the millersburg ferry (m=13000.0 kg loaded) puts its engines in full reverse and stops in 65 seconds. if the speed before brakin
1 answer:
The braking force is -400 N
Explanation:
We can solve this problem by using the impulse theorem, which states that the impulse applied on the ferry (the product of force and time) is equal to its change in momentum:
where in this problem, we have:
F is the force applied by the brakes
is the time interval
m = 13,000 kg is the mass of the ferry
u = 2.0 m/s is the initial velocity
v = 0 is the final velocity
And solving for F, we find the force applied by the brakes:
where the negative sign indicates that the direction is backward.
Learn more about impulse:
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Given :
Initial velocity, u = -6 m/s.
Time taken, t = 4 seconds.
Acceleration due to gravity, .( Here negative sign means downward direction )
To Find :
Velocity after 4 seconds.
Solution :
By equation of motion.
v = u + at
Here , a = g.
v = u + gt
v = -6 + (-9.8)×4
v = -6 + (-39.2)
v = -45.2 m/s
Therefore, velocity after 4 seconds is -45.2 m/s.
Hence, this is the required solution.
Answer:
a.) At their highest point, both of them have the same amount of mechanical energy.
Explanation:
Grasshopper A jumps upwards with speed v while grasshopper B jumps at an angle of 66 degree with horizontal at same speed
so vertical speed of B is given
So when both the grasshopper will jump upwards they both will attain different maximum heights
Since initial speed of both grasshopper is same so their mechanical energy is always conserved
So mechanical energy of grasshopper A is given as
Mechanical energy of grasshopper B is given as
So we have
a.) At their highest point, both of them have the same amount of mechanical energy.
Refer to the diagram shown below.
Neglect wind resistance, and use g = 9.8 m/s².
The pole vaulter falls with an initial vertical velocity of u = 0.
If the velocity upon hitting the pad is v, then
v² = 2*(9.8 m/s²)*(4.2 m) = 82.32 (m/s)²
v = 9.037 m/s
The pole vaulter comes to res after the pad compresses by 50 cm (or 0.5 m).
If the average acceleration (actually deceleration) is (a m/s²), then
0 = (9.037 m/s)² + 2*(a m/s²)*(0.5 m)
a = - 82.32/(2*0.5) = - 82 m/s²
Answer: - 82 m/s² (or a deceleration of 82 m/s²)
Neglect wind resistance, and use g = 9.8 m/s².
The pole vaulter falls with an initial vertical velocity of u = 0.
If the velocity upon hitting the pad is v, then
v² = 2*(9.8 m/s²)*(4.2 m) = 82.32 (m/s)²
v = 9.037 m/s
The pole vaulter comes to res after the pad compresses by 50 cm (or 0.5 m).
If the average acceleration (actually deceleration) is (a m/s²), then
0 = (9.037 m/s)² + 2*(a m/s²)*(0.5 m)
a = - 82.32/(2*0.5) = - 82 m/s²
Answer: - 82 m/s² (or a deceleration of 82 m/s²)