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3 years ago

11

Which equation below represents a generic equation suggested by a graph showing a hyperbola?

1 answer:

ruslelena [56]3 years ago

8 0

Answer:

y = k/x

Explanation:

y = k/x is a graph of a hyperbola that has been rotated about the origin.

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An asteroid is discovered in a nearly circular orbit around the Sun, with an orbital radius that is 3.570 times Earth's. What is

inysia [295]

Explanation:

Given:

$r_a = 3.570R_E$

$R_E = 1.499×10^{11}\:\text{m}$

$M_S = 1.989×10^{30}\:\text{kg}$

$G = 6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2$

Let $m_s$ = mass of the asteroid and $r_a$ = orbital radius of the asteroid around the sun. The centripetal force $F_c$ is equal to the gravitational force $F_G:$

$F_c = F_G \Rightarrow m_a\dfrac{v_a^2}{r_a} = G\dfrac{m_aM_S}{r_a^2}$

or

$\dfrac{4\pi^2 r_a}{T^2} = G\dfrac{M_S}{r_a^2}$

where

$v = \dfrac{2\pi r_a}{T}$

with T = period of orbit. Rearranging the variables, we get

$T^2 = \dfrac{4\pi^2 r_a^3}{GM_S}$

Taking the square root,

$T = 2\pi \sqrt{\dfrac{r_a^3}{GM_S}}$

$\:\:\:\:=2\pi \sqrt{\dfrac{(3.57(1.499×10^{11}\:\text{m}))^3}{(6.674×10^{-11}\:\text{N-m}^2\text{/kg}^2)(1.989×10^{30}\:\text{kg})}}$

$\:\:\:\:= 2.13×10^8\:\text{s} = 6.75\:\text{years}$

3 0

3 years ago

A girl is running in a long distance race. As she runs, her respiration rate 10 points increases. Her body cells must process en

Alik [6]

A. Internal stimuli
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4 0

3 years ago

A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the

Anvisha [2.4K]

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

$U=\dfrac{1}{2}kx^2$ ...(I)

Using gravitational potential energy

$U' =mgh$ ....(II)

Using energy of conservation

$E_{i}=E_{f}$

$U_{i}+U'_{i}=U_{f}+U'_{f}$

$\dfrac{1}{2}kx^2+0=0+mgh$

$h=\dfrac{kx^2}{2mg}$

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

$h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}$

$h=11.653\ m$

Hence, The maximum height above the point of release is 11.653 m.

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3 years ago

Determine the answer to the equation 30 km/h × 17 h =

Jobisdone [24]

30 km/h * 17 h = 30*17 km/h *h
= 510 km

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3 years ago

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3 years ago