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3 years ago

How do the orbits of the planets farthest from the sun compare to the orbits of the planets closest to the sun?

1 answer:

6 0

Closer orbit less distance to travel round shorter time taken for complete orbit. further orbit further to travel longer time for complete orbit

You might be interested in

you push a 51 kg box with a force of 485 N. the friction force on the box is 232 N. calculate the acceleration of the crate.

vichka [17]

Answer:

a = 4.96 m/s²

Explanation:

Given,

The mass of the box, m = 51 Kg

The magnitude of the applied force, Fₐ = 485 N

The friction force on the box, Fₓ = 232 N

The net force acting on the box is,

F = Fₐ - Fₓ

Substituting the given values in the above equation

F = 485 - 232

= 253 N

The acceleration of the crate is given by

a = F/m

= 253 / 51

= 4.96 m/s²

Hence, the acceleration of the crate is, a = 4.96 m/s²

3 0

3 years ago

A body of mass m1 = 1.5 kg moving along a directed axis in the positive sense with a velocity

larisa86 [58]

Answer:

3.71 m/s in the negative direction

Explanation:

From collisions in momentum, we can establish the formula required here which is;

m1•u1 + m2•v2 = m1•v1 + m2•v2

Now, we are given;

m1 = 1.5 kg

m2 = 14 kg

u1 = 11 m/s

v1 = -1 m/s (negative due to the negative direction it is approaching)

u2 = -5 m/s (negative due to the negative direction it is moving)

Thus;

(1.5 × 11) + (14 × -5) = (1.5 × -1) + (14 × v2)

This gives;

16.5 - 70 = -1.5 + 14v2

Rearranging, we have;

16.5 + 1.5 - 70 = 14v2

-52 = 14v2

v2 = - 52/14

v2 = 3.71 m/s in the negative direction

8 0

3 years ago

Which of these chemical equations demonstrates the law of conservation of mass as it applies to the combustion of ethane? A. 2 C

WINSTONCH [101]

A. 2 C₂ H₆ + 7 O₂ → 4 C O₂ + 6 H₂ O

according to law of conservation of mass , the total mass of reactants side must be same as the total mass of product side. so we need to check if each atom in the equation has same number on both side of the equation or not.

in this equation , we have

4 atoms of carbon left and 4 atoms of carbon on right

12 atoms of hydrogen on left and 12 atoms of hydrogen on right

14 atoms of oxygen on left and 14 atoms of oxygen on right

3 0

3 years ago

A 5.0-kilogram sphere, starting from rest, falls freely 22 meters in 3.0 seconds near the surface of a planet. Compared to the a

Whitepunk [10]

Answer:

C) one-half as great

Explanation:

We can calculate the acceleration of gravity in that planet, using the following kinematic equation:

$\Delta x=v_0t+\frac{gt^2}{2}$

In this case, the sphere starts from rest, so $v_0=0$ . Replacing the given values and solving for g':

$g'=\frac{2\Delta x}{t^2}\\g'=\frac{2(22m)}{(3s)^2}\\g'=4.89\frac{m}{s^2}$

The acceleration due to gravity near Earth's surface is $g=9.8\frac{m}{s^2}$ . So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.

5 0

3 years ago

An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f

konstantin123 [22]

A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force $F=m \frac{v^2}{r}$ , directed toward the center of the circle (so, horizontally)
- the weight of the plane: $W=mg$ , downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is $r= \frac{260 m}{2} = 130 m$ , so the centripetal force acting on the plane is
$F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N$
On the vertical axis, we have two forces: the weight
$W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N$
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to $a=12 m/s^2$ , we can find the magnitude of this other force F by using Newton's second law:
$F+mg=ma$
$F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N$

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
$R= \sqrt{(F_c^2+(W+F)^2}=$
$= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2} =2.23 \cdot 10^6 N$

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
$\tan \theta = \frac{-(W+F)}{F_c}= -0.5$
And so, the angle is
$\theta = \arctan (-0.5)=-26.8 ^{\circ}$

7 0

3 years ago