s

8 0
4 years ago
A roadway for stunt drivers is designed for racecars moving at a speed of 40 m/s. A curved section of the roadway is a circular
Fynjy0 [20]

Answer:

Bank angle = 35.34o

Explanation:

Since the road is frictionless,

Tan (bank angle) = V^2/r*g

Where V = speed of the racing car in m/s, r = radius of the arc in metres and g = acceleration due to gravity in m/s^2

Tan ( bank angle) = 40^2/(230*9.81)

Tan (bank angle) = 0.7091

Bank angle = tan inverse (0.7091)

Bank angle = 35.34o

3 0
3 years ago
A billiard ball is moving in the x-direction at 30.0 cm/s and strikes another billiard ball moving in the y-direction at 40.0 cm
PSYCHO15rus [73]

Explanation:

Given that,

Initial speed of the billiard ball 1, u = 30i cm/s

Initial speed of another billiard ball 2, u' = 40j cm/s

After the collision,

Final speed of first ball, v = 50 cm/s

Final speed of second ball, v' = 0 (as it stops)

Let us consider that both balls have same mass i.e. m

Initial kinetic energy of the system is :

K_i=\dfrac{1}{2}mu^2+\dfrac{1}{2}mu'^2\\\\K_i=\dfrac{1}{2}m(u^2+u'^2)\\\\K_i=\dfrac{1}{2}m((30)^2+(40)^2)\\\\K_i=1250m\ J

Final kinetic energy of the system is :

K_f=\dfrac{1}{2}mv^2+\dfrac{1}{2}mv'^2\\\\K_f=\dfrac{1}{2}m(v^2+v'^2)\\\\K_f=\dfrac{1}{2}m((50)^2+(0)^2)\\\\K_f=1250m\ J

The change in kinetic energy of the system is equal to the difference of final and initial kinetic energy as :

\Delta K=K_f-K_i\\\\\Delta K=1250m-1250m\\\\\Delta K=0

So, the change in kinetic energy of the system as a result of the collision is equal to 0.    

7 0
4 years ago
The lens of the eye focuses light on the
pav-90 [236]

Answer

Cornea

Explanation:

The lens is composed of transparent, flexible tissue and is located directly behind the iris and the pupil. It is the second part of your eye, after the cornea, that helps to focus light and images on your retina.

4 0
4 years ago
Read 2 more answers
A person pushes a block 4 m with a force of 25 N. How much work is being done?
Elina [12.6K]

Answer:

<h2>100 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 25 × 4

We have the final answer as

<h3>100 J</h3>

Hope this helps you

3 0
3 years ago
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