Can someone please help solve this

an object of mass 1 g is hung from a spring and set in oscillatory motion .At t=0 the displacement is 43.75cm and the accelerati

charle [14.2K]

The spring constant is $4.0\cdot 10^{-5} N/m$

Explanation:

For an object in a simple harmonic motion, the acceleration of the object is related to the displacement by

$a=-\omega^2 x$

where

a is the acceleration

$\omega$ is the angular frequency

x is the displacement

The angular frequency is defined as

$\omega=\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

Substituting the second equation into the first one, we get

$a=-\frac{k}{m}x$

In this problem we have

m = 1 g = 0.001 kg

And at t=0,

x = 43.75 cm

a = -1.754 cm/s

Therefore, we can re-arrange the equation above to find the spring constant:

$k=-\frac{ma}{x}=-\frac{(0.001)(-1.754)}{43.75}=4.0\cdot 10^{-5} N/m$

#LearnwithBrainly

6 0

3 years ago

a baseball is hit straight up into the air. If the initial velocity was 20 m/s, how high will the ball go?How long will it be un

AleksAgata [21]

1) The motion of the ball is an uniformly accelerated motion, with constant acceleration equal to $g=-10 m/s^2$ (the negative sign means it is directed towards the ground).

For an uniformly accelerated motion, we can use the following relationship:
$2gh=v_f^2-v_i^2$
where h is the maximum height reached by the ball, $v_i = 20 m/s$ is its initial velocity and $v_f$ the velocity of the ball when it reaches the maximum height. But $v_f=0$ (when the ball reaches the maximum height, it stops before going down, so its velocity at that moment is zero), so we can use the relationship to calculate h, the maximum height:
$h=- \frac{v_i^2}{2g} =- \frac{20 m/s)^2}{2 \cdot (-10 m/s^2)} =20 m$

2) We can find the time the ball takes to return to the ground by requiring that the space covered by the ball returns to zero:
$S(t)=0$
where for an uniformly accelerated motion,
$S(t)=v_i t + \frac{1}{2} gt^2 =0$
By solving this, we have two solutions: one is t=0, which corresponds to the moment the player hits the ball, the second one is
$t=- \frac{2 v_i}{g}=- \frac{2 \cdot 20 m/s}{-10 m/s^2}=4 s$
so, the ball returns to the ground after 4 s.

3) The velocity of the ball when it returns to the ground is given by:
$v(t) = v_i +gt=20 m/s + (-10 m/s^2)(4 s)=-20 m/s$
where the negative sign means the ball is going downwards.

8 0

3 years ago

In the modern atomic model the number of ________ in a neutral atom is the same as the number of protons.

MrRissso [65]

In the modern atomic model the number of electrons in a neutral atom is the same as the number of protons.

An atom contain, three fundamental particles, electron, proton, and neutron, neutron is charge less, there the charge of the atom is determined by the number of proton, and number of electron, Atoms are electrically neutral, it is only possible when number of electrons in an atom is equal to the number of proton.

3 0

3 years ago

If you lose control of your vehicle and collide with a fixed object, such as a tree, at 60 m.p.h., the force of impact is the sa

My name is Ann [436]

You can compare the velocity of the car, 60 mph, with the velocity that a mass would acquire when falls from certain height.

First, convert 60 mph to m/s:

60 miles/h * 1.60 km/mile * 1000 m/km * 1h/3600s = 26.67 m/s

Second, calculate from what height a body in free fall reachs 26.67 m/s velocity when hits the floor.

free fall => Vf^2 = 2g*H => H = Vf^2 / (2g)

H = (26.67m/s)^2 / (2*9.8 m/s) = 36.2 m

If you consider that the height between the floors of a building is approximately 3.6 m, you get 36.2 m / 3.6 m/floor = 10 floors.

Then, you conclude that the force of impact is the same as driving you vehicle off a 10 story building.

7 0

4 years ago

The electromagnetic interaction _______.A. applies only to charges at rest B. applies only to charges in motion C. is responsibl

Bond [772]

Answer:

C. is responsible for sliding friction and contact forces

8 0

4 years ago