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3 years ago

54 km/h is equal to

Physics

1 answer:

pogonyaev3 years ago

7 0

$\Huge{\underline{\underline{\mathfrak{\green{AnsWeR:}}}}}$

1 km/h = 5/18 m/s

54 km/h = 54×5/18 m/s

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What would be the speed of an object just before hitting the ground if dropped 91.5 meters

Aleks04 [339]

Answer:

about 42.35 m/s

Explanation:

Use the equation for accelerated motion (g), and with zero initial velocity that doesn't include time:

$v_f^2=v_i^2+2\,a\,\Delta x$

which for our case would reduce to:

$v_f^2=v_i^2+2\,a\,\Delta x\\v_f^2=0+2\,9.8\,(91.5)\\v_f^2= 1793.4\\v_f=\sqrt{1793.4} \\v_f \approx 42.35$

then the velocity just before hitting would be about 42.35 m/s

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3 years ago

Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each othe

deff fn [24]

Answer:

$\theta_2 - \theta_1 = 156.93 degree$

Explanation:

As we know that the displacement of the particle from the mean position is 1/5 times of its amplitude

so we have

$y = A sin\omega t$

$y = \frac{A}{5}$

so now we have

$\frac{A}{5} = A sin\omega t$

now we have

$\theta_1 = 11.53 degree$

so the phase other particle in opposite direction is given as

$\theta_2 = 180 - 11.53 = 168.46 degree$

so we have phase difference given as

$\theta_2 - \theta_1 = 168.46 - 11.53$

$\theta_2 - \theta_1 = 156.93 degree$

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3 years ago

How are Babys Born?

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Answer:

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3 years ago

Read 2 more answers

You and a friend each hold a lump of wet clay. Each lump has a mass of 30 grams. You each toss your lump of clay into the air, w

Vesna [10]

Answer:

$\ \text{m/s}$

Explanation:

$u_1$ = Velocity of one lump = $3x+3y-3z$

$u_2$ = Velocity of the other lump = $-4x+0y-4z$

m = Mass of each lump = $30\ \text{g}$

The collision is perfectly inelastic as the lumps stick to each other so we have the relation

$mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=\ \text{m/s}$

The velocity of the stuck-together lump just after the collision is $\ \text{m/s}$ .

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3 years ago

How do you solving kinematic equations for horizontal projectiles?

daser333 [38]

See projectiles are very simple unless you understand its core concepts....projectile is nothing just mixture of upward motion and horizontal motion....
THE KEY IS FORGET THE NAME PROJECTILE...ITS JUST HORIZONTAL MOTION + VERTICAL MOTION

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2 years ago