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3 years ago

10

Suppose the Sun moved twice the distance from Earth. Describe immediate and future changes to our planet. Justify your reasoning

1 answer:

Reptile [31]3 years ago

5 0

Our year would now be 2.8 times longer, we would also be receiving only 1/4 of the energy from the sun that we currently do. This means that we’d now be out beyond the orbit of Mars and right at the edge of the asteroid belt, and things would rapidly get very cold with temperatures expected to drop by around 50 degrees Celsius on average, and that’s with our current atmospheric composition which would not be stable in the new conditions. And also, any living thing on earth would die.

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What happens to the force between two charges if the magnitude of both charges is doubled and the distance between them is doubl

vlabodo [156]

Answer:the force will remain same

Explanation:

because force is equal to the ratio of magnitude and distance

4 0

3 years ago

1. It’s fall and time for the corn maze and bonfire and you just can’t wait. On your way to the farm though a turkey flies out i

Reptile [31]

Acceleration = (change in velocity ( final speed - starting speed))/ (time)

Acceleration = (18-30)/10.5

Acceleration = -12/10.5

Acceleration = -1.14 m/s^2

Distance = 30m/s x 10.5s + 1/2(1.14)(10.5)^2

Distance = 252.2 meters

8 0

3 years ago

A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg

Kitty [74]

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

$p_i=mv+0=mv$

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

$p_f=mv'+MV'$

Equating initial and the final momenta we get

$mv=mv'+MV'\\\\m(v-v')=MV'.....i$

Now since the surface is frictionless thus the energy is also conserved thus

$E_i=\frac{1}{2}mv^2$

Similarly the final energy becomes

$E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2$ \

Equating initial and final energies we get

$\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)$

Solving i and ii we get

$v+v'=V'$

Using this in equation i we get

$v'=\frac{v(m-M)}{(M-m)}=-v$

Thus putting v = -v' in equation i we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

4 0

3 years ago

A projectile is launched with an initial velocity of (40 m/s), at an angle of (30°) above

zlopas [31]

Answer:

330.5 m

Explanation:

In this case, the object is launched horizontally at 30° with an initial velocity of 40 m/s .

The maximum height will be calculated as;

$h=\frac{v^2_isin^2\alpha }{2g}$

where ∝ is the angle of launch = 30°

vi= initial launch velocity = 40 m/s

g= 10 m/s²

h= 40²*sin²40° / 2*10

h={1600*0.4132 }/ 20

h= 661.1/2 = 330.5 m

8 0

3 years ago

I NEED HELP ASAP

Levart [38]

Answer:

ig d. transmits only green light.....

8 0

1 year ago