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Svetlanka [38]
3 years ago
10

A boat is moving along the water. The forward thrust from the engines is 35,000 N. Water resistance is 20,000 N and air resistan

ce 10,000 N. What is the resultant force on the boat?
Physics
1 answer:
dedylja [7]3 years ago
7 0

Answer:

The resultant force on the boat is 5,000 N

Explanation:

Given;

the forward thrust on the boat, Ft = 35,000 N

water resistance on the boat, Fw = 20,000 N

Air resistance on the boat, Fa = 10,000 N

The forces opposing the boat motion will be negative, while the force pushing the boat forward will be positive

The resultant force on the boat, Fr = Ft - Fw - Fa

Fr = 35,000 N - 20,000 N - 10,000 N

Fr = 35,000 N - 30,000 N

Fr = 5,000 N

Therefore, the resultant force on the boat is 5,000 N

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Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid ra
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Explanation:

It is given that,

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3 years ago
The driver of a car going 86.0 km/h suddenly sees the lights of a barrier 44.0 m ahead. It takes the driver 0.75 s to apply the
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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:

s=ut+0.5at^2

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, a=-10.0 m/s^2

\Rightarrow s=23.9 m/s \times 0.75s+0.5\times (-10.0 m/s^2)\times (0.75 s)^2=15.11 m

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 s=44.0 m

The acceleration is same, a=-10.0 m/s^2

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:

v^2-u^2=2as

0-u^2=-2 \time 10.0m/s^2 \times 44.0 m\Rightarrow u=\sqrt{880 m^2/s^2}=29.6 m/s

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