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MaRussiya [10]
3 years ago
7

Of orbital radius, orbital velocity, orbital circumference, or radial area swept through, which of these stays constant accordin

g to one of Kepler’s Laws?
a. orbital distance
b. radial area swept through
c. none of these
d. orbital velocity
Physics
1 answer:
kow [346]3 years ago
5 0

Answer:

dyow

Explanation:

sorrrey

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What's 600,000,000 divided by 3,000.000,000,000?
Murljashka [212]

Answer:0.000002

Explanation: I Looked It Up lol

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Cora, an electrician, wraps a copper wire with a thick plastic coating. What is she most likely trying to do?
alukav5142 [94]
The correct answer among the choices given is the last option. Cora wrapping the copper wire with a thick plastic coating keeps a current from passing out a wire. The plastic wire here serves as an insulator. An insulator is a material that prevents electricity or current to flow out the circuit. In order to lessen the loss of energy.

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3 years ago
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During a hurricane, the atmospheric pressure inside a house may blow off the roof because of the reduced pressure outside. If ai
m_a_m_a [10]

Answer:

817.5 Pa

Explanation:

From Bernoulli's equation, considering thst there is no height difference then

P1+½d(v1)²=P2+½d(v2)²

P1-P2=½d(v2²-v1²)

∆P=½d(v2²-v1²)

Where P represent pressure, d is density and v is velocity. Subscripts 1 and 2 represent inside and outside. ∆P is tge change in pressure

Given the speed at roof top as 128 km/h, we convert it to m/s as follows

128*1000/3600=35.555555555555=35.56 m/s

Velocity at the bottom of roof is 0 m/s

Density is given as 1.293 kg/m³

∆P=½*1.293*(35.56²-0)=817.5 Pa

5 0
3 years ago
A sphere of mass m and radius r is released from rest at the top of a curved track of height H. The sphere travels down the curv
iren2701 [21]

Explanation:

<em>(a) On the dots below, which represent the sphere, draw and label the forces (not components) that are exerted on the sphere at point A and at point B, respectively.  Each force must be represented by a distinct arrow starting on and pointing away from the dot.</em>

At point A, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing left, and static friction force Fs pushing down.

At point B, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing down, and static friction force Fs pushing right.

<em>(b) i. Derive an expression for the speed of the sphere at point A.</em>

Energy is conserved:

PE = PE + KE + RE

mgH = mgR + ½mv² + ½Iω²

mgH = mgR + ½mv² + ½(⅖mr²)(v/r)²

mgH = mgR + ½mv² + ⅕mv²

gH = gR + ⁷/₁₀ v²

v² = 10g(H−R)/7

v = √(10g(H−R)/7)

<em>ii. Derive an expression for the normal force the track exerts on the sphere at point A.</em>

Sum of forces in the radial (-x) direction:

∑F = ma

N = mv²/R

N = m (10g(H−R)/7) / R

N = 10mg(H−R)/(7R)

<em>(c) Calculate the ratio of the rotational kinetic energy to the translational kinetic energy of the sphere at point A.</em>

RE / KE

= (½Iω²) / (½mv²)

= ½(⅖mr²)(v/r)² / (½mv²)

= (⅕mv²) / (½mv²)

= ⅕ / ½

= ⅖

<em>(d) The minimum release height necessary for the sphere to travel around the loop and not lose contact with the loop at point B is Hmin.  The sphere is replaced with a hoop of the same mass and radius.  Will the value of Hmin increase, decrease, or stay the same?  Justify your answer.</em>

When the sphere or hoop just begins to lose contact with the loop at point B, the normal force is 0.  Sum of forces in the radial (-y) direction:

∑F = ma

mg = mv²/R

gR = v²

Applying conservation of energy:

PE = PE + KE + RE

mgH = mg(2R) + ½mv² + ½Iω²

mgH = 2mgR + ½mv² + ½(kmr²)(v/r)²

mgH = 2mgR + ½mv² + ½kmv²

gH = 2gR + ½v² + ½kv²

gH = 2gR + ½v² (1 + k)

Substituting for v²:

gH = 2gR + ½(gR) (1 + k)

H = 2R + ½R (1 + k)

H = ½R (4 + 1 + k)

H = ½R (5 + k)

For a sphere, k = 2/5.  For a hoop, k = 1.  As k increases, H increases.

<em>(e) The sphere is again released from a known height H and eventually leaves the track at point C, which is a height R above the bottom of the loop, as shown in the figure above.  The track makes an angle of θ above the horizontal at point C.  Express your answer in part (e) in terms of m, r, H, R, θ, and physical constants, as appropriate.  Calculate the maximum height above the bottom of the loop that the sphere will reach.</em>

C is at the same height as A, so we can use our answer from part (b) to write an equation for the initial velocity at C.

v₀ = √(10g(H−R)/7)

The vertical component of this initial velocity is v₀ sin θ.  At the maximum height, the vertical velocity is 0.  During this time, the sphere is in free fall.  The maximum height reached is therefore:

v² = v₀² + 2aΔx

0² = (√(10g(H−R)/7) sin θ)² + 2(-g)(h − R)

0 = 10g(H−R)/7 sin²θ − 2g(h − R)

2g(h − R) = 10g(H−R)/7 sin²θ

h − R = 5(H−R)/7 sin²θ

h = R + ⁵/₇(H−R)sin²θ

4 0
3 years ago
State the relative position for the sun and earth during the lunar eclipse (ASAP)
Mademuasel [1]

Answer:

Solar eclipses result from the Moon blocking the Sun relative to the Earth; thus Earth, Moon and Sun all lie on a line. Lunar eclipses work the same way in a different order: Moon, Earth and Sun all on a line. In this case the Earth's shadow hides the Moon from view.

8 0
3 years ago
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