Answer:
33.6 Ns backward.
Explanation:
Impulse: This can be defined as the product of force and time. The S.I unit of impulse is Ns.
From Newton's second law of motion,
Impulse = change in momentum
I = mΔv................................. Equation 1
Where I = impulse, m = mass of the skater, Δv = change in velocity = final velocity - initial velocity.
Given: m = 28 kg, t = 0.8 s, Δv = -1.2-0 = -1.2 m/s (Note: the initial velocity of the skater = 0 m/s)
Substituting into equation 1
I = 28(-1.2)
I = -33.6 Ns
Thus the impulse = 33.6 Ns backward.
Answer:27 km per hour West + 17 km per hour North
Answer:
Direction of ship: 9.45° West of North
Ship's relative speed: 7.87m/s
Explanation:
A. Direction of ship: since horizontal of the velocity of boat relative to the ground is 0
Vx=0
Therefore, -VsSin∅+VcCos∅40°
Sin∅ = Vc/Vs × Cos 40°
Sin∅ = 1.5/7 ×Cos40°
Sin∅= 0.164
∅= Sin-¹ (0.164)
∅= 9.45° W of N
B. Ship's relative speed:
Vy= VsCos∅ + Vcsin40°
= 7Cos9.45° + 1.5sin40°
= 7×0.986 + 1.5×0.642
= 7.865
= 7.87m/s
Answer:
Its called the hydrosphere UwU!
Explanation:
Part a
Answer: NO
We need to calculate the distance traveled once the brakes are applied. Then we would compare the distance traveled and distance of the barrier.
Using the second equation of motion:

where s is the distance traveled, u is the initial velocity, t is the time taken and a is the acceleration.
It is given that, u=86.0 km/h=23.9 m/s, t=0.75 s, 

Since there is sufficient distance between position where car would stop and the barrier, the car would not hit it.
Part b
Answer: 29.6 m/s
The maximum distance that car can travel is 
The acceleration is same, 
The final velocity, v=0
Using the third equation of motion, we can find the maximum initial velocity for car to not hit the barrier:

Hence, the maximum speed at which car can travel and not hit the barrier is 29.6 m/s.