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3 years ago

PLEASE ANSWER!!!!

Physics

1 answer:

const2013 [10]3 years ago

3 0

Answer:

B) 8.0 m

Explanation:

First of all, we can find the frequency of the wave, which is equal to the number of waves that pass a given point per second. Therefore:

$f=\frac{N}{t}=\frac{5.0}{4.0 s}=1.25 Hz$

which means 1.25 waves/second.

Then we can find the wavelength of the water waves, which is given by:

$\lambda=\frac{v}{f}$

where

v = 10.0 m/s is the speed of the wave

f = 1.25 Hz is the wave frequency

Substituting, we find

$\lambda=\frac{10.0 m/s}{1.25 Hz}=8.0 m$

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The average velocity of an object over 6.0 seconds interval is 2 m/s what is the total distance traveled and M by the object doi

lina2011 [118]

Answer:

<h2>The answer is 12 m</h2>

Explanation:

The distance covered by an object given it's velocity and time taken can be found by using the formula

distance = velocity × time

From the question we have

distance = 2 × 6

We have the final answer as

Hope this helps you

8 0

3 years ago

The relationship between mass and inertia is described by newton's second law of motion. true or false

Mariana [72]

False. Inertia and mass is not described in Newton’s second law of motion but in Newton’s first law of motion. Newton’s first law of motion or sometimes referred to as the law of inertia. In Newton’s first law indicates that an object at rest will remain at rest unless acted by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

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4 years ago

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$