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AURORKA [14]
3 years ago
12

A sailor drops a wrench from the top of a sailboat's vertical mast while the boat is moving rapidly and steadily straight forwar

d. Where will the wrench hit the deck?
(A) ahead of the base of the mast
(B) at the base of the mast
(C)behind the base of the mast
(D)on the windward side of the base of the mast
(E)None of the above choices
Physics
1 answer:
Aleks04 [339]3 years ago
6 0

Answer:

B

Explanation:

The sailor, the boat and the wrench are all moving at he same constant rate, so the wrench will appear to fall straight down. This due to that fact there is no relative motion among them and all are at rest w.r.t to one another. Hence the correct answer would be B.

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Which of the following statements about X-rays and radio waves is not true? Which of the following statements about X-rays and r
Lapatulllka [165]

Answer:

X-rays travel through space faster than radio waves.

Explanation:

Electromagnetic waves consist of oscillations of the electric and the magnetic field in a plane perpendicular to the direction of motion the wave.

All electromagnetic waves travel in a vacuum always at the same speed, the speed of light, whose value is:

c=3.0\cdot 10^8 m/s

Electromagnetic waves are classified into 7 different types, according to their wavelength/frequency. From shortest to longest wavelength (and so, from highest to lowest frequency), we have:

Gamma rays

X rays

Ultraviolet

Visible light

Infrared radiation

Microwaves

Radio waves

Now we can analyze the 4 statements:

X-rays and radio waves are both forms of light, or electromagnetic radiation --> TRUE. They are both types of electromagnetic waves.

X-rays have higher frequency than radio waves. --> TRUE, as we can see from the table above.

X-rays have shorter wavelengths than radio waves. --> TRUE, as we can see from the table above.

X-rays travel through space faster than radio waves. --> FALSE: all electromagnetic waves travel in space at the same speed, the speed of light.

8 0
3 years ago
A ball is at rest at the top of a hill until a boy kicked it with his foot. What is the force that causes motion in this scenari
vazorg [7]
The boy’s foot causes the motion. His foot is the one that causes the ball to roll down the hill.
3 0
3 years ago
In what climates would you expect soil to form fastest? Why?
enyata [817]

Answer:

Soils develop faster in warm, moist climates and slowest in cold or arid ones. Rainfall is one of the most important climate factors in soil formation.

Please mark me brainliest!

Hope this helped God bless :)

5 0
3 years ago
A parked car begins to roll down a hill, what can you conclude from that observation?
Fynjy0 [20]

Answer:

its The rolling friction is greater than the force of the car’s weight against the hill.

and A force was required to start the car rolling.

Explanation:

3 0
3 years ago
Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin
tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of m_2 would be four times that of m_1.

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2.

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is T, then the angular velocity of the SHM would be

\displaystyle \omega = \frac{2\pi}{T}.

Assume that the mass starts with a zero displacement and a positive velocity. If A represent the amplitude of the SHM, then the displacement of the mass at time t would be:

\mathbf{x}(t) = A\sin(\omega\cdot t).

The velocity of the mass at time t would be:

\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t).

The acceleration of the mass at time t would be:

\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t).

Let m represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time t would be:

\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t),

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant k will be equal to:

\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}.

Since \displaystyle \omega = \frac{2\pi}{T}, it can be concluded that:

\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}.

For the first mass m_1, if the time period is T_1, then the spring constant would be:

\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2.

Similarly, for the second mass m_2, if the time period is T_2, then the spring constant would be:

\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2.

Since the two springs are the same, the two spring constants should be equal to each other. That is:

\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2.

Simplify to obtain:

\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2.

6 0
3 years ago
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