Answer:
Average speed: 86 km/h
Explanation:
Driving from San Antonio to Houston:
1st. half time: 54km/h
2nd. half time: 118 km/h
Average speed = [tex] \frac{54 \frac{km}{h}+ 118 \frac{km}{h} }{2}=86 \frac{km}{h} [\tex]
Driving way back:
1st. half time: 54km/h
2nd. half time: 118 km/h
Average speed = [tex] \frac{54 \frac{km}{h}+ 118 \frac{km}{h} }{2}=86 \frac{km}{h} [\tex]
As in both routes we have the same average speed, then the average speed for the whole trip is 86 km/h
Answer:
0.196 m
Explanation:
Given in the question that,
time taken by the dolphin to go back to water = 0.2 sec
To solve the question we will use Newton's Law of motion
<h3>S = ut + 0.5(a)t²</h3>
here S is distance covered
u is initial speed
a = acceleration due to gravity
t = time taken
Plug value in the equation above
S = 0(0.2) + 0.5(-9.8)(0.2)²
S = 0.5(-9.8)(0.2)²
S = -0.196 m
Negative sign represent direction
(Assuming that dolphin have a vertical straight jump not a projectile motion)
<h2>
Answer: A. land-based</h2>
Explanation:
An artificial satellite is one that is launched into space to orbit the Earth (or another body of the solar system) for various purposes.
In this sense, the characteristics of an artificial satellite will depend on its purpose and functionality. From there we can list: meteorological satellites (for weather) , telecommunications satellites, remote sensing satellites, global positioning systems satellites, environmental satellites, research satellites, among others.
In addition, their orbits can be classified according to their height and inclination, depending on the use they have.
On the other hand, one of the main conditions for a satellite to be considered as such is that it must be kept orbiting. This means, it must not touch land during its useful life, even if it remains in constant contact with its earth based control stations.
Therefore,<u> a land-based is not a type of satellite.</u>
The relationship between the initial velocity, final velociy, distance, and deceleration can be expressed in the following equation.
2(a)(0.270 m) = 0² - (5.70 m/s)²
The value of a (which is the deceleration) is 0.06 m/s². Thus, the answer is that the deceleration value is approximately 0.06 m/s².
