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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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An object accelerates from rest and travels 53 m west in 5.2 s. Determine the acceleration

zmey [24]

Answer:

20.4m/s²

Explanation:

Given parameters:

Initial velocity = 0m/s

Distance = 53m

Time = 5.2s

Unknown:

Acceleration = ?

Solution:

This is a linear motion and we use the right motion equation;

S = ut + $\frac{1}{2}$ at²

S is the distance

u is the initial velocity

a is the acceleration

t is the time

Insert the parameters and solve;

53 = (0x 5.2) + $\frac{1}{2}$ x a x 5.2

53 = 2.6a

a = $\frac{53}{2.6}$ = 20.4m/s²

6 0

3 years ago

1. The large wheels on the train have a mass of 0.5 kg, while the small wheels have a mass of 0.15 kg. The train body, and child

blsea [12.9K]

Answer: 60.975 Joules

KE = 1/2 mv^2

=> 1/2 * 40.65 * 3

=> 60.975 J

5 0

2 years ago

As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, afte

Klio2033 [76]

Answer:

$r^2 = \frac{ 2 k \ Ze^2}{ 2m}$

Explanation:

For this exercise we must use the principle of conservation of energy

starting point. The proton very far from the nucleus

Em₀ = K = ½ m v²

final point. The point where the proton is stopped (v = 0)

Em_f = U = q V

where the potential is

V = k Ze / r²

Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching

Energy is conserved

Em₀ = Em_f

½ m v² = e ( $k \frac{Ze}{r^2}$ )

$r^2 = \frac{ 2 k \ Ze^2}{ 2m}$

with this expression we can find the closest approach distance (r)

3 0

3 years ago

A pendulum of length L is suspended from the ceiling of an elevator. When the elevator is at rest the period of the pendulum is

Juliette [100K]

Answer:

Explanation:

When the pendulum falls freely the net acceleration due to gravity is zero.

As we know that the time period of simple pendulum is inversely proportional to the square root of acceleration due to gravity, thus the time period becomes infinity.

6 0

3 years ago

A house is lifted from its foundations onto a truck for relocation. The house is pulled upward by a net force of 2850 N. This fo

Gwar [14]

Answer:

m = 95000 kg

Explanation:

Given that,

Net force acting on the house, F = 2850 N

Initial speed, u = 0

Final speed, v = 15 cm/s = 0.15 m/s

We need to find the mass of the house. Let the mass be m. We know that the net force is given by :

F = ma

Where

a is the acceleration of the house.

So,

$F=m\dfrac{v-u}{t}\\\\m=\dfrac{Ft}{(v-u)}\\\\m=\dfrac{2850\times 5}{(0.15-0)}\\\\m=95000\ kg$

So, the mass of the house is equal to 95000 kg.

3 0

3 years ago