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2 years ago

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A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. Its speed is 38 m/s when i

t reaches ground level. What was its speed when its height was half that of its starting point

1 answer:

inysia [295]2 years ago

5 0

Explanation:

look !

speed= 38m/s

start from rest= 0

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The preamble rock is filled with air spaces

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The equation below is for potassium oxide.

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The answer is C 4:1.

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3 years ago

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A sinusoidal electromagnetic wave is propagating in vacuum. At a given point P and at a particular time, the electric field is i

olasank [31]

Answer:

a

The direction of the wave propagation is the negative z -axis

b

The amplitude of electric and magnetic field are $A_E= 3.35*10^5 V/m$ ,

$A_M= 1.12 *10^{-3} T$ respectively

Explanation:

According to right hand rule, your finger (direction of electric field) would be pointing in the positive x-axis i.e towards your right let your palms be face toward the direction of the magnetic field i.e negative y-axis (toward the ground ) Then anywhere your thumb stretched out is facing is the direction of propagation of the wave here in this case is the negative z -axis

The Intensity of the wave is mathematically represented as

$I = \frac{1}{2} c \epsilon _O E_{rms}^2$

Given that $I = 7.43 \frac{kW}{cm^2} = 7.43 \frac{*10^3}{*10^-{4} }= 7.43*10^7 \frac{W}{m^2}$

Making $E_{rms}$ the subject we have

$E_{rms} = \sqrt{\frac{I}{0.5*c*\epsilon_o} }$

Substituting values as given on the question

$E_{rms} = \sqrt{\frac{7.43 *10^7[\frac{W}{m^2} ]}{0.5 * 3.08*10^8 *8.85*10^{-12}} }$

$= 2.37*10^5 \ V/m$

The amplitude of the electric field is mathematically represented as

$A_E = \sqrt{2} * E_{rms}$

$= \sqrt{2} * 2.37*10^5$

$A_E= 3.35*10^5 V/m$

The amplitude of the magnetic field is mathematically represented as

$A_M = \frac{A_E}{c}$

Substituting value

$A_M = \frac{3.35 *10^5}{3.0*10^8}$

$A_M= 1.12 *10^{-3} T$

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3 years ago

Some element can be either solid or liquid. At the melting point, the liquid has 8 × 10-22 J more enthalpy per atom than the sol

OlgaM077 [116]

Answer:

481.76 J/mol

133.33 K

Explanation:

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

Change in enthalpy is given by

$\Delta H=8\times 10^{-22}\times 6.022\times 10^{23}\\\Rightarrow \Delta H=481.76\ J/mol$

Entropy is given by

$\Delta S=6\times 10^{-24}\times 6.022\times 10^{23}\\\Rightarrow \Delta S=3.6132\ J/mol K$

Latent heat of fusion is given by

$L_f=\Delta H\\\Rightarrow L_f=481.76\ J/mol$

The latent heat of fusion is 481.76 J/mol

Melting point is given by

$T_m=\dfrac{L_f}{\Delta S}\\\Rightarrow T_m=\dfrac{8\times 10^{-22}\times 6.022\times 10^{23}}{6\times 10^{-24}\times 6.022\times 10^{23}}\\\Rightarrow T_f=133.33\ K$

Melting occurs at 133.33 K

3 0

2 years ago

A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She th

Andre45 [30]

The horizontal distance covered by the ball before hitting the water is 70.4 m

Explanation:

The motion of the ball is the motion of a projectile, so it consists of two independent motions:

A uniform motion along the horizontal (x) direction
A uniformly accelerated motion along the vertical (y) direction

We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:

$s=u_y t + \frac{1}{2}at^2$

where:

s = -16.5 m is the vertical displacement of the ball (it is negative because we take upward as positive direction)

$u_y$ is the initial vertical velocity of the ball, which is given by

$u_y = u sin \theta$

where

u = 23.5 m/s is the initial velocity

$\theta=33.5^{\circ}$ is the angle of projection

Substituting,

$u_y=(23.5)(sin 33.5^{\circ})=13.0 m/s$

$a=g=-9.8 m/s^2$ is the acceleration of gravity, downward

Substituting everything into the equation we get:

$-16.5=13.0t-4.9t^2\\4.9t^2-13.0t-16.5=0$

Solving the equation for t, we find the time of flight of the ball:

t = -0.94 s

t = 3.59 s

We ignore the 1st solution since it is negative, so the ball reaches the water after 3.59 seconds.

Now we analyze the horizontal motion of the ball. The horizontal velocity is constant and it is:

$v_x=u cos \theta=(23.5)(cos 33.5^{\circ})=19.6 m/s$

Therefore, the horizontal distance covered in a time t is

$d=v_x t$

And substituting t = 3.59 s, we find

$d=(19.6)(3.59)=70.4 m$

So, the horizontal distance covered by the ball before hitting the water is 70.4 m.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

4 0

3 years ago