Spring stretches 14 cm when an object weighing 28 n is hung from it. what is the spring constant

In the picture below identify the type of wave in the top picture. *

Lerok [7]

Answer; transverse wave
Explanation;

In transverse waves the particles vibrate perpendicular to the wave's motion.

4 0

3 years ago

A 1.2 m long wave travels 11.2 m to a wall and back again in 4.0 s. What is

algol [13]

Answer:

f = 4.67 Hz

Explanation:

We can approximate the wave as a traveling wave, therefore the speed of the wave is constant

v = d / t

the total distance remember, in going to the wall and back is

d = 2 11.2 = 22.4 me

we substitute

v = 22.4 / 4.0

v = 5.6 m / s

now we can use the relationship between the speed of the wave, its wavelength and u frequency

v =λ f

f = v /λ

f = 5.6 / 1.2

f = 4.67 Hz

3 0

3 years ago

A sound wave can be considered as a displacement wave or a pressure wave? What phase difference exists between the displacement

netineya [11]

Where the displacement is a maximum the pressure is a minimum.
Where the displacement is zero, the pressure is a maximum.

3 0

4 years ago

a 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.88 m/s left. afterwards, the 282 kg car

Vaselesa [24]

The momentum of the 155 kg car afterwards is 469.7 kg m/s to the right

Explanation:

We can solve the problem by using the law of conservation of momentum: the total momentum of the system is conserved before and after the collision, so we can write:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 282 kg$ is the mass of the bumper car

$u_1 = 3.50 m/s$ is the initial velocity of the bumper car (we take the right as positive direction)

$v_1 = 0.800 m/s$ is the final velocity of the bumper car

$m_2 = 155 kg$ is the mass of the second bumper car

$u_2 = -1.88 m/s$ is the initial velocity of the second car (moving to the left)

$v_2$ is the final velocity of the second car

Solving for $v_2$ ,

$v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(282)(3.50)+(155)(-1.88)-(282)(0.800)}{155}=3.03 m/s$

where the positive sign means the direction is to the right.

And now we can find the momentum of the 155 kg afterwards, which is

$p_2 = m_2 v_2 = (155)(3.03)=469.7 kg m/s$ (to the right)

Learn more about momentum:

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#LearnwithBrainly

7 0

3 years ago

What is a magnetic field?

monitta

A magnetic field is charged electricity acting in the forced of magnatism

4 0

3 years ago