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Oliga [24]
3 years ago
13

When Pluto was classified as a planet it was known as a oddball planet why? Why is it less if an oddball now?

Physics
1 answer:
hichkok12 [17]3 years ago
7 0

Answer:

a strange world that has baffled scientists ever since it was discovered in 1930. It is not the large gas giant that one might expect to find in the outer reaches of the solar system.

Explanation:

Explanation

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A 160.-kilogram space vehicle is traveling along a
mario62 [17]

If the object is moving in a straight line at a constant speed, then that's
the definition of zero acceleration.  It can only happen when the sum of
all forces (the 'net' force) on the object is zero.

And it doesn't matter what the object's mass is.  That argument is true
for specks of dust, battleships, rocks, stars, rock-stars, planets, and
everything in between.

4 0
3 years ago
1.)State which definition of power below is correct.
Natalija [7]

C

kW, watts, kilowatts, W

6 0
3 years ago
When did robert fulton invent the steamboat
bekas [8.4K]

Answer:

1807

Explanation:

Robert Fulton (1765–1815) was an American engineer and inventor who is widely known for developing a commercially successful steamboat called Clermont. In 1807, that steamboat took passengers from New York City to Albany and back again, a round trip of 300 miles, in 62 hours.

7 0
3 years ago
The weight of a hydraulic barber's chair with a client is 2100 N. When the barber steps on the input piston with a force of 44 N
guajiro [1.7K]

Answer:

\frac{r_1}{r_2}=6.9

Explanation:

According to Pascal's Law, the pressure transmitted from input pedal to the output plunger must be same:

P_1 = P_2\\\\\frac{F_1}{A_1}=\frac{F_2}{A_2}\\\\\frac{F_1}{F_2}=\frac{A_1}{A_2}\\\\\frac{F_1}{F_2}=\frac{\pi r_1^2}{\pi r_2^2}\\\\\frac{F_1}{F_2}=\frac{r_1^2}{r_2^2}

where,

F₁ = Load lifted by output plunger = 2100 N

F₂ = Force applied on input piston = 44 N

r₁ = radius of output plunger

r₂ = radius of input piston

Therefore,

\frac{r_1^2}{r_2^2}=\frac{2100\ N}{44\ N}\\\\\frac{r_1}{r_2}=\sqrt{\frac{2100\ N}{44\ N}} \\\\\frac{r_1}{r_2}=6.9

4 0
3 years ago
You just calibrated a constant volume gas thermometer. The pressure of the gas inside the thermometer is 294.0 kPa when the ther
Travka [436]

Answer: 361° C

Explanation:

Given

Initial pressure of the gas, P1 = 294 kPa

Final pressure of the gas, P2 = 500 kPa

Initial temperature of the gas, T1 = 100° C = 100 + 273 K = 373 K

Final temperature of the gas, T2 = ?

Let us assume that the gas is an ideal gas, then we use the equation below to solve

T2/T1 = P2/P1

T2 = T1 * (P2/P1)

T2 = (100 + 273) * (500 / 294)

T2 = 373 * (500 / 294)

T2 = 373 * 1.7

T2 = 634 K

T2 = 634 K - 273 K = 361° C

5 0
3 years ago
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