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.
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Answer:
100/10 = 10 , 10 × 10 = 100÷20 = 5
I'm pretty sure its wrong
Answer:
The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J
Explanation:
The given variables are
Work done = 550 J
Volume change = V₂ - V₁ = -0.5V₁
Thus the product of pressure and volume change = work done by gas, thus
P × -0.5V₁ = 500 J
Hence -PV₁ = 1000 J
also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂
Also to compress the gas by a factor of 11 we have
P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work
Answer:
Explanation:
Impulse = change in momentum
mv - mu , v and u are final and initial velocity during impact at surface
For downward motion of baseball
v² = u² + 2gh₁
= 2 x 9.8 x 2.25
v = 6.64 m / s
It becomes initial velocity during impact .
For body going upwards
v² = u² - 2gh₂
u² = 2 x 9.8 x 1.38
u = 5.2 m / s
This becomes final velocity after impact
change in momentum
m ( final velocity - initial velocity )
.49 ( 5.2 - 6.64 )
= .7056 N.s.
Impulse by floor in upward direction
= .7056 N.s
Refer to the diagram shown below.
When an athlete is in motion, he/she exerts a vertical force (the person's weight, W) on the ground. The ground exerts an equal and opposite force, N, the normal reaction on the athlete, so that W = N.
At the same time, the ground exerts a horizontal force, F, o n the athlete so that he/she does not slip.
The magnitude of the horizontal force is
F = μN = μW
where μ = the dynamic coefficient of friction.
Answer:
The horizontal force is μW,
where
W = the weight of the athlete and,
μ = the dynamic coefficient of friction.
