Tarzan, whose mass is 96 kg, is hanging at rest from a tree limb. Then he lets go and falls to the ground. Just before he lets go, his center of mass is at a height 2.1 m above the ground and the bottom of his dangling feet are at a height 1.3 above the ground. When he first hits the ground he has dropped a distance 1.3, so his center of mass is (2.1 - 1.3) above the ground. Then his knees bend and he ends up at rest in a crouched position with his center of mass a height 0.4 above the ground.Consider the point particle system. What is the speed v at the instant just before Tarzan's feet touch the ground?

Answers

Answer 1

Answer:

Answer:

5.05 m/s

Explanation:

The distance from the bottom of his feet to his center of mass is (when is hanging at rest) is 2.1 - 1.3 = 0.8 m. Assume he keeps the posture, as soon as his feet touches the ground, his center of mass is 0.8 m above the ground. This would mean that he has traveled a distance of 2.1 - 0.8 = 1.3 m vertically. Using the law of energy conservation for potential and kinetic energy, also let the ground be ground 0 for potential energy, we have the following mechanical conservation energy:

$mgH = mgh + mv^2/2$

Since he was hanging at rest, his initial kinetic energy at H = 2.1m must be 0. Let g = 9.81m/s2 and m be his mass, we can calculate for his velocity v at h = 0.8 m. First start by dividing both sides by m

$gH = gh + v^2/2$

$v^2 = 2g(H - h)$

$v^2 = 2*9.81(2.1 - 0.8) = 25.506$

$v = √(25.506) = 5.05 m/s$

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Use a(t) =−32 feet per second squared as the acceleration due to gravity. a ball is thrown vertically upward from the ground with an initial velocity of 56 feet per second. for how many seconds will the ball be going upward?

Answers

Since the ball is moving by uniformly accelerated motion, its vertical velocity at time t is given by
$v(t)= v_0 - a t$
where we took upward as positive direction, and where $v_0$ is the initial velocity, a the acceleration and t the time.

The instant at which $v(t)=0$ is the instant when the ball reverses its velocity (from upward to downward). This means that the difference between the time t at which v(t)=0 and the instant t=0 is the total time during which the ball was going upward:
$0=v_0 - at$
By plugging numbers into the equation, we find
$t= (v_0)/(a)= (56 ft/s)/(32 ft/s^2)=1.75 s$

A person is pushing a lawnmower of mass m D 38 kg and with h D 0:75 m, d D 0:25 m, `A D 0:28 m, and `B D 0:36 m. Assuming that the force exerted on the lawnmower by the person is completely horizontal and that the mass center of the lawnmower is at G, and neglecting the rotational inertia of the wheels, determine the minimum value of this force that causes the rear wheels (labeled A) to lift off the ground. In addition, determine the corresponding acceleration of the mower.

Answers

Answer:

The acceleration of the mower will be "4.7 m/s²".

Explanation:

Balance of vertical force will be:

⇒ $Ra + Rb = mg$

For wheel to take off at A,

⇒ $Ra = 0$

Hence,

$Rb=mg$

Balancing moments about G will be:

⇒ $F* h = Rb* LB$

As we know,

Force, F = $(Rb* LB )/(h)$

On putting the values, we get

⇒ = $(38* 9.81* 0.36)/(0.75)$

⇒ = $178.9 \ N$

Now,

Acceleration, a = $(F)/(m)$

⇒ = $(178.9)/(38)$

⇒ = $4.7 \ m/s^2$

What is the revolution ratio of two spur gears with one having 12 teeth and the other having 36 teeth

Answers

3:1 is the ratio i hope this belps

22. What force is necessary to accelerate a 2500kg car from rest to 20m/s over 10s?(6 Points)
2N
250N
5000N
50000N

Answers

Answer:

50000N

Explanation:

Force = mass × acceleration

= 2500 × 20

= 50000N

Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal energy dissipated from brakes in a 1600 kg car that descends a 15 ∘ hill. The car begins braking when its speed is 95 km/h and slows to a speed of 40 km/h in a distance of 0.34 km measured along the road.

Answers

Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

= mg x loss of height

= 1600 x 9.8 x 340 sin 15

= 1379816 J

Sum of Loss of potential energy and Loss of kinetic energy

= 1379816 + 458400

= 1838216 J

This is the work done by the friction . So this is heat generated.

Final answer:

To calculate the thermal energy dissipated from the brakes of a car, use the equation Q = Mgh/10, where Q is the energy transferred to the brakes, M is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill. The temperature change of the brakes can then be calculated using the equation Q = mc∆T, where m is the mass of the brakes and c is its specific heat capacity.

Explanation:

The thermal energy dissipated from the brakes of a car can be calculated by converting the gravitational potential energy lost by the car into internal energy of the brakes. By using the equation Q = Mgh/10, where Q is the energy transferred to the brakes, M is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill, we can calculate the thermal energy dissipated. From there, the temperature change of the brakes can be calculated using the equation Q = mc∆T, where m is the mass of the brakes and c is its specific heat capacity.

Learn more about Thermal energy dissipation here:

brainly.com/question/16043710

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An implanted pacemaker supplies the heart with 72 pulses per minute, each pulse providing 6.0 V for 0.65 ms. The resistance of the heart muscle between the pacemaker’s electrodes is 550 Ω. Find (a) the current that flows during a pulse, (b) the energy delivered in one pulse, and (c) the average power supplied by the pacemaker.