Note:
The height of a high bar from the floor is h = 2.8 m (or 9.1 ft).
It is not provided in the question, so the standard height is assumed.
g = 9.8 m/s², acceleration due to gravity.
Note that the velocity and distance are measured as positive upward.
Therefore the floor is at a height of h = -2.8 m.
First dismount:
u = 4.0 m/s, initial upward velocity.
Let v = the velocity when the gymnast hits the floor.
Then
v² = u² - 2gh
v² = 16 - 2*9.8*(-2.8) = 70.88
v = 8.42 m/s
Second dismount:
u = -3.0 m/s
v² = (-3.0)² - 2*9.8*(-2.8) = 63.88 m/s
v = 7.99 m/s
The difference in landing velocities is 8.42 - 7.99 = 0.43 m/s.
Answer:
First dismount:
Acceleration = 9.8 m/s² downward
Landing velocity = 8.42 m/s downward
Second dismount:
Acceleration = 9.8 m/s² downward
Landing velocity = 7.99 m/s downward
The landing velocities differ by 0.43 m/s.
The roadway with the highest number of hazards is <span>city streets</span>
The answer is letter a. It is best to slow down in situations of heavy rain or flooded road as skid could be the result if you lose out of control because the driver isn't slowing down. That is why it is being said that tires can ride on a thin film of water skis as it could skid if it has lost control if the driver hadn't slowed down.
For this case, what we can do is use the Pythagorean theorem to find the magnitude of the displacement of the car.
We have then
From here, we clear the value of d.
We have then:
Rewriting:
Answer:
The magnitude of the car's displacement is:
d = 20 miles
Given that,
Distance in south-west direction = 250 km
Projected angle to east = 60°
East component = ?
since,
cos ∅ = base/hypotenuse
base= hyp * cos ∅
East component = 250 * cos 60°
East component = 125 km
