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3 years ago

With a diameter that's 11 times larger than Earth's, _______ is the largest planet.

Physics

2 answers:

IrinaK [193]3 years ago

8 0

<span>With a diameter that's 11 times larger than Earth's, JUPITER is the largest planet.</span>

Svetradugi [14.3K]3 years ago

5 0

With a diameter that's 11 times larger than Earth's, Jupiter is the largest planet.

Jupiter is the biggest planet in the solar system. No wonder ancient astronomers named it after Jupiter, King of the Roman gods. More than 1,000 planets the size of Earth could fit inside Jupiter. A tunnel through Jupiter would be 11 times longer than a tunnel through Earth. A trip around Jupiter equator would take 6 times longer than a trip around the Earth's equator.

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In one experiment, the students allow the block to oscillate after stretching the spring a distance A. If the potential energy s

atroni [7]

Answer:

K = m g (A - A2)

Explanation:

In a block spring system the total energy is the sum of the potential energy plus the kinetic energy, for maximum elongation all the energy is potential

Em = U₀ = m g A

For when the system is at an ele

Elongation A2 less than A, energy has two parts

Em = K + U₂

K = Em –U₂

We substitute

K = m g A - m gA2

K = m g (A - A2)

3 0

3 years ago

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An object initially at rest experiences an acceleration of 0.281 m/s2 to the South for a time of 5.44 seconds. It then increases

andre [41]

Answer:

12.0 meters

Explanation:

Given:

v₀ = 0 m/s

a₁ = 0.281 m/s²

t₁ = 5.44 s

a₂ = 1.43 m/s²

t₂ = 2.42 s

Find: x

First, find the velocity reached at the end of the first acceleration.

v = at + v₀

v = (0.281 m/s²) (5.44 s) + 0 m/s

v = 1.53 m/s

Next, find the position reached at the end of the first acceleration.

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²

x = 4.16 m

Finally, find the position reached at the end of the second acceleration.

x = x₀ + v₀ t + ½ at²

x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²

x = 12.0 m

5 0

3 years ago

Greyhound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but

maw [93]

${\large{\bold{\rm{\underline{Given \; that}}}}}$

★ A grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

${\large{\bold{\rm{\underline{To\; find}}}}}$

★ The speed of the hound and the hare

${\large{\bold{\rm{\underline{Solution}}}}}$

★ The speed of the hound and the hare = 25:18

${\large{\bold{\rm{\underline{Full \; Solution}}}}}$

$\dashrightarrow$ As it's given that a grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

So firstly let us assume a metres as the distance covered by the hare in one leap.

Ok now let's talk about 5 leaps,.! As it's cleared that the hare cover the distance of 5a metres.

But 3 leaps of the hound are equal to 5 leaps of the hare.

Henceforth, (5/3)a meters is the distance that is covered by the hound.

Now according to the question,

Hound pursues a hare and takes 5 leaps for every 6 leaps of the hare..! (Same interval)

Now the distance travelled by the hound in it's 5 leaps..!

(5/3)a × 5

25/3a metres

Now the distance travelled by the hare in it's 6 leaps..!

6a metres

Now let us compare the speed of the hound and the hare. Let us calculate them in the form of ratio..!

25/3a = 6a

25/3 = 6

25:18

5 0

3 years ago

A car is driving away from a crosswalk. The formula d = t 2 + 2 t expresses the car's distance from the crosswalk in feet, d , i

Ede4ka [16]

Answer:

1) No, the car does not travel at constant speed.

2) V = 9 ft/s

3) No, the car does not travel at constant speed.

4) V = 5.9 ft/s

Explanation:

In order to know if the car is traveling at constant speed we need to derive the given formula. That way we get speed as a function of time:

V(t) = 2*t + 2 Since the speed depends on time, the speed is not constant at any time.

For the average speed we evaluate the formula for t=2 and t=5:

d(2) = 8 ft and d(5) = 35 ft

$V_{2-5}=\frac{d(5)-d(2)}{5-2}=9 ft/s$

Again, for the average speed we evaluate the formula for t=1.8 and t=2.1:

d(1.8) = 6.84 ft and d(2.1) = 8.61 ft

$V_{1.8-2.1}=\frac{d(2.1)-d(1.8)}{2.1-1.8}=5.9 ft/s$

4 0

3 years ago

What type of pollution did the Clean Water Act succeed in limiting?

Ilya [14]

Sewage. If thats not it, then I need to see your choices. :)

4 0

3 years ago

Read 2 more answers