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Fynjy0 [20]
3 years ago
5

A 8.00-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical

oscillations having a period of 1.70 s. Find the force constant of the spring.
Physics
1 answer:
Alex_Xolod [135]3 years ago
8 0

Answer:

109.32 N/m

Explanation:

Given that

Mass of the hung object, m = 8 kg

Period of oscillation of object, T = 1.7 s

Force constant, k = ?

Recall that the period of oscillation of a Simple Harmonic Motion is given as

T = 2π √(m/k), where

T = period of oscillation

m = mass of object and

k = force constant if the spring

Since we are looking for the force constant, if we make "k" the subject of the formula, we have

k = 4π²m / T², now we go ahead to substitute our given values from the question

k = (4 * π² * 8) / 1.7²

k = 315.91 / 2.89

k = 109.32 N/m

Therefore, the force constant of the spring is 109.32 N/m

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lbvjy [14]

Answer:

their electrical potential energy decreases. True, because of the negative sign as the distance decreases, it becomes more negative.

The kinetic energy increases. True, the force is attractive as the distance decreases the force increases, so acceleration and therefore the speed

Explanation:

The only force between the proton and the electron is electric

      Fe = k q1q2 / r2

      Fe = - k e2 / r2

We can see that it is an attractive force (negative sign)

The electric power energy is

       U = k q1 q2 / r

       U = -k e2 / r

 

The kinetic energy is

       K = ½ m v2

With the expressions for each term we can analyze the sentences :

Their electric potential energy increases. False,

their electrical potential energy decreases. True, because of the negative sign as the distance decreases, it becomes more negative.

The kinetic energy increases. True, the force is attractive as the distance decreases the force increases, so acceleration and therefore the speed

Kinetic energy decreases False

The acceleration decreases.  False, as the force increases so does the acceleration

7 0
4 years ago
) a 1.0 kilogram laboratory cart moving with a velocity of 0.50 meter per second due east collides with and sticks to a similar
ZanzabumX [31]
Momentum would be the same before and after the collision
 Before the collision:
 Momentum of the single cart: 1 * 0.50 = 0.50
 After the collision
 velocity = 0.25m / s
 1 * 0.25 + 1 * 0.25 =
 0.25 * (1 + 1) =
 0.25 * 2 =
 0.50
 Now new momentum will be 0.5
 answer
 the same before and after the collision
4 0
4 years ago
Read 2 more answers
You lift a 10-N physics book up in the air a distance of 1.0 m, at a constant velocity
Anarel [89]

The book is lifted upward, but gravity points down, so the work done by gravity must be negative (so you can eliminate options 1 and 3).

The force exerted on the book by gravity has magnitude

<em>F</em> = <em>mg</em> = (10 N) (9.80 m/s^2) = 9.8 N ≈ 10 N

You raise the book 1.0 m in the opposite direction, so the work done is

<em>W</em> = (10 N) (-1.0 m) = -10 J

5 0
3 years ago
a pillow , a textbook and a paper airplane are dropped from the top of a tall building at the same time. consider what you have
MAVERICK [17]

A textbook would hit the ground first


Factors:

-Textbook weighs most

-Pillow is flat and fluffy not very aerodynamic) also is very light

-Paper airplane will glide to the ground do to its wings and will hit the ground last

3 0
3 years ago
Read 2 more answers
A uniform plank of length 5.0 m and weight 225 N rests horizontally on two supports, with 1.1 m of the plank hanging over the ri
lawyer [7]

Answer:

x = 0.6034 m

Explanation:

Given

L = 5 m

Wplank = 225 N

Wman = 522 N

d = 1.1 m

x = ?

We have to take sum of torques about the right support point.  If the board is just about to tip, the normal force from the left support will be going to zero.  So the only torques come from the weight of the plank and the weight of the man.

∑τ = 0  ⇒     τ₁ + τ₂ = 0  

Torque come from the weight of the plank = τ₁

Torque come from the weight of the man = τ₂

⇒  τ₁ = + (5 - 1.1)*(225/5)*((5 - 1.1)/2) - (1.1)*(225/5)*((1.1)/2) = 315 N-m (counterclockwise)

⇒  τ₂ = Wman*x = 522 N*x   (clockwise)

then

τ₁ + τ₂ = (315 N-m) + (- 522 N*x) = 0

⇒  x = 0.6034 m

7 0
3 years ago
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