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Lunna [17]
2 years ago
9

Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 77.3 N, Jill pull

s with 61.7 N in a direction 45° to the left, and Jane pulls in a direction 45° to the right with 147 N. (Since the donkey is involved with such uncoordinated people, who can blame it for being stubborn?) Find the magnitude of the net force the people exert on the donkey. Magnitude of the net force: _____N What is the direction of the net force? Express this as the angle from straight ahead between 0° and 90°, with a positive sign for angles to the left and a negative sign for angles to the right.
Direction of the net force: _____°
Physics
1 answer:
DiKsa [7]2 years ago
3 0

Answer:

F_{net} = 232.8 N

towards right so it is -15 degree

Explanation:

Net force in forward direction due to all three is given as

F_x = F_1 + F_2cos45 + F_3cos45

here we know that

F_1 = 77.3 N

F_2 = 61.7 N

F_3 = 147 N

F_x = 77.3 + 61.7 cos45 + 147 cos45

F_x = 224.9 N

Similarly in Y direction we will have

F_y = F_3 sin45 - F_2 sin45

F_y = (147 - 61.7)sin45

F_y = 60.3 N

Now the net force on the donkey is given as

F_{net} = \sqrt{F_x^2 + F_y^2}

F_{net} = \sqrt{224.9^2 + 60.3^2}

F_{net} = 232.8 N

Now direction of force is given as

tan\theta = \frac{F_y}{F_x}

tan\theta = \frac{60.3}{224.9}

\theta = 15^o towards right so it is -15 degree

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You have designed a machine that requires 1000 J of work from a motor for every 800 J of useful work done by the machine. What i
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80%

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Explain the connection of heat to change in temperature AND states of matter
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2 years ago
A 62.0-kg athlete leaps straight up into the air from a trampoline with an initial speed of 9.6 m/s. The goal of this problem is
pochemuha

Answer:

2856.96 J

0

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\frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f

6.78822 m/s

Explanation:

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g = Acceleration due to gravity = 9.81 m/s²

h = Height

The athlete only interacts with the gravitational potential energy. Air resistance is neglected.

At height y = 0

Kinetic energy

K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 62\times 9.6^2\\\Rightarrow K=2856.96\ J

At height y = 0 the potential energy is 0 as

P=mgy\\\Rightarrow P=mg0=0

At maximum height her velocity becomes 0 so the kinetic energy becomes zero.

As the the potential and kinetic energy are conserved

The general equation

K_i+P_i=K_f+P_f\\\Rightarrow \frac{1}{2}mv_i^2+mgh_i=\frac{1}{2}mv_f^2+mgh_f

Half of maximum height

\\\Rightarrow mgh_i+\frac{1}{2}mv_f^2=mg\frac{h_i}{2}+\frac{1}{2}mv^2\\\Rightarrow gh_i=g\frac{h_i}{2}+\frac{1}{2}v^2\\\Rightarrow g\frac{h_i}{2}=\frac{1}{2}v^2\\\Rightarrow v=\sqrt{gh}

h_i=\frac{v_i^2}{2g}

v=\sqrt{gh}\\\Rightarrow v=\sqrt{g\times \frac{v_i^2}{2g}}\\\Rightarrow v=\sqrt{\frac{v_i^2}{2}}\\\Rightarrow v=\sqrt{\frac{9.6^2}{2}}\\\Rightarrow v=6.78822\ m/s

The velocity of the athlete at half the maximum height is 6.78822 m/s

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