Answer:
The International Space Station move at 7.22 km/s.
Explanation:
Orbital speed of satellite is given by
, where G is gravitational constant, M is mass of Earth and r is the distance to satellite from centre of Earth.
r = R + h = 6350 + 1400 = 7750 km = 7.75 x 10⁶ m
G = 6.673 x 10⁻¹¹ Nm²/kg²
M = 5.98 x 10²⁴ kg
Substituting

The International Space Station move at 7.22 km/s.
Answer:
The order of increasing energy is as follows
"microwave < infrared < visible < ultraviolet"
Option (A) is correct.
Explanation:
Given:
Arrange the following spectral regions in order of increasing energy: infrared, microwave, ultraviolet, visible.
From the formula of energy in terms of frequency.

Where
planck constant,
frequency of light.
From above formula we can conclude that higher frequency means higher energy.
In our case ultraviolet has higher frequency and microwave has lower frequency.
So ultraviolet has higher energy and microwave has lower energy.
microwave < infrared < visible < ultraviolet
Therefore, the order of increasing energy is as follows
"microwave < infrared < visible < ultraviolet"
Answer:
You will reach both your arms out to break your fall and save your head.
Explanation:
It common sense you don't want your head injured. Do you?
Answer:
Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.
Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.
Explanation:
Part a
When Road is Level
The stopping sight distance is given as

Here
- SSD is the stopping sight distance which is to be calculated.
- u is the speed which is given as 60 mi/hr
- t is the perception-reaction time given as 2.5 sec.
- a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
- G is the grade of the road, which is this case is 0 as the road is level
Substituting values

So the minimum stopping sight distance is 563.36 ft.
Part b
When Road has a maximum grade of 4%
The stopping sight distance is given as

Here
- SSD is the stopping sight distance which is to be calculated.
- u is the speed which is given as 60 mi/hr
- t is the perception-reaction time given as 2.5 sec.
- a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
- G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade
For upgrade of 4%, Substituting values

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>
For downgrade of 4%, Substituting values

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>
As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.
I believe the answer is C