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castortr0y [4]
4 years ago
15

A rope swing is hung from a tree right at the edge of a small creek. The rope is 5.0 m long; the creek is 3.0 m wide.

Physics
2 answers:
irina [24]4 years ago
8 0

Answer:

1) T = 4.5 s

2) T = 4.5 s

3) v = 9.9 m/s

Explanation:

We can use the equation

T = 2π√(L/g)

1) T = 2π√(5m/9.81 m/s²) = 4.5 s

2) T = 2π√(L/g)

T = 2π√(5m/9.81 m/s²) = 4.5 s

3) v = √(2gR)

v = √(2(9.81 m/s²)(5m))

v = 9.9 m/s

Flauer [41]4 years ago
8 0

Answer:

2.24

2.24

6.3 m/s

Explanation:

- The given problem can be modeled like a swinging pendulum.

- We will use already derived expressions for SHM of a simple pendulum, so for the time period we have the expression:

                                       T = 2*pi * sqrt ( L / g )

                                       T = 2*pi*sqrt (5.0/9.81)

                                       T = 4.4857 s

- The entire cycle takes 4.4857 s, to complete. However, we are asked to find the half cycle or that is the first time he reaches his starting point.

- Hence, 1/2 T                 t= 2.24 s

- We can see that the time period of the SHM is independent of the mass of the object, hence the answer to second question is also t = 2.24

c)

- In this case we will have to determine the angle that the rope makes with the vertical position:

- We are told its right across the creek, so the angle can be computed:

                                        cos(Q) = 3 / 5

- Now the maximum velocity of a pendulum is given by:

                                        v_max = sqrt(2*g*L*(1-cosQ))

- plug in the values:        v_max = sqrt(2*9.81*5*(1-3/5))

                                        v_max = 6.26 m/s

Hence option A is 6.3 m/s is correct.

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muminat

Answer:

here on the surface of moon the gravity decreases so the frequency of oscillation will also decrease

Explanation:

As we know that frequency of oscillation of the simple pendulum is given as

f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}

here we know

g = acceleration due to gravity at the surface of the planet

L = length of the pendulum

Now we know that the pendulum is on the surface of earth then gravity is given as

g = 9.81 m/s^2

now when same pendulum is taken to the surface of the moon then we have

g_{moon} = \frac{g}{6}

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8 0
3 years ago
Consider the following electron configurations to answer the question:
Anika [276]

Answer:

(ii) 1s2 2s2 2p6 3s2

Explanation:

Electron Affinity is the energy change that occur when an atom gains an electron.

                                  X₍₉₎ + e⁻  →  X⁻           ΔE = E<em>ea</em>

ΔE is change in energy

E<em>ea is electron affinity</em>

Often, electron affinity has negative energy values. The more negative the electron affinity, the easier it is to add an electron to a particular atom. Electron affinity increases across the period in the periodic table. However, there are few exceptions:

1. The electron affinities of group 18 (8A) elements are greater than zero. This is because the atom has a filled valence shell, an addition of electron causes the electron to move to a higher energy shell.

2. The electron affinities of group 2 (2A) elements are more positive because addition of an electron requires it to reside in the previously unoccupied p sub-shell.

3. The electron affinities of group 15 (5A) elements are more positive because addition of an electron requires it to be put in an already occupied orbital.

Applying these consideration to the elements given in the question:

(i) The sum of the electron in 1s²2s² 2p⁶ 3s¹ = 2+2+6+1 = 11 Sodium (Na)

(ii) The sum of the electron in 1s² 2s² 2p⁶ 3s² = 2+2+6+2 = 12 Magnesium (Mg)

(iii) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p¹ = 2+2+6+2+1 = 13 Aluminium (Al)

(iv) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p⁴ = 2+2+6+2+4 = 16 Sulfur (S)

(v) The sum of the electron in 1s² 2s² 2p⁶ 3s² 3p⁵ = 2+2+6+2+5 = 17 Chlorine (Cl)

The atom that is expected to have a positive electron affinity is Magnesium which is a group 2A element with electron configuration of 1s² 2s² 2p⁶ 3s².

4 0
4 years ago
Which of the following is an example of an insulator?
charle [14.2K]

Answer:

Hey mate

Answer is glass

7 0
3 years ago
Read 2 more answers
How would the terminal velocity of a piece of tissue paper compare to the terminal velocity of a rock?
matrenka [14]

Answer: Rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.  

Explanation: Terminal velocity is defined as the final velocity attained by an object falling under the gravity. At this moment weight is balanced by the air resistance or drag force and body falls with zero acceleration i.e. with a constant velocity.

Case 1: Terminal velocity of a piece of tissue paper.

The weight of tissue paper is very less and it experiences an air resistance while falling downward under the effect of gravity.

Downward gravitational force, F = mg

Upward air resistance or friction or drag force will be f_{1}

So, paper will attain terminal velocity when mg =  f_{1}

Case 2: Rock is very heavy and require larger air resistance to balance the weight of rock relative to the tissue paper case.

Downward force on rock, F = Mg

Drag force = f_{2}

Rock will attain terminal velocity when Mg = f_{2}

Mg > mg

so, f_{2} > f_{1}

And rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.  

5 0
4 years ago
What is the centripetal acceleration of a point on the perimeter of a bicycle wheel of diameter70 cm when the bike is moving 8.0
sergeinik [125]

Answer:

The acceleration of a point on the wheel is 11.43 m/s² acting radially inward.

Explanation:

The centripetal acceleration acts on a body when it is performing a circular motion.

Here, a point on the bicycle is performing circular motion as the rotation of the wheel produces a circular motion.

The centripetal acceleration of a point moving with a velocity  and at a distance of  from the axis of rotation is given as:

a = v2/r

Here, V = 8m/ s,r = 0.70 m

∴ a = 8/0.70 = 11.43m/ s2

Therefore, the acceleration of a point on the wheel is 11.43 m/s² acting radially inward.
Hope it Helps!
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4 0
2 years ago
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