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Nikitich [7]
3 years ago
13

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.14 m. The mug

slides off the counter and strikes the floor 0.60 m from the base of the counter.
Physics
1 answer:
defon3 years ago
7 0
Say the horizontal component of the velocity is vx and the vertical is vy.
Initially at t=0 (as the mug leaves the bar) the components are v0x and v0y.
Obviously (I hope!) v0y = 0.

The equations for horizontal and vertical projectile motion (with the positive direction up) are

x = x0 + v0x t
y = y0 + v0y t - 1/2 g t^2 = y0 - 1/2 g t^2

Now choose the origin to be the end of the counter. x0=0 and y0=0. The equations simplify to

x = v0x t
y = - 1/2 g t^2

You know that x = 1.20m when y = -0.88m
From the y equation (and g=10 m/s^2) you can calculate the time that the mug hits the floor.
t = 0.420s
From the x equation we get the initial horizontal velocity
v0x = x/t = 1.2/0.42 = 2.86 m/s

(b) x-component of velocity is constant since there are no horizontal forces so vx = 2.86 m/s
y-component is given by v = u+at with u=0 and a=-g
vy = -gt = -4.2m/s

Now tan(angle) = vy/vx so angle = arctan(vy/vx)
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An artificial satellite in a low orbit circles the earth every 98 minutes. what is its angular speed in rad/s?
postnew [5]
To finish one orbit it will take 98 x 60 seconds. So; <span>(2 x pi)/(98 x 60) = 1.07 x 10^-3 rad/sec. </span><span>
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8 0
4 years ago
(Fill In The Blanks Plz)
Mice21 [21]

Answer:

No. of Molecules = 18.06 x 10²³ molecules

No. of Atoms of P = 3 atoms

No. of Atoms of Na = 9 atoms

Total No. of Atoms = 24 atoms

Explanation:

<u>FOR NUMBER OF MOLECULES</u>:

No.\ of\ Molecules = (No.\ of\ Moles)(Avigadro's\ Number)\\No.\ of\ Molecules = (3)(6.02\ x\ 10^{23}\ molecules})\\

<u>No. of Molecules = 18.06 x 10²³ molecules</u>

<u></u>

<u>FOR NUMBER OF P ATOMS</u>:

No.\ of\ Atoms\ of\ P = (No.\ of\ Moles)(N.\ of atoms of P)\\No.\ of\ Atoms\ of\ P = (3)(1\ atom})

<u>No. of Atoms of P = 3 atoms</u>

<u></u>

<u>FOR NUMBER OF Na ATOMS</u>:

No.\ of\ Atoms\ of\ Na = (No.\ of\ Moles)(N.\ of atoms of Na)\\No.\ of\ Atoms\ of\ Na = (3)(3\ atom})

<u>No. of Atoms of Na = 9 atoms</u>

<u></u>

<u>FOR TOTAL NUMBER OF ATOMS</u>:

Total\ No.\ of\ Atoms = (No.\ of\ Moles)(N.\ of\ atoms\ of\ Na + No.\ of\ atoms\ of\ P + No.\ of\ atoms\ of\ O)\\ Total\ No.\ of\ Atoms\ = (3)(3\ atoms + 1\ atom + 4\ atoms}) = (3)(8\ atoms)

<u>Total No. of Atoms = 24 atoms</u>

<u></u>

<u></u>

8 0
3 years ago
Which is the best estimated pressure, represented in units of GPa, of the lower mantle?
IrinaVladis [17]

Answer : The best pressure is 110 GPa of the lower mantle.

Explanation : We know that the pressure in the lower mantle starts at 24 GPa and reach a 136 GPa at the boundary.

So, the best pressure is 110 GPa because 220 GPa, a 305 GPa and 360 GPa are the higher value of  the maximum value 136 GPa of the lower mantle.

4 0
4 years ago
Read 2 more answers
Can someone help me with this please
andrezito [222]
Carbon: C, 12.011, 6, 12
Oxygen: O, 8, 8, 8, 16
Boron: B, 10.811, 5, 5, 11
3 0
3 years ago
An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electro
valkas [14]

Answer:

Final velocity of electron, v=6.45\times 10^5\ m/s    

Explanation:

It is given that,

Electric field, E = 1.55 N/C

Initial velocity at point A, u=4.52\times 10^5\ m/s

We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

v^2=u^2+2as........(1)

a is the acceleration, a=\dfrac{F}{m}

We know that electric force, F = qE

a=\dfrac{qE}{m}

Use above equation in equation (1) as:

v^2=u^2+\dfrac{2qEs}{m}

v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m

v = 647302.09 m/s

or

v=6.45\times 10^5\ m/s

So, the final velocity of the electron when it reaches point B is 6.45\times 10^5\ m/s. Hence, this is the required solution.

3 0
3 years ago
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