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motikmotik
3 years ago
14

One end of the rope is vibrated to produce a wave with a wavelength of .25 m. The Frequency

Physics
1 answer:
Sphinxa [80]3 years ago
7 0

Answer:

Wave speed = 1.25 m/s

Explanation:

Given that the

wavelength = 0.25 m.

Frequency F = 5.0 Hz

The speed of a wave is the product of wave frequency and the wavelength. That is

V = F λ

Where

V = wave speed (m/s)

λ = wavelength (m).

F = frequency (Hz).

Using the formula above gives:

V = 0.25 × 5

V = 1.25 m/s

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A 1kg cart slams into a stationary 1kg cart at 2 m/s. The carts stick together and move forward at a speed of 1 m/sl. Determine
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Answer:

No, it is not conserved

Explanation:

Let's calculate the total kinetic energy before the collision and compare it with the total kinetic energy after the collision.

The total kinetic energy before the collision is:

K_i = K_1 + K_2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2=\frac{1}{2}(1 kg)(2 m/s)^2+\frac{1}{2}(1 kg)(0)^2=2 J

where m1 = m2 = 1 kg are the masses of the two carts, v1=2 m/s is the speed of the first cart, and where v2=0 is the speed of the second cart, which is zero because it is stationary.

After the collision, the two carts stick together with same speed v=1 m/s; their total kinetic energy is

K_f = \frac{1}{2}(m_1+m_2)v^2=\frac{1}{2}(1 kg+1kg)(1 m/s)^2=1 J

So, we see that the kinetic energy was not conserved, because the initial kinetic energy was 2 J while the final kinetic energy is 1 J. This means that this is an inelastic collision, in which only the total momentum is conserved. This loss of kinetic energy does not violate the law of conservation of energy: in fact, the energy lost has simply been converted into another form of energy, such as heat, during the collision.

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A 2498 kg car is moving at 17.1 m/s slams on its brakes and slows to 2.6 m/s. What is the magnitude (absolute value) of the impu
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Answer:

<em>J=36221 Kg.m/s</em>

Explanation:

<u>Impulse-Momentum Theorem</u>

These two magnitudes are related in the following way. Suppose an object is moving at a certain speed v_1 and changes it to v_2. The impulse is numerically equivalent to the change of linear momentum. Let's recall the momentum is given by

p=mv

The initial and final momentums are, respectively

p_1=mv_1,\ p_2=mv_2

The change of momentum is

\Delta p=p_2-p_1=m(v_2-v_1)

It is numerically equal to the Impulse J

J=\Delta p

J=m(v_2-v_1)

We are given

m=2498\ kg,\ v_1=17.1\ m/s,\ v_2=2.6\ m/s

The impulse the car experiences during that time is

J=2498(2.6-17.1)=2498(-14.5)

J=-36221 Kg.m/s

The magnitude of J is

J=36221 Kg.m/s

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