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salantis [7]
3 years ago
10

What is the difference between a bound orbit and an unbound orbit around the sun?

Physics
1 answer:
Anit [1.1K]3 years ago
8 0

Answer:

Explanation:

The difference between a bound orbit and an unbound orbit around the sun is that:

An object on a bound orbit pursues the same way around the Sun again and again, while an object on an unbound orbit moves toward the Sun only a single time and afterward stays away forever &  never returns.

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A5 kg frisbee is thrown from rest to a final speed of 12 m/s. What is the impulse of the frisbee?
lesya692 [45]

Answer:

60kgm/s

Explanation:

Given parameters:

Mass of frisbee  = 5kg

Final speed  = 12m/s

Unknown:

Impulse of the frisbee  = ?

Solution:

The impulse of the frisbee is the same as the change in momentum.

It is given as:

 Impulse  = mass (Final velocity  -  Initial velocity)

 Impulse  = 5(12  - 0)  = 60kgm/s

3 0
3 years ago
When pressing the accelerator on a car, the ___________ energy of the fuel is transformed into the _____________ energy that mak
lesya [120]
B. Chemical; Mechanical
5 0
3 years ago
Read 2 more answers
The site from which an airplane takes off is the origin. The X axis points east, the y axis points straight up. The position and
Contact [7]

Answer:

d = 3.5*10^4 m

Explanation:

In order to calculate the displacement of the airplane you need only the information about the initial position and final position of the airplane. THe initial position is at the origin (0,0,0) and the final position is given by the following vector:

\vec{r}=(1.21*10^3\hat{i}+3.45*10^4\hat{j})m

The displacement of the airplane is obtained by using the general form of the Pythagoras theorem:

d=\sqrt{(x-x_o)^2+(y-y_o)^2}   (1)

where x any are the coordinates of the final position of the airplane and xo and yo the coordinates of the initial position. You replace the values of all variables in the equation (1):

d=\sqrt{(1.12*10^3-0)^2+(3.45*10^4-0)^2}=3.45*10^4m

hence, the displacement of the airplane is 3.45*10^4 m

6 0
3 years ago
If an object travels 15 metres in 5 seconds, what is the slope of its distance-time graph?
SOVA2 [1]

Answer:

3 m/s

Explanation:

The slope is distance divided by time, also known as 'speed'.

15 meters / 5 seconds = 3 meters per second.

8 0
2 years ago
A cannon ball is fired directly upward with a velocity of 160 m/s. How long does it take to fall back to the ground? s How fast
Andrej [43]
To answer this problem, we will use the equations of motions.

Part (a):
For the ball to start falling back to the ground, it has to reach its highest position where its final velocity will be zero.
The equation that we will use here is:
v = u + at where
v is the final velocity = 0 m/sec
u is the initial velocity = 160 m/sec
a is acceleration due to gravity = -9.8 m/sec^2 (the negative sign is because the ball is moving upwards, thus, its moving against gravity)
t is the time that we want to find.
Substitute in the equation to get the time as follows:
v = u + at
0 = 160 - 9.8t
9.8t = 160
t = 160/9.8 = 16.3265 sec
Therefore, the ball would take 16.3265 seconds before it starts falling back to the ground

Part (b):
First, we will get the total distance traveled by the ball as follows:
s = 0.5 (u+v)*t
s = 0.5(160+0)*16.3265
s = 1306.12 meters
The equation that we will use to solve this part is:
v^2 = u^2 + 2as where
v is the final velocity we want to calculate
u is the initial velocity of falling = 0 m/sec (ball starting falling when it reached the highest position, So, the final velocity in part a became the initial velocity here)
a is acceleration due to gravity = 9.8 m/sec^2 (positive as ball is moving downwards)
s is the distance covered = 1306.12 meters
Substitute in the above equation to get the final velocity as follows:
v^2 = u^2 + 2as
v^2 = (0)^2 + 2(9.8)(1306.12)
v^2 = 25599.952 m^2/sec^2
v = 159.99985 m/sec
Therefore, the velocity of the ball would be 159.99985 m/sec when it hits the ground.
6 0
3 years ago
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