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3 years ago

The fact that an object placed in water displaces some of that water known as the "Ripple Effect"

Physics

2 answers:

Softa [21]3 years ago

8 0

Answer:

True

Hope this helps!

Explanation:

Igoryamba3 years ago

3 0

True, for sure. When anything hits a still body of water it, the water will move in all directions of the object hitting it.

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What is the slit spacing of a diffraction necessary for a 600 nm light to have a first order principal maximum at 25.0°?

Sauron [17]

Explanation:

Given that,

Wavelength of light, $\lambda=600\ nm=6\times 10^{-7}\ m$

Angle, $\theta=25^{\circ}$

We need to find the slit spacing for diffraction. For a diffraction, the first order principal maximum is given by :

$d\sin\theta=n\lambda$

n is 1 here

d is slit spacing

$d=\dfrac{\lambda}{\sin\theta}\\\\d=\dfrac{6\times 10^{-7}}{\sin(25)}\\\\d=1.41\times 10^{-6}\ m\\\\d=1.41\ \mu m$

So, the slit spacing is $1.41\ \mu m$ .

6 0

3 years ago

A child does 350J of work while pulling a box from the ground up to his tree house with a rope. The tree house is 5.2 m above th

Advocard [28]

I think the answer to that questions is B.

3 0

3 years ago

A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi

nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

$W = F * d$

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

$d=h*N$ , for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

$V_f^2-V_i^2 = 2a\Delta X$

Where,

$V_f =$ Final velocity

$V_i =$ Initial Velocity

a = Acceleration

$\Delta X =$ Displacement

PART A) For the particular case of work we know then that,

$W = F*d$

$W = m*g*(h*N)$

$W = 50*9.8*(0.3*30)$

$W = 4.41kJ$

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

$V_f^2-V_i^2 = 2a\Delta X$

Here,

$V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow$ 3 steps in one second

$v_f = 0$

Replacing,

$V_f^2-V_i^2 = 2a\Delta X$

$0-0.9^2=2a(30*0.3)$

Re-arrange for a,

$a = -\frac{0.9^2}{2*30*0.3}$

$a = -45*10^{-3}m/s^2$

At this point we can calculate the time, which is,

$t = \frac{\Delta V}{a}$

$t = \frac{0-0.9}{-45*10^{-3}}$

$t = 20s$

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

6 0

3 years ago

A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent

Julli [10]

Answer:

The kinetic energy of the merry-go-round is $\bf{475.47~J}$ .

Explanation:

Given:

Weight of the merry-go-round, $W_{g} = 826~N$

Radius of the merry-go-round, $r = 1.17~m$

the force on the merry-go-round, $F = 57.8~N$

Acceleration due to gravity, $g= 9.8~m.s^{-2}$

Time given, $t=3.47~s$

Mass of the merry-go-round is given by

$m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg$

Moment of inertial of the merry-go-round is given by

$I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}$

Torque on the merry-go-round is given by

$\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m$

The angular acceleration is given by

$\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}$

The angular velocity is given by

$\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}$

The kinetic energy of the merry-go-round is given by

$E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J$

5 0

3 years ago

20. A mass of gas has a volume of 4 m3, a temperature of 290 K, and an absolute pressure of 475 kPa. When the gas is allowed to

aliina [53]

$Given:\\V_1=4m^3\\T_1=290K\\p_1=475kPa\\\\V_2=6.5m^3\\T_2=277K\\\\Find:\\p_2=?\\\\Solution:\\\\ \frac{pV}{T} =const.\\\\ \frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \\\\\frac{p_1V_1T_2}{T_1}=p_2V_2\\\\\frac{p_1V_1T_2}{T_1V_2}=p_2\\\\p_2=p_1 \frac{V_1}{V_2} \frac{T_2}{T_1} \\\\\\p_2=475kPa\cdot \frac{4m^3}{6.5m^3} \cdot \frac{277K}{290K} \approx 279.2kPa\\\\Correct\;is\;answer\;\;C$

7 0

2 years ago