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3 years ago

What is true about pluto?

Physics

2 answers:

Lapatulllka [165]3 years ago

8 0

Answer:

Explanation:

Pluto is a dwarf planet that lies in the Kuiper Belt, an area full of icy bodies and other dwarf planets out past Neptune. Pluto is very small, only about half the width of the United States and its biggest moon Charon is about half the size of Pluto

Elodia [21]3 years ago

3 0

Answer: Pluto is not always the planet farthest from the sun (apex)

Explanation:

You might be interested in

Can and object have a negative position and a positive velocity? Or vice versa, a positive position and a negative velocity? Exp

zavuch27 [327]

Imagine a ball is moving on the following horizontal line.

. . . . . . . . . . . . . . . . . . . O. . . . . . . . . . . . . . . . . .

Take right as positive. O is the starting point of the ball. Denote the ball by o.

. . . . . . . . . . . . . . . . . . . O. . . . . . . ... . . o . . . . . .

Assume the ball is moving to the right. It has positive displacement since it is on the right of O, and positive velocity since its positive displacement is increasing.

.ñ

. . . . . . . . . . . . . . . . . . . O. . . . o . . . . . . . . . . . . .

Now the ball is returning to O. It still has positive displacement since its current position is still on the right of O. However, its velocity is negative since its positive displacement is decreasing and the direction of the velocity vector points left, which is the negative side.

By now you should be able to come up with a scenario where the ball has negative displacement and positive velocity.

You can observe the same phenomenon in daily life. Say, as a stretched spring bounces to its starting position, if we let the returning direction be positive, the string has negative displacement since it is on the negative direction, but has positive velocity. Bungee jump can also used to illustrate the phenomenon.

4 0

2 years ago

What are two ways air resistance can be increased?

elena-14-01-66 [18.8K]

The increase in speed leads to an increase in the amount of air resistance. Eventually, the force of air resistance becomes large enough to balances the force of gravity. At this instant in time, the net force is 0 Newton; the object will stop accelerating. The object is said to have reached a terminal velocity.

7 0

3 years ago

A daring stunt woman sitting on a tree limb wishes to drop vertically onto a horse gallop ing under the tree. The constant speed

Shkiper50 [21]

Answer:

0.67 seconds

8.576 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.23=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2.23\times 2}{9.8}}\\\Rightarrow t=0.67\ s$

Time taken by the stunt woman to drop to the saddle is 0.67 seconds which is the time she will stay in the air.

Speed of the horse = 12.8 m/s

Distance = Speed × Time

⇒Distance = 12.8×0.67

⇒Distance = 8.576 m

Hence, the distance between the horse and stunt woman should be 8.576 m when she jumps.

3 0

3 years ago

A horizontal uniform bar of mass 3 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the

kirill115 [55]

Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂ : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F : Force ( N)

d : Horizontal distance of the force to the point 2 ( N*m

Data

mb = 3 kg : mass of the bar

mm = 1.5 kg : mass of the monkey

L = 3m : lengt of the bar

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the bar

T₁ : Tension in string 1 (vertical upward)

T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

Wm: Weihgt of the monkey (vertical downward)

Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances from forces the point 2

d₁₂ = (3-0.6) m = 2.4m : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium of moments at the point 2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

5 0

2 years ago

The center of mass is

PolarNik [594]

D is the best answer. In many physics problems we treat an extended object as if it were a point with the same mass located at the center of mass.

5 0

3 years ago