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3 years ago

15

Suppose for an unexplained reason a star near Earth started shrinking. At what stage is the gravitational force between Earth an

d the star the greatest? Stage 1 Stage 2 Stage 3 Stage 4

2 answers:

tatiyna3 years ago

3 0

the answer is A- stage 1

DENIUS [597]3 years ago

3 0

stage 1 is the correct answer!

You might be interested in

Please help, I really don't know

zubka84 [21]

circular motion.

cent acc = r omega^2 ... omega is ang vel ... omega=2pi/T ,,,

9.8=rx(2pi/T)^2

if r is known, solve for T

7 0

3 years ago

Consider a small child sliding down two differently shaped slides that have the same coefficient of kinetic friction. How must t

timama [110]

Answer:

Having a bigger angle above the horizontal

Explanation:

Applying the energy conservation theorem:

$K_1+U_1+W_{ext}=K_2+U_2$

The kinetic energy is reduced because of the work done by the friction force.

The friction force is given by:

$F_f=F_N*\µ$

so the friction force depends on the Normal force, because the slide has an angle the normal force is given by:

$F_N=m.g*cos(\theta)$

So when the angle of the slide is bigger, the friction force decreases, for example:

for 45 degrees:

$F_N=m.g*cos(45)\\F_N=0.70(m.g)\\$

for 75 degrees:

$F_N=m.g*cos(75)\\F_N=0.26(m.g)\\$

as you can see if the angle is bigger above the horizontal, the friction force is reduced and so the work done by that force. We didn't have to change the height of the slide, so the potential gravitational energy remains the same.

6 0

4 years ago

A feather is dropped on the moon from a height of 1.40 meters. The acceleration due to gravity on the moon is 1.67 m/s2. Determi

RoseWind [281]

Distance = (1/2) (acceleration) (time)²

1.4m = (0.835 m/s²) (time)²

(time)² = (1.4/0.835) s²

<em>time = 1.295 s</em>

7 0

3 years ago

If a small rock is dropped from a height of 3.1 m how fast will it be moving when it reaches the ground 0.80 seconds later

garik1379 [7]

X=1/2 at^2
3.1=1/2 a *0.64
a=9.68
v=at
v=0.8*9.6875=7.75

5 0

3 years ago

An electron emitted in the beta decay of bismuth-210 has a mean kinetic energy of 390 keV. (a) Find the de Broglie wavelength of

Sauron [17]

Explanation:

Given that,

The mean kinetic energy of the emitted electron, $E=390\ keV=390\times 10^3\ eV$

(a) The relation between the kinetic energy and the De Broglie wavelength is given by :

$\lambda=\dfrac{h}{\sqrt{2meE}}$

$\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}$

$\lambda=1.96\times 10^{-12}\ m$

(b) According to Bragg's law,

$n\lambda=2d\ sin\theta$

n = 1

For nickel, $d=0.092\times 10^{-9}\ m$

$\theta=sin^{-1}(\dfrac{\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})$

$\theta=0.010^{\circ}$

As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.

4 0

3 years ago