Which of the following rivers flows north in eastern Florida?

How does electroscope work ?

s2008m [1.1K]

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6 0

2 years ago

Supposing you were in space in a weightless environment, would it require a force to set an object in motion

Jobisdone [24]

Yes. If your smartphone was floating in front of your face, motionless
relative to you, it would require a force to start it moving toward you or
away from you.

But there's no minimum force required. ANY force, no matter how small,
even smaller than the smallest force that you can imagine, would set it in
motion.

The thing is, though, that the smaller the force acting on it, the smaller
acceleration it would get, and the slower it would move away from where
it is.

So if, say, you wanted to send it across the crew compartment and over
to your sleeping bag on the wall, and you had all day to watch it mope
along over there, you might breathe on it, and the force of your breath
would set it in slow motion in that direction. But if you wanted to throw it
at your crewmate, you'd need to give it more force.

7 0

2 years ago

An archer standing on a 15 degree slope shoots an arrow at an angle of 26 degrees above the horizontal. How far below its origin

aivan3 [116]

Answer:

The arrow will hit 112.07 m from the point of release.

Explanation:

The equation for the position of an object in a parabolic movement is as follows:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

x0 = initial horizontal position

v0 = initial velocity

α = launching angle

y0 = initial vertical position

t = time

g = acceleration due to gravity

We know that at the final time the y-component of the vector "r" (see figure") is -1.88 m. The x-component of that vector will be the horizontal distance traveled by the arrow. Using the equation of the y-component of "r", we can obtain the final time and with that time we can calculate the value of the x-component (horizontal distance).

Then:

y = y0 + v0 · t · sin α + 1/2 · g · t²

Since the origin of the frame of reference is located at the point where the arrow is released, y0 = 0. Notice that the angle α = 26° + 15° = 41° ( see figure)

-1.88 m = 33 m/s · sin 41° · t - 1/2 · 9.8 m/s² · t² (g is downward)

0 = -4.9 m/s² · t² + 33 m/s · sin 41° · t + 1.88 m

Solving the quadratic equation:

t = 4.5 s ( the negative value is discarded)

Now, with this time we can calculate the horizontal distance:

x = x0 + v0 · t · cos α (x0 = 0, the same as y0)

x = 33 m/s · 4.5 s · cos 41° = 112.07 m

6 0

3 years ago

Help me please, it's so hard

fomenos

Answer:

4.58×10²³ atoms

5.94×10⁻²¹ J

1340 m/s

Explanation:

Use ideal gas law to find moles of gas.

PV = nRT

(1.266 atm × 101300 Pa/atm) (4/3 π (0.15 m)³) = n (8.31451 J/mol/K) (14 + 273) K

n = 0.760 mol

Use Avogadro's number to find number of atoms.

(0.760 mol) (6.02214×10²³ atom/mol) = 4.58×10²³ atoms

Average kinetic energy per molecule is:

KE = 3/2 kT

KE = 3/2 (1.38066×10⁻²³ J/K) (14 + 273) K

KE = 5.94×10⁻²¹ J

RMS speed of each atom is:

KE = 1/2 mv²

5.94×10⁻²¹ J/atom = 1/2 (0.004 kg/mol) (1 mol / 6.02214×10²³ atom) v²

v = 1340 m/s

8 0

3 years ago

Read 2 more answers

A car moves in a straight line such that for a

jeyben [28]

Answer:

s = 90 m

a = 56 m/s²

Explanation:

I will ASSUME that your equation is silly as it reduces to V = 11t which is constant, and that you mean V = 9t² + 2t

Position is the integral of differential velocities

s = $\int\limits^3_0 {9t^2 + 2t} \, dt$

s = 3t³ + t² | from 0 to 3

s = 3(3)³ + 3² - (0) = 90 m

acceleration is the derivative of velocity

a = v' = 18t + 2

a(3) = 18(3) + 2 = 56 m/s²

3 0

3 years ago