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Gennadij [26K]
3 years ago
9

Use the worked example above to help you solve this problem. An airboat with mass 4.30 102 kg, including the passenger, has an e

ngine that produces a net horizontal force of 8.02 102 N, after accounting for forces of resistance. (a) Find the acceleration of the airboat. m/s2 (b) Starting from rest, how long does it take the airboat to reach a speed of 15.1 m/s?
Physics
1 answer:
umka21 [38]3 years ago
3 0

Answer:

a) a = 1,865 m / s²  and  b)  t = 8.1 s

Explanation:

a) Let's use Newton's second law to find acceleration, we can work the equation in scalar form because displacement and force have the same direction

         F = m .a

         a = F / m

         a = 8.02 10² /4.3 10²

         a = 1,865 m / s²

b) We use kinematic relationships in one dimension

        vf = vo + at

        vf = 0 + a t

         t = vf / a

          t = 15.1 / 1.865

          t = 8.1 s

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To calculate the force of impact F, first lets calculate the acceleration a of the ball: 

a=v/t where v is the velocity of the ball and t is time

a=32/0.8=40 m/s²

To get the force F we need the Newtons second law:

F=m*a where m is the mass of the ball and a is the acceleration.

F=m*a= 0.2*40 = 8 N

So the impact force is F= 8 N.
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Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The object
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Explanation:

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4 years ago
You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo
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Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

\Delta E_k=W=W_g-W_p

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}

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c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

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Explanation:

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