Which sentence best describes the axis of rotation of Earth?

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac

Sedbober [7]

Answer:

v = 1.2 m/s

Explanation:

The wavelength of the waves is given as the horizontal distance between the crests:

λ = wavelength = 5.5 m

Now, the time period is given as the time taken by boat to move from the highest point again to the highest point. So it will be equal to twice the time taken by the boat to travel from highest to the lowest point:

T = Time Period = 2(2.3 s) = 4.6 s

Now, the speed of the wave is given as:

$v = f\lambda$

where,

v= speed of wave = ?

f = frequency of wave = $\frac{1}{T} = \frac{1}{4.6\ s} = 0.217\ Hz$

Therefore,

$v = (0.217\ Hz)(5.5\ m)\\$

5 0

3 years ago

In moving out of a dormitory at the end of the semester, a student does 1.20 x 104 J of work. In the process, his internal energ

Zielflug [23.3K]

Answer:

(a) W=1.20×10⁴J

(b) U= -5.46×10⁴J

(c) Q= -4.26×10⁴J

Explanation:

Given that student does 1.20×10⁴J work

(a) W=1.20×10⁴J

Work done by student,so positive sign

During the process, his internal energy decreases by 5.46×10⁴J.

(b) U= -5.46×10⁴J.

As the Energy decreases therefore negative sign

For (c) Q

We know the formula

$Q=W+U\\Q=1.2*10^{4}+(-5.46*10^{4} )\\ Q=-4.26*10^{4}J$

8 0

3 years ago

A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm

Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

$F = - k * \Delta x (1)$

where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
Solving for k:

$k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)$

b)

Assuming no friction present, total mechanical energy mus keep constant.
When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

$U = \frac{1}{2}* k* (\Delta x)^{2} (3)$

Replacing k and Δx by their values, we get:

$U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)$

c)

When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
So, we can write the following expression:

$\frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2} = \frac{1}{2}*k*\Delta x^{2} (5)$