Answer:
The Answer is gonna be According D.separate atoms from their electrons
Answer:
0.51 m
Explanation:
Using the principle of conservation of energy, change in potential energy equals to the change in kinetic energy of the spring.
Kinetic energy, KE=½kx²
Where k is spring constant and x is the compression of spring
Potential energy, PE=mgh
Where g is acceleration due to gravity, h is height and m is mass
Equating KE=PE
mgh=½kx²
Making x the subject of formula

Substituting 9.81 m/s² for g, 1300 kg for m, 10m for h and 1000000 for k then

Answer:
98.13m
Explanation:
Complete question
Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building
CHECK THE ATTACHMENT
From the figure, using trigonometry
Tan(θ ) = opposite/adjacent
Where Angle (θ )= 63°
Opposite= X = height of the building
Adjacent= 50 m
Then substitute the values we have
Tan(63)= X/50
1.9626= X/50
X= 1.9626 × 50
X= 98.13m
Hence, the height of the building is 98.13m
Answer:
distance = 3 + 4 = 7 mi
displacement is √(3² + 4²) = 5 mi
Explanation:
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.