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MissTica
3 years ago
14

Suppose you wish to whirl a pail full of water in a vertical circle at a constant speed without spilling any of its contents (ev

en at the top of the circle!). If your arm is 0.65 m long (from shoulder to fist) and the distance from the handle to the surface of the water is 19.0 cm, what minimum speed is required?
Physics
1 answer:
Yanka [14]3 years ago
6 0

Answer:

V = 2.87 m/s

Explanation:

The minimum speed required would be that at which the acceleration due to gravity is negated by the centrifugal force on the water.

Thus, we simply need to set the centripetal acceleration equal to gravity and solve for the speed V using the following equation:

Centripetal acceleration = V^2 / r

where r is the distance of water from the pivot or shoulder.

For our case, r will be 0.65 + 0.19 = 0.84 m

and solving the above equation we get:

9.81 = V^2 / 0.84

V^2 = 8.2404

V = 2.87 m/s

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A model rocket is launched straight upward with an initial speed of 52.0 m/s. It accelerates with a constant upward acceleration
JulsSmile [24]

Explanation:

(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.

(b) We must separate the motion into two parts, when the rocket's engines is on  and when the rocket's engines is off.

First we must find the rocket speed when the engines stop:

v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}

This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:

v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m

So, the maximum height reached by the rocket is:

h=y_1+y_2\\h=160m+154.28m=314.28m

(c) In the first part we have:

v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s

And in the second part:

t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s

So,  the time it takes to reach the maximum height is:

t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s

(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:

v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}

t_4=\frac{v_f-v_0}{g}\\t_4=\frac{78.48\frac{m}{s}-0}{9.8\frac{m}{s^2}}\\t_4=8.01s

So, the total time is:

t=t_3+t_4\\t=8.60s+8.01s\\t=16.61s

7 0
4 years ago
Explain how power is determined.
ivanzaharov [21]
It is the amount of work you have done in a certain amount of time. <span />
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3 years ago
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Characteristics we use to tell the difference between kinds of matter are called___ properties
d1i1m1o1n [39]

Answer:

I believe it's called qualitative

Explanation:

8 0
3 years ago
A cannonball is fired at a 45.0° angle and an initial velocity of 670 m/s. Assume no air resistance. How long until the cannonba
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Answer:

96.7 s

Explanation:

Time of flight in projectile can be calculated thus:

T = 2 × u × sin ϴ/ g

Where;

T = time of flight (s)

u = initial velocity (m/s)

ϴ = Angle of projectile (°)

g = acceleration due to gravity (9.8m/s²)

Based on the provided information; u = 670m/s, ϴ = 45°

Hence, using T = 2.u.sin ϴ/ g

T = 2 × 670 × sin 45° ÷ 9.8

T = 1340 × 0.7071 ÷ 9.8

T = 947.52 ÷ 9.8

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What happened to the kelp forest when the otter was hunted to near extinction?
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Explanation:

If sea otters are hunted and their population is brought to extinction then this will cause major harm the ecosystem of the kelp forest and it will disturb the ecological balance in the kelp forest. The herbivorous animals consumed by the sea otters will increase in population and they will consume a lot of vegetation in the forest. The kelp forest which forms the coastline will not remain effective in providing protection against the storms to the neighboring areas.

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3 years ago
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