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3 years ago

In the oscillating spring ball system, where is the velocity of the ball the greatest?

Physics

1 answer:

nlexa [21]3 years ago

6 0

The speed of an object in a mass-spring system is given under the function

$v = \pm \sqrt{\frac{k}{m}(A^2-x^2)}$

Here,

m = mass

k = Spring constant

A = Amplitude

x = Position

When the position is at the equilibrium point (x = 0), the speed will be maximum, and could even be expressed as

$v_{max}= A\sqrt{\frac{x}{m}}$

So the correct answer is B.

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Jack and Jill are maneuvering a 3300 kg boat near a dock. Initially the boat's position is < 2, 0, 3 > m and its speed is

Paraphin [41]

Answer:

The workdone by Jack is $W_{jack} = -1050J$

The workdone by Jill is $W_{Jill} = 0J$

The final velocity is $v = 1.36 m/s$

Explanation:

From the question we are given that

The mass of the boat is $m_b = 3300kg$

The initial position of the boat is $P_i = (2 \r i + 0 \r j + 3\r k)m$

The Final position of the boat is $P_f = (4\r i + 0 \r j + 2\r k )\ m$

The Force exerted by Jack $\r F = (-420\r i + 0 \r j + 210\r k) \ N$

The Force exerted by Jill $\r F_{Jill} =(180 \r i + 0\r j + 360\r k)$

Now to obtain the displacement made we are to subtract the final position from the initial position

$\r P = P_f - P_i$

$= (4\r i + 0\r j + 2 \r k) - (2\r i + 0\r j + 3\r k )$

$= (2\r i + 0\r j -\r k )m$

Now that we have obtained the displacement we can obtain the Workdone

which is mathematically represented as

$W =\r F * \r P$

The amount of workdone by jack would be

$W_{jack} =\r F * \r P$

$= [(-420\r i +0\r j +210\r k)(2\r i + 0\r j - \r k)]$

$= (-420) (2) + (210)(-1)$

$= -840 - 210$

$=-1050J$

The amount of workdone by Jill would be

$W_{Jill} =\r F * \r P$

$= [(180 \r i + 0\r j + 360\r k)(2\r i +0\r j -\r k)]$

$= (180 )(2) +(360)(-1)$

$=0J$

According to work energy theorem the Workdone is equal to the kinetic energy of the boat

$W = K.E = \frac{1}{2} m *[v^2 - (1.1)^2]$

$-1050 = 0.5*3300 [*v^2- (1.1)^2]$

$-1050 = 1650 [v^2 -1.21]$

$0.6363 = v^2 -1.21$

$v^2 = 0.6363+1.21$

$v^2 =1.846$

$v = 1.36\ m/s$

6 0

4 years ago

The general electron configuration for atoms of the halogen group is

m_a_m_a [10]

Answer:

$ns^2np^5$

Explanation:

Halogen group:This group contains Cl,F,Br and I

There are 7 electrons in the outermost shell of each element of halogen family.

Each element of this family required one electron to fill their complete outermost shell.

We know that

$F=9=1s^22s^22p^5$

Cl=17= $1s^22s^22p^63s^23p^5$

Therefore, the general configuration for atoms of the halogen group is given by

$ns^2np^5$

8 0

4 years ago

In order to make an electric current flow it is necessary to have

ASHA 777 [7]

In order to make an electric current flow it is necessary to have a source of the current or source of free electrons and also the circuit should be a close one or complete. When there is a break in a circuit, no current would flow in the whole system.

8 0

3 years ago

Read 2 more answers

a dragster gets to a peed of 112m/s over a distance of 300m from rest. Determine the acceleration of the dragster.

madreJ [45]

A dragster gets to a speed of 112 m/s over a distance of 300 m from the rest, then the acceleration will be 20.9 m/s².

<h3>What is Acceleration?</h3>

The rate of change in an object's velocity with respect to time is known as acceleration in mechanics. The vector quantity of accelerations. The direction of the net force that is acting on an object determines its acceleration.

Since acceleration has both a magnitude and a direction, it is a vector quantity. Velocity is a vector quantity as well. The definition of acceleration is the change in velocity vector over a time interval divided by the time interval.

According to the question, the given values are :

Final velocity, v = 112 m/s.

Initial velocity, u = 0 m/s

Distance, s = 300 m.

Use the equation of motion,

v² - u² = 2as

112 ² - 0 = 2 (a)(300)

a = 12544/600

a = 20.9 m/s².

Hence, the acceleration of the dragster will be 20.9 m/s².

To get more information about Acceleration :

brainly.com/question/20382454

#SPJ1

3 0

1 year ago

10 The magnitude J of the current density in a certain lab wire with a circular cross section of radius R = 2.00 mm is given by

rewona [7]

Answer:

I = 0.002593 A = 2.593 mA

Explanation:

Current density = J = (3.00 × 10⁸)r² = Br²

B = (3.00 × 10⁸) (for ease of calculations)

The current through outer section is given by

I = ∫ J dA

The elemental Area for the wire,

dA = 2πr dr

I = ∫ Br² (2πr dr)

I = ∫ 2Bπ r³ dr

I = 2Bπ ∫ r³ dr

I = 2Bπ [r⁴/4] (evaluating this integral from r = 0.900R to r = R]

I = (Bπ/2) [R⁴ - (0.9R)⁴]

I = (Bπ/2) [R⁴ - 0.6561R⁴]

I = (Bπ/2) (0.3439R⁴)

I = (Bπ) (0.17195R⁴)

Recall B = (3.00 × 10⁸)

R = 2.00 mm = 0.002 m

I = (3.00 × 10⁸ × π) [0.17195 × (0.002⁴)]

I = 0.0025929449 A = 0.002593 A = 2.593 mA

Hope this Helps!!!

4 0

3 years ago