The average speed of the car for the entire trip can be calculate by using:
where S is the total distance covered by the car, and t is the total time taken.
The total distance travelled by the car is:
while the total time taken is:
so, the average speed of the car is:
so, the correct answer is (3) 85 km/h.
As Potential energy =mgh
m= 0.95kg
h=3 meter
g = 9.8 m/sec^2. ( acceleration due to gravity)
So P.E =(0.95)(9.8)(3)kgm^2/s^2
P.E =27.93 joules
To stop instantly, you would need infinite deceleration. This in turn, requires infinite force, as demonstrable with this equation:F=ma<span>So when you hit a wall, you do not instantly stop (e.g. the trunk of the car will still move because the car is getting crushed). In a case of a change in momentum, </span><span><span>m<span>v⃗ </span></span><span>m<span>v→</span></span></span>, we can use the following equation to calculate force:F=p/h<span>However, because the force is nowhere close to infinity, time will never tend to zero either, which means that you cannot come to an instantaneous stop.</span>
Answer:
Explanation:
Initial moment of inertia of the earth I₁ = 2/5 MR² , M is mss of the earth and R is the radius . If ice melts , it forms an equivalent shell of mass 2.3 x 10¹⁹ Kg
Final moment of inertia I₂ = 2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²
For change in period of rotation we shall apply conservation of angular momentum law
I₁ ω₁ = I₂ ω₂ , ω₁ and ω₂ are angular velocities initially and finally .
I₁ / I₂ = ω₂ / ω₁
I₁ / I₂ = T₁ / T₂ , T₁ , T₂ are time period initially and finally .
T₂ / T₁ = I₂ / I₁
(2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²) / 2/5 MR²
1 + 5 / 3 x 2.3 x 10¹⁹ / M
= 1 + 5 / 3 x 2.3 x 10¹⁹ / 5.97 x 10²⁴
= 1 + .0000064
T₂ = 24 (1 + .0000064)
= 24 hours + .55 s
change in length of the day = .55 s .
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.
