Tidal waves are caused by winds and high tides, and a tidal range changes. there you go
By definition we have that the final speed is:
Vf² = Vo² + 2 * a * d
Where,
Vo: Final speed
a: acceleration
d: distance.
We cleared this expression the acceleration:
a = (Vf²-Vo²) / (2 * d)
Substituting the values:
a = ((0) ^ 2- (60) ^ 2) / ((2) * (123) * (1/5280))
a = -77268 mi / h ^ 2
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:
First you must make a free body diagram and see the acceleration of the car:
g = 32.2 feet / sec ^ 2
a = -77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2
a = -31.48 feet / sec ^ 2
A = a + g * sin (θ) = -31.48 + 32.2 * sin17.0
A = -22.07 feet / sec ^ 2
Clearing the braking distance:
Vf² = Vo² + 2 * a * d
d = (Vf²-Vo²) / (2 * a)
Substituting the values:
d = ((0) ^ 2- (60 * (5280/3600)) ^ 2) / (2 * (- 22.07))
d = 175.44 feet
answer:
its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet
By equation of motion :
Now, it is given that it stops after 5 seconds of motion and time at that point is 12:00.
So, time when it started is 11 : 59 : 55.
Therefore, distance travelled is 10 m.
Hence, this is the required solution.
Answer: 909 m/s
Explanation:
Given
Mass of the bullet, m1 = 0.05 kg
Mass of the wooden block, m2 = 5 kg
Final velocities of the block and bullet, v = 9 m/s
Initial velocity of the bullet v1 = ? m/s
From the question, we would notice that there is just an object (i.e the bullet) moving before the collision. Also, even after the collision between the bullet and wood, the bullet and the wood would move as one object. Thus, we would use the conservation of momentum to solve
m1v1 = (m1 + m2) v, on substituting, we have
0.05 * v1 = (0.05 + 5) * 9
0.05 * v1 = 5.05 * 9
0.05 * v1 = 45.45
v1 = 45.45 / 0.05
v1 = 909 m/s
Thus, the original velocity of the bullet was 909 m/s
